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Solving
Polynomial
 Equations
๏ Solve by Factoring
๏ Identify solutions
  on Graph
๏ Solve by Graphing
Solving Polynomial Equations
★   Solving a polynomial equation is the same as solving a
    quadratic equation, except that the quadratic might be
    replaced by a different kind of polynomial (such as a
    cubic or a quartic).  
Solving Polynomial Equations
★   Solving a polynomial equation is the same as solving a
    quadratic equation, except that the quadratic might be
    replaced by a different kind of polynomial (such as a
    cubic or a quartic).  
★   There are 3 ways to solve Polynomial Equations
Solving Polynomial Equations
★   Solving a polynomial equation is the same as solving a
    quadratic equation, except that the quadratic might be
    replaced by a different kind of polynomial (such as a
    cubic or a quartic).  
★   There are 3 ways to solve Polynomial Equations
    (1) Using factoring and the zero product property
Solving Polynomial Equations
★   Solving a polynomial equation is the same as solving a
    quadratic equation, except that the quadratic might be
    replaced by a different kind of polynomial (such as a
    cubic or a quartic).  
★   There are 3 ways to solve Polynomial Equations
    (1) Using factoring and the zero product property
    (2) Using the graphing calculator to graph
Solving Polynomial Equations
★   Solving a polynomial equation is the same as solving a
    quadratic equation, except that the quadratic might be
    replaced by a different kind of polynomial (such as a
    cubic or a quartic).  
★   There are 3 ways to solve Polynomial Equations
    (1) Using factoring and the zero product property
    (2) Using the graphing calculator to graph
    (3) Using Synthetic Division (separate notes)
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.



★   It is possible for a polynomial equation to have fewer
    solutions (or none at all).
★   The degree of the polynomial gives you the maximum
    number of solutions that are theoretically possible.
★   Some solutions may be irrational or even imaginary.
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.        ( )      3
                    f x = x + x − x +12




★   It is possible for a polynomial equation to have fewer
    solutions (or none at all).
★   The degree of the polynomial gives you the maximum
    number of solutions that are theoretically possible.
★   Some solutions may be irrational or even imaginary.
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.       ( )       3
                    f x = x + x − x +12


     Degree is 3. Therefore 3 or fewer solutions possible.
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.        ( )      3
                    f x = x + x − x +12


     Degree is 3. Therefore 3 or fewer solutions possible.

★   It is possible for a polynomial equation to have fewer
    solutions (or none at all).
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.        ( )      3
                    f x = x + x − x +12


     Degree is 3. Therefore 3 or fewer solutions possible.

★   It is possible for a polynomial equation to have fewer
    solutions (or none at all).
★   The degree of the polynomial gives you the maximum
    number of solutions that are theoretically possible.
Total Number of Solutions
★   The maximum number of solutions that a polynomial
    equation can have is equal to the degree of the
    polynomial.        ( )      3
                    f x = x + x − x +12


     Degree is 3. Therefore 3 or fewer solutions possible.

★   It is possible for a polynomial equation to have fewer
    solutions (or none at all).
★   The degree of the polynomial gives you the maximum
    number of solutions that are theoretically possible.
★   Some solutions may be irrational or even imaginary.
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier.


★   Factor completely!
★   Set each factor equal to 0 and solve.
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                           GCF                (      )
                                            3x x 2 − 4 = 0

★   Factor completely!
★   Set each factor equal to 0 and solve.
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                         GCF               (      )
                                         3x x 2 − 4 = 0

★   Factor completely!
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                         GCF                  (      )
                                            3x x 2 − 4 = 0

★   Factor completely!   Keep going here.
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                         GCF                   (         )
                                            3x x 2 − 4 = 0

★   Factor completely!   Keep going here.
                                            3x ( x − 2 ) ( x + 2 ) = 0
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                           GCF                  (         )
                                             3x x 2 − 4 = 0

★   Factor completely!    Keep going here.
                                             3x ( x − 2 ) ( x + 2 ) = 0
★   Set each factor equal to 0 and solve.
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                           GCF                      (         )
                                                 3x x 2 − 4 = 0

★   Factor completely!    Keep going here.
                                                 3x ( x − 2 ) ( x + 2 ) = 0
★   Set each factor equal to 0 and solve.
                         3x = 0   ( x − 2) = 0      ( x + 2) = 0
Solving by Factoring
★   When using factoring, remember to always look for a
    GCF (Greatest Common Factor) first. This will help
    make the remaining factoring easier. 3x 3 − 12x = 0

                           GCF                      (         )
                                                 3x x 2 − 4 = 0

★   Factor completely!    Keep going here.
                                                 3x ( x − 2 ) ( x + 2 ) = 0
★   Set each factor equal to 0 and solve.
                         3x = 0   ( x − 2) = 0      ( x + 2) = 0
                          x=0        x=2                x = −2
Example: Solve by Factoring
     4    2
    x − 6x = 27
Example: Solve by Factoring
     4    2
    x − 6x = 27   Degree is 4 so can up
                     to 4 solutions.
Example: Solve by Factoring
        4       2
       x − 6x = 27   Degree is 4 so can up
   4        2           to 4 solutions.
  x − 6x − 27 = 0
Example: Solve by Factoring
               4                2
              x − 6x = 27               Degree is 4 so can up
          4             2                  to 4 solutions.
  x − 6x − 27 = 0
 (x   2
                   )(
          −9 x +3 =0        2
                                    )
Example: Solve by Factoring
                   4                2
                  x − 6x = 27               Degree is 4 so can up
              4             2                  to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                        )
 2
x −9=0
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                     Degree is 4 so can up
              4             2                        to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                     Degree is 4 so can up
              4             2                        to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2
     x =9
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                     Degree is 4 so can up
              4             2                        to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2
     x =9
     x=± 9
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                     Degree is 4 so can up
              4             2                        to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2
     x =9
     x=± 9
     x = ±3
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                     Degree is 4 so can up
              4             2                        to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2                                     2
     x =9                               x = −3
     x=± 9
     x = ±3
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                      Degree is 4 so can up
              4             2                         to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2                                     2
     x =9                               x = −3
     x=± 9                              x = ± −3
     x = ±3
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                      Degree is 4 so can up
              4             2                         to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )
 2                                      2
x −9=0                                  x +3= 0
      2                                     2
     x =9                               x = −3
     x=± 9                              x = ± −3
     x = ±3                             x = ±i 3
Example: Solve by Factoring
                   4                2
                  x − 6x = 27                      Degree is 4 so can up
              4             2                         to 4 solutions.
      x − 6x − 27 = 0
     (x   2
                       )(
              −9 x +3 =0        2
                                            )        Solutions:

 2
x −9=0                                  2
                                        x +3= 0
                                                          {±3, ±i 3}
      2                                     2
     x =9                               x = −3
     x=± 9                              x = ± −3
     x = ±3                             x = ±i 3
Try this: Solve by Factoring
    3    2
   x + 3x − x − 3 = 0
Try this: Solve by Factoring
    3    2
   x + 3x − x − 3 = 0   Degree is 3 so can up
                           to 3 solutions.
Try this: Solve by Factoring
        3       2
        x + 3x − x − 3 = 0    Degree is 3 so can up
                                 to 3 solutions.
 (          )
     x + 3x + ( −x − 3) = 0
      3     2
Try this: Solve by Factoring
         3        2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2
Try this: Solve by Factoring
         3        2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2


      2
     x −1= 0
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2


      2
     x −1= 0              x+3= 0
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2


      2
     x −1= 0              x+3= 0
                           x = −3
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2


      2
     x −1= 0              x+3= 0
        2
       x =1                x = −3
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2


          (           )
           x − 1 ( x + 3) = 0
            2


      2
     x −1= 0              x+3= 0
        2
       x =1                x = −3
        x = ±1
Try this: Solve by Factoring
          3       2
        x + 3x − x − 3 = 0        Degree is 3 so can up
                                     to 3 solutions.
 (            )
     x + 3x + ( −x − 3) = 0
      3      2


     x ( x + 3) − 1( x + 3) = 0
      2
                                     Solutions:
          (           )
           x − 1 ( x + 3) = 0
            2
                                          {1, −1, −3}
      2
     x −1= 0              x+3= 0
        2
       x =1                x = −3
        x = ±1
Try this: Solve by Factoring
    3   2
   x + x − 4x = 0
Try this: Solve by Factoring
    3   2
   x + x − 4x = 0   Degree is 3 so can up
                       to 3 solutions.
Try this: Solve by Factoring
        3       2
   x + x − 4x = 0       Degree is 3 so can up
                           to 3 solutions.
    (       2
   x x +x−4 =0      )
Try this: Solve by Factoring
          3       2
      x + x − 4x = 0      Degree is 3 so can up
                             to 3 solutions.
      (
   x x +x−4 =02
                      )
x=0
Try this: Solve by Factoring
          3
      x + x − 4x = 0  2       Degree is 3 so can up
                                 to 3 solutions.
      (
   x x +x−4 =02
                          )
                  2
x=0               x +x−4=0
Try this: Solve by Factoring
          3
      x + x − 4x = 0  2       Degree is 3 so can up
                                 to 3 solutions.
      (
   x x +x−4 =02
                          )
                  2           Not factorable so use
x=0               x +x−4=0    the quadratic formula
Try this: Solve by Factoring
          3
      x + x − 4x = 0    2                        Degree is 3 so can up
                                                    to 3 solutions.
      (
   x x +x−4 =02
                                )
                    2                            Not factorable so use
x=0               x +x−4=0                       the quadratic formula
                  a = 1; b = 1; c = −4
                        −1 ± 12 − 4 (1) ( −4 )
                  x=
                                2 (1)
Try this: Solve by Factoring
          3
      x + x − 4x = 0    2                        Degree is 3 so can up
                                                    to 3 solutions.
      (
   x x +x−4 =02
                                )
                    2                            Not factorable so use
x=0               x +x−4=0                       the quadratic formula
                  a = 1; b = 1; c = −4
                        −1 ± 12 − 4 (1) ( −4 )
                  x=
                                2 (1)

                       −1 ± 17
                    x=
                           2
Try this: Solve by Factoring
          3
      x + x − 4x = 0    2                        Degree is 3 so can up
                                                    to 3 solutions.
      (
   x x +x−4 =02
                                )
                    2                            Not factorable so use
x=0               x +x−4=0                       the quadratic formula
                  a = 1; b = 1; c = −4
                        −1 ± 12 − 4 (1) ( −4 )
                  x=
                                2 (1)            Solutions:
                       −1 ± 17                        −1 ± 17 
                                                              
                    x=                               0,       
                           2
                                                     
                                                          2   
                                                               
Try this: Solve by Factoring
 f ( x ) = x + 64
          3
Try this: Solve by Factoring
 f ( x ) = x + 64
          3
                    Degree is 3 so can up to 3 solutions.
Try this: Solve by Factoring
 f ( x ) = x + 64
            3
                    Degree is 3 so can up to 3 solutions.
                3
  0 = x + (4)
        3
Try this: Solve by Factoring
 f ( x ) = x + 64
            3
                     Degree is 3 so can up to 3 solutions.
                3
  0 = x + (4)
        3
                    Sum of cubes. Apply the formula.
Try this: Solve by Factoring
  f ( x ) = x + 64
              3
                        Degree is 3 so can up to 3 solutions.
                  3
    0 = x + ( 4 ) Sum of cubes. Apply the formula.
          3


0 = ( x + 4 ) ( x − 4x + 16 )
                 2
Try this: Solve by Factoring
  f ( x ) = x + 64
              3
                        Degree is 3 so can up to 3 solutions.
                  3
    0 = x + ( 4 ) Sum of cubes. Apply the formula.
          3


0 = ( x + 4 ) ( x − 4x + 16 )
                 2


x+4=0
Try this: Solve by Factoring
  f ( x ) = x + 64
              3
                        Degree is 3 so can up to 3 solutions.
                  3
    0 = x + ( 4 ) Sum of cubes. Apply the formula.
          3


0 = ( x + 4 ) ( x − 4x + 16 )
                 2

                     2
x+4=0              x − 4x + 16 = 0
Try this: Solve by Factoring
   f ( x ) = x + 64
                3
                         Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2

                      2
 x+4=0              x − 4x + 16 = 0
−4 = x
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                      2
 x+4=0              x − 4x + 16 = 0 use completing
                                           the square.
−4 = x
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                       2
 x+4=0                x − 4x + 16 = 0 use completing
                                           the square.
                                 2
−4 = x            x − 4x + ( −2 ) = −16 + 4
                    2
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                       2
 x+4=0                x − 4x + 16 = 0 use completing
                                           the square.
                                  2
−4 = x            x − 4x + ( −2 ) = −16 + 4
                    2
                               2
                      ( x − 2 ) = −12
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                       2
 x+4=0                x − 4x + 16 = 0 use completing
                                           the square.
                                  2
−4 = x            x − 4x + ( −2 ) = −16 + 4
                    2
                               2
                      ( x − 2 ) = −12
                          x − 2 = ± −12
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                       2
 x+4=0                x − 4x + 16 = 0 use completing
                                           the square.
                                  2
−4 = x            x − 4x + ( −2 ) = −16 + 4
                    2
                               2
                      ( x − 2 ) = −12
                          x − 2 = ± −12
                              x = 2 ± 2i 3
Try this: Solve by Factoring
    f ( x ) = x + 64
                3
                          Degree is 3 so can up to 3 solutions.
                    3
     0 = x + ( 4 ) Sum of cubes. Apply the formula.
            3


 0 = ( x + 4 ) ( x − 4x + 16 )
                  2
                                        Not factorable so
                       2
 x+4=0                x − 4x + 16 = 0 use completing
                                           the square.
                                  2
−4 = x            x − 4x + ( −2 ) = −16 + 4
                    2
                               2
                      ( x − 2 ) = −12
 Solutions:               x − 2 = ± −12
     {−4, 2 ± 2i 3}           x = 2 ± 2i 3
Practice Time!
★   Follow this link to practice solving polynomial equations
    using Factoring.
Solutions by Observing the Graph
★   The degree of the function tells you the maximum
    number of solutions possible.
Solutions by Observing the Graph
★   The degree of the function tells you the maximum
    number of solutions possible.
★   The real solutions are where the function crosses or
    touches the x-axis.
Solutions by Observing the Graph
★   The degree of the function tells you the maximum
    number of solutions possible.
★   The real solutions are where the function crosses or
    touches the x-axis.
★   The graph below has 4 solutions because it crosses the
    x-axis in 4 places. Notice 2 are positive real numbers
    and 2 are negative real numbers.
Example: Find all real zeros on the graph.
Example: Find all real zeros on the graph.
Example: Find all real zeros on the graph.
Example: Find all real zeros on the graph.
Example: Find all real zeros on the graph.
Example: Find all real zeros on the graph.
★   The real zeros for the graph below are   {−3, −1,1, 2}
You try: Find all real zeros on the graph.
You try: Find all real zeros on the graph.
You try: Find all real zeros on the graph.
You try: Find all real zeros on the graph.
You try: Find all real zeros on the graph.
★   The real zeros for the graph below are   {−1, 2, 5}
Practice Time!
★   Follow this link to practice solving polynomial equations
    using Factoring.
Solving by Graphing in Calculator
★   Graph the left side of the equation in Y1.
Solving by Graphing in Calculator
★   Graph the left side of the equation in Y1.
★   Graph the right side of the equation in Y2.
Solving by Graphing in Calculator
★   Graph the left side of the equation in Y1.
★   Graph the right side of the equation in Y2.
★   Find all the points the two graphs intersect. The x-
    coordinate is the solution.
Solving by Graphing in Calculator
★   Graph the left side of the equation in Y1.
★   Graph the right side of the equation in Y2.
★   Find all the points the two graphs intersect. The x-
    coordinate is the solution.
★   If you are given a function such as f(x) = x2 - 1, use zero
    for f(x). So Y1 = 0 and Y2 = x2 - 1. The find all the
    intersections.
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
  Solutions:
       {−2,1}
Example: Solve by Graphing
                                      2
    f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
                                2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
  Solutions:
       {−2,1}
                  Notice if you used the zero product
                  property, x = 1 would have occurred
                twice. We say 1 has multiplicity of 2.
Example: Solve by Graphing
       3    2
     4x − 8x = x − 2
Example: Solve by Graphing
            3       2
          4x − 8x = x − 2
      3         2
Y1 = 4x − 8x
Y2 = x − 2
Example: Solve by Graphing
            3       2
          4x − 8x = x − 2
      3         2
Y1 = 4x − 8x
Y2 = x − 2
Example: Solve by Graphing
            3       2
          4x − 8x = x − 2
      3         2
Y1 = 4x − 8x
Y2 = x − 2
Example: Solve by Graphing
            3       2
          4x − 8x = x − 2
      3         2
Y1 = 4x − 8x
Y2 = x − 2
Example: Solve by Graphing
            3       2
          4x − 8x = x − 2
      3         2
Y1 = 4x − 8x
Y2 = x − 2
Example: Solve by Graphing
              3       2
            4x − 8x = x − 2
        3         2
Y1 = 4x − 8x
Y2 = x − 2

   Solutions:
        1 1 
       − , , 2 
        2 2 
The end.

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Solve Polynomial Equations 3 Ways

  • 1. Solving Polynomial Equations ๏ Solve by Factoring ๏ Identify solutions on Graph ๏ Solve by Graphing
  • 2. Solving Polynomial Equations ★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).  
  • 3. Solving Polynomial Equations ★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).   ★ There are 3 ways to solve Polynomial Equations
  • 4. Solving Polynomial Equations ★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).   ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property
  • 5. Solving Polynomial Equations ★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).   ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph
  • 6. Solving Polynomial Equations ★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).   ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph (3) Using Synthetic Division (separate notes)
  • 7. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ★ It is possible for a polynomial equation to have fewer solutions (or none at all). ★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible. ★ Some solutions may be irrational or even imaginary.
  • 8. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 ★ It is possible for a polynomial equation to have fewer solutions (or none at all). ★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible. ★ Some solutions may be irrational or even imaginary.
  • 9. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.
  • 10. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible. ★ It is possible for a polynomial equation to have fewer solutions (or none at all).
  • 11. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible. ★ It is possible for a polynomial equation to have fewer solutions (or none at all). ★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.
  • 12. Total Number of Solutions ★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible. ★ It is possible for a polynomial equation to have fewer solutions (or none at all). ★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible. ★ Some solutions may be irrational or even imaginary.
  • 13. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. ★ Factor completely! ★ Set each factor equal to 0 and solve.
  • 14. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! ★ Set each factor equal to 0 and solve.
  • 15. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely!
  • 16. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! Keep going here.
  • 17. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0
  • 18. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0 ★ Set each factor equal to 0 and solve.
  • 19. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0 ★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0
  • 20. Solving by Factoring ★ When using factoring, remember to always look for a GCF (Greatest Common Factor) first. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0 ★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0 ★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0 x=0 x=2 x = −2
  • 21. Example: Solve by Factoring 4 2 x − 6x = 27
  • 22. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up to 4 solutions.
  • 23. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0
  • 24. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 )
  • 25. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 x −9=0
  • 26. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0
  • 27. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 x =9
  • 28. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 x =9 x=± 9
  • 29. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 x =9 x=± 9 x = ±3
  • 30. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ±3
  • 31. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3
  • 32. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2 x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
  • 33. Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) Solutions: 2 x −9=0 2 x +3= 0 {±3, ±i 3} 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
  • 34. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0
  • 35. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions.
  • 36. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2
  • 37. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2
  • 38. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2
  • 39. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0
  • 40. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0
  • 41. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 x = −3
  • 42. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3
  • 43. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
  • 44. Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 Solutions: ( ) x − 1 ( x + 3) = 0 2 {1, −1, −3} 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
  • 45. Try this: Solve by Factoring 3 2 x + x − 4x = 0
  • 46. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions.
  • 47. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( 2 x x +x−4 =0 )
  • 48. Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) x=0
  • 49. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 x=0 x +x−4=0
  • 50. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so use x=0 x +x−4=0 the quadratic formula
  • 51. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so use x=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1)
  • 52. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so use x=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) −1 ± 17 x= 2
  • 53. Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so use x=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) Solutions: −1 ± 17  −1 ± 17    x= 0,  2   2  
  • 54. Try this: Solve by Factoring f ( x ) = x + 64 3
  • 55. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions.
  • 56. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3
  • 57. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3 Sum of cubes. Apply the formula.
  • 58. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2
  • 59. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 x+4=0
  • 60. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 2 x+4=0 x − 4x + 16 = 0
  • 61. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 2 x+4=0 x − 4x + 16 = 0 −4 = x
  • 62. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. −4 = x
  • 63. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2 −4 = x x − 4x + ( −2 ) = −16 + 4 2
  • 64. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2 −4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12
  • 65. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2 −4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12
  • 66. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2 −4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12 x = 2 ± 2i 3
  • 67. Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2 −4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 Solutions: x − 2 = ± −12 {−4, 2 ± 2i 3} x = 2 ± 2i 3
  • 68. Practice Time! ★ Follow this link to practice solving polynomial equations using Factoring.
  • 69. Solutions by Observing the Graph ★ The degree of the function tells you the maximum number of solutions possible.
  • 70. Solutions by Observing the Graph ★ The degree of the function tells you the maximum number of solutions possible. ★ The real solutions are where the function crosses or touches the x-axis.
  • 71. Solutions by Observing the Graph ★ The degree of the function tells you the maximum number of solutions possible. ★ The real solutions are where the function crosses or touches the x-axis. ★ The graph below has 4 solutions because it crosses the x-axis in 4 places. Notice 2 are positive real numbers and 2 are negative real numbers.
  • 72. Example: Find all real zeros on the graph.
  • 73. Example: Find all real zeros on the graph.
  • 74. Example: Find all real zeros on the graph.
  • 75. Example: Find all real zeros on the graph.
  • 76. Example: Find all real zeros on the graph.
  • 77. Example: Find all real zeros on the graph. ★ The real zeros for the graph below are {−3, −1,1, 2}
  • 78. You try: Find all real zeros on the graph.
  • 79. You try: Find all real zeros on the graph.
  • 80. You try: Find all real zeros on the graph.
  • 81. You try: Find all real zeros on the graph.
  • 82. You try: Find all real zeros on the graph. ★ The real zeros for the graph below are {−1, 2, 5}
  • 83. Practice Time! ★ Follow this link to practice solving polynomial equations using Factoring.
  • 84. Solving by Graphing in Calculator ★ Graph the left side of the equation in Y1.
  • 85. Solving by Graphing in Calculator ★ Graph the left side of the equation in Y1. ★ Graph the right side of the equation in Y2.
  • 86. Solving by Graphing in Calculator ★ Graph the left side of the equation in Y1. ★ Graph the right side of the equation in Y2. ★ Find all the points the two graphs intersect. The x- coordinate is the solution.
  • 87. Solving by Graphing in Calculator ★ Graph the left side of the equation in Y1. ★ Graph the right side of the equation in Y2. ★ Find all the points the two graphs intersect. The x- coordinate is the solution. ★ If you are given a function such as f(x) = x2 - 1, use zero for f(x). So Y1 = 0 and Y2 = x2 - 1. The find all the intersections.
  • 88. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)
  • 89. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1)
  • 90. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1)
  • 91. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1)
  • 92. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1)
  • 93. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1}
  • 94. Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1) Y1 = 0 2 Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1} Notice if you used the zero product property, x = 1 would have occurred twice. We say 1 has multiplicity of 2.
  • 95. Example: Solve by Graphing 3 2 4x − 8x = x − 2
  • 96. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2
  • 97. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2
  • 98. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2
  • 99. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2
  • 100. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2
  • 101. Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2 Y1 = 4x − 8x Y2 = x − 2 Solutions:  1 1  − , , 2   2 2 

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