Math 4 Topics

1. THE FACTOR
THEOREM AND
ReyITnaSl dCoO BN.V PERanStEino,
T2

2. Objectives: • To identify whether a
given factor is a factor of
a polynomial function.
• To determine the factor of
a polynomial function.
• To find f(x) when the roots
or the zeros are given.

3. Just for a moment!
Is 3 is a factor of 27? If
yes, what makes it a
factor of 27?

4. Just for a moment!
Consider this pie graph below. Is each part
of the figure is equally divided? What does
it says?

5. Questions to answers:
Is factor a divisor?
Is quotient a factor?
If P(x) is divided by (x –c)
then what can you say
about the remainder so
that (x – c ) is a factor
of P(x)?

6. Let us discuss the
foTlhleo Rweimnagind:er Theorem states that when
the polynomial P(x) is divided by x – c, the
remainder is P(c).
Example: When P(x) = x3 – x2 – 4x + 4 is
by
divided by x – 2, the remainder is 0.
That is, P(2) = 0.
remainder
theorem

7. Let us discuss the
foTlhleo Rweimnagind:er Theorem states that when
the polynomial P(x) is divided by x – c, the
remainder is P(c).
Example: When P(x) = x3 – x2 – 4x + 4 is
by synthdeitvici ded by x – 2, we have;
division
22 11 --11 --44 44
22 22 --44
11 11 --22 00
remainder

8. Let us discuss the
following:
22 11 --11 --44 44
22 22 --44
11 11 --22 00
Notice that P(c) = 0, using synthetic division
P(x) = (x – c) ● Q(x) + R becomes
P(x) = (x – c) ● Q(x) + 0
P(x) = (x – c) ● Q(x).
remainder

9. Remember that: FACTOR THEOREM
Let P(x) be a polynomial. If P(c) = 0, where c is a real number, then (x
– c) is a factor of P(x). Conversely, if (x – c) is a factor of P(x), then P(c) = 0.
Since the theorem has a converse, the proof consists of two parts.
a.) If (x – c) is a factor of P(x), then P(c) = 0.
b.) If P(c) = 0, then (x – c) is a factor of P(x).

10. Remember that: Proof:
(a)
Suppose (x – c) is a factor of P(x), then P(x) = (x – c) ● Q(x). Since
the equation is an identity and is true for any value of x, then it must be true
for x = c. Then;
P(c) = (c – c) ● Q(x)
P(c) = 0 ● Q(x)
P(c) = 0

11. Remember that: Proof:
(b)
Suppose P(c) = 0. By remainder theorem, when P(x) is divided by (x –
c), the remainder (R) = P(c) = 0. Then;
P(x) = (x – c) ● Q(x) + 0
P(x) = (x – c) ● Q(x)
Therefore, (x – c) is a factor of P(x).

12. Illustrative Examples: 1. Show that x + 1 is a factor of 2x3 + 5x2 – 3.
Solution:
Let P(x) = 2x3 + 5x2 – 3
P(-1) = 2(-1)3 + 5(-1)2 – 3
P(-1) = -2 + 5 – 3
P(-1) = 0
By Factor theorem, x + 1 is a factor of
2x3 + 5x2 – 3.

13. Illustrative Examples: 2. Show that x - 2 is a factor of
x4 + x3 – x2 – x - 18.
Solution:
Let P(x) = x4 + x3 – x2 – x - 18
P(2) = (2)4 + (2)3 – (2)2 – (2) – 18
P(2) = 16 + 8 – 4 – 2 – 18
P(2) = 0
By Factor theorem, x – 2 is a factor of
x4 + x3 – x2 – x - 18.

14. Illustrative Examples: 3. Find a polynomial function of minimum degree whose
zeros are –2, 1, –1.
Solution:
By factor theorem, the polynomial must have the
following as factors, (x + 2) (x – 1) and (x + 1)
Thus; P(x) = (x + 2)(x – 1)(x + 1)
= (x + 2)(x2 – 1)
= x3 + 2x2 – x – 2

15. Test Yourself:
A - Use the factor theorem to determine whether the
first polynomial is a factor of the second.
1. (x + 1); x3 + x2 + x + 1
2. (x + 2); x8 + 2x7 + x + 2
3. (a – 1); a3 – 2a2 + a – 2
4. (x – 2); 4x3 – 3x2 – 8x + 4
5. (y – 2); 3y4 – 6y3 – 5y + 10

16. Test Yourself:
B – Find a polynomial function with integral
coefficients that has the given numbers as roots.
1.) 0, 1, - 2
2.) 2, -1 , -2
3.) 1, 1, 3
4.) 1/2 , 1, -1, 2
5.) 0, 1/2, -1, 2

17. Exercises:
1.) Find the value of k so that polynomial x – 2
is the factor of 2x3 – kx - 3.
2.) A. Tell whether the second polynomial is a
factor of the first .
a. P(x) = 3x3 – 8x2 + 3x + 2; (x – 2)
b. P(x) = 2x4 + x3 + 2x + 1; (x + 1)
c. P(x) = x3 + 4x2 + x – 6; (x + 3)
d. G(x) = 4x3 – 6x2 + 2x + 1; (2x – 1)
e. H(x) = x3 – 6x2 + 3x + 10; (x – 1)

18. >Let’s play< Determine the value of k which is necessary to
meet the given condition.
(x – 2) is a factor of 3x3 – x2 – 11x + k
(x + 3) is a factor of 2x5 + 5x4 + 3x3 + kx2 – 14x + 3
(x + 1) is a factor of –x4 + kx3 – x2 + kx + 10
(x + 2) is a factor of x3 + x2 + 5x + k
(x – 1) is a factor of x3 – x2 – 4x + k
(x – 5) is a factor of x3 – 3x2 – kx - 5
(x + 1) is a factor of 3x3 + kx2 – x – 2
(x + 4) is a factor of kx3 + 4x2 – x - 4
( x + 5) is a factor of kx2 + 4x - 5
(x – 2) is factor of x3 + 3x2 – kx + 2