1.
Block 2
Polynomials
(Approximate Roots)

2.
What is to be learned?
• How to find approximate roots.

3.
x2
+ 3x + 1 = 0
Will not factorise
→ Use Big Nasty Formula
x3
+ 2x2
– 5 = 0
Using Big L?
No whole number root

4.
Solving Graphically
x2
– 5x + 6 = 0
y = x2
– 5x + 6
2 3
solutions
x = 2 , x = 3
solutions occur
at y = 0

5.
Solving Graphically
x3
+ 2x2
– 5 = 0
y = x3
+ 2x2
– 5
1 2
solution between 1 and 2
1< x < 2
y = 13
+ 2(1)2
– 5 = -2
y = 23
+ 2(2)2
– 5 = 11
x = 1
x = 2
(1 , -2)
(2 , 11)

6.
For “exact” root, y = 0
For approximate roots, get as close as
you can to y = 0
Looking for x value, when y → 0
Root somewhere between positive and
negative value. (of y!)
1 2
(1 , -2)
(2 , 11)

7.
x = 1.5,
x = 1.3,
x = 1.2,
x = 1.25,
x = 1 ,
x3
+ 2x2
– 5 = 0
y = x3
+ 2x2
– 5
x = 2 , 1< x < 2
better estimate than x = 2
1< x < 1.5
1< x < 1.3
1.2< x < 1.3
1.2< x < 1.25
calculate x (to 1 d.p.)
x = 1.2 to 1 d.p.
y = 2.875y = 2.875
y = 0.577y = 0.577
y =y = --0.3920.392
y = 0.078y = 0.078
y = -2y = -2
y = 11y = 11

8.
Finding Approximate roots of
Polynomials
For “exact” root, y = 0
For approximate roots, get as close as
you can to y = 0
Looking for x value, when y → 0
Root (x =) between positive and
negative value of y.

9.
x = 2.5,
x = 2.3,
x = 2.2,
x = 2.25,
x = 2 , y = -2
x3
– x2
– 6 = 0
y = x3
– x2
– 6
x = 3 , y = 12 2< x < 3
better estimate than x = 3
2< x < 2.5
2< x < 2.3
2.2< x < 2.3
2.2< x < 2.25
calculate x (to 1 d.p.)
x = 2.2 to 1 d.p.
root between x = 2 and x = 3
y = 3.375y = 3.375
y = 0.877y = 0.877
y =y = --0.3920.392
y = 0.078y = 0.078
(negative)
(positive)

10.
x = 1.5,
x = 1.7,
x = 1.6,
x = 1.65,
x = 1 ,
x3
– x2
– 2 = 0
y = x3
– x2
– 2
x = 2 , 1< x < 2
1.5 < x < 2
1.5 < x < 1.7
1.6< x < 1.7
1.65< x < 1.7
calculate x (to 1 d.p.)
x = 1.7 to 1 d.p.
y = -0.875y = -0.875
y = 0.023y = 0.023
y =y = --0.4640.464
y = -0.230y = -0.230
y = -2y = -2
y = 2y = 2
has a root between
x = 1 and x = 2
Key Question