Polynomials

The ROOTS/ZEROES of P(x) are the values of x when P(x) = 0- If n is EVEN, then there may be NO real roots- If n is ODD, then there is AT LEAST ONE real root
A polynomial of degree n can have at most, n distinct solutions/roots
DOUBLE ROOTS can be found using the DERIVATIVE. If P(x) is a double root at x=a, then P(a) = P’(a) = 0

Graphs of polynomials
If n is EVENWhere Pn is POSITIVE (Pn > 0)
The curve starts from 2nd to 1st Quadrant.


Where Pn is NEGATIVE (Pn < 0)
The curve starts from 3rd to 4th quadrant


If n is ODDWhere Pn is POSITIVE (Pn > 0)
The curve starts from 3rd to 1st Quadrant



Where Pn is NEGATIVE (Pn < 0)
The curve starts from 2nd to 4th quadrantCurve is CONTINUOUS
EVEN fn does NOT have to cut x-axis
ODD fn MUST cut x-axis AT LEAST ONCE.
Long division

Division follows the format of:
P(x) = A(x). Q(x) + R(x) where

P(x) is the polynomial.
A(x) is the divisor
Q(x) is the quotient
R(x) is the



HSC Questions

Prediction Questions
1. Consider the polynomial equation  EMBED Equation.3 
 x3 – 20×2 – 99x + 81 = 0
a) Show that this equation has at least one real root in the interval -1

b) Draw a sketch of the graph of y= f(x), showing the coordinates of its points of intersection with the axes and all stationary points.
c) Apply Newton’s method once in order to approximate a root of f(x) = 0, beginning with an initial approximation of x1=1
d) Willy chose an initial approximation of x1=0.49, and used Newton’s Method a number of times in order to approximate a root of f(x) =0. State why this is NOT a good approximation. (It is not necessary to do further calculations)

3. The coefficient S and T are such that the polynomial x4 + Sx3 + Tx2 + 27x – 18 is exactly divisible by x2-3x+2. Find S and T, and completely factorise the polynomial.

Prediction Question 1 Solution
1. a) Let P(x) =  EMBED Equation.3 
x3 – 20 x2 – 99x + 81
P(-1) =  EMBED Equation.3 
(-1)3 – 20 (-1)2 – 99(-1) + 81 = 126.66

a) Given only one value, long division should thus be used.
 EMBED Equation.3 

8×3 + 16×2
- 4×2 – 18x
- 4×2 – 8x
- 10x – 20
- 10x – 20
0
.’. 8×3 + 12×2 – 18x – 20  EMBED Equation.3 
(x + 2)(8×2 - 4x -10)

For roots, factorise 8×2 - 4x -10 by letting F(x) = 0
8×2 - 4x -10 = 0 … 2 is common, so factorise out
4×2 – 2x – 5 = 0
Using quadratic formula: .’. Roots are -2, EMBED Equation.3 

x = EMBED Equation.3 

=  EMBED Equation.3 

= EMBED Equation.3 

= EMBED Equation.3 
b)


Prediction Question 3 Solution

P(2) = 24 + S(2)4 + T(2)2 + 27(2) – 18 … (1)
= 16 + 8S + 4T + 54 – 18
= 52 + 8S + 4T
= 0

P(1) = 14 + S(1)3 + T(1)2 + 27(1) -18 … (2)
= 1 + S + T + 9
= 10 + S + T
= 0

52 + 8S + 4 (- S – 10) = 0 … Sub (2) into (1)
52 + 8S – 4S – 40 = 0
4S + 12 = 0
4S = -12
S = -3

T = - (-3) – 10 … Sub value of S into T
 = 3 – 10
= -7

.’. S = -3 and T = -7








Method ONE
- Halving the interval –

If f(x) is a continuous

Where x=a is close to the root

More accurate but needs to be a continuous function and near the root!
It may also not work in some cases.

a1 = a – f(a)
f’(a)

- Three possible situations of halving the root –

1.

b

a a + b
2

2.
a + b b
2

a

3.

b

a + b
a 2

- When Newton’s Method does not work –

a1

a a1 a a

Q(x) +

The marking crit. tells you some commonly made mistakes AND has some useful suggestions.

In-depth explanation:
0. In THIS case, long division will hinder you and it turns into a mess of algebra.
1. Instead, factorise the divisor (remember, that’s the quadratic in this case!)
2. Substitute in the values into P(x)
3. Now use substitution with simultaneous equations.
4. After finding one value,


h`(Ö

y

(-3/2, 7)

-2 0

 EMBED Equation.3 
  EMBED Equation.3 
 x

-18

(1/2, -25)

c) f(x) = 8×3 + 12×2 – 18x – 20
f’(x) = 24×2 +24x – 18

x2 = x1 –  EMBED Equation.3 

= 1 -  EMBED Equation.3 

= 1.6

d) It is not close to the actual root itself.