Polynomials

Higher Unit 2
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Higher

Outcome 1

What is a polynomials
Evaluating / Nested / Synthetic Method
Factor Theorem
Factorising higher Orders
Factors of the form (ax + b)
Finding Missing Coefficients
Finding Polynomials from its zeros
Exam Type Questions

Polynomials
Higher

Definition

Outcome 1


A polynomial is an expression with several terms.
These will usually be different powers of a particular letter.
The degree of the polynomial is the highest power that
appears.

Examples
3x4 – 5x3 + 6x2 – 7x - 4

Polynomial in x of degree 4.

7m8 – 5m5 – 9m2 + 2

Polynomial in m of degree 8.

w13 – 6

Polynomial in w of degree 13.

NB: It is not essential to have all the powers from the highest
down, however powers should be in descending order.

Disguised Polynomials

Higher

Outcome 1
(x + 3)(x – 5)(x + 5)= (x + 3)(x2 – 25)= x3 + 3x2 – 25x - 75
So this is a polynomial in x of degree 3.
Coefficients
In the polynomial 3x4 – 5x3 + 6x2 – 7x – 4 we say that
the coefficient of x4 is 3
the coefficient of x3 is -5
the coefficient of x2 is 6
the coefficient of x is -7
and the coefficient of x0 is -4
(NB: x0 = 1)
In w13 – 6 , the coefficients of w12, w11, ….w2, w are all zero.

Evaluating Polynomials
Outcome 1


Higher

Suppose that g(x) = 2x3 - 4x2 + 5x - 9
Substitution Method
g(2) = (2 X 2 X 2 X 2) – (4 X 2 X 2 ) + (5 X 2) - 9
= 16 – 16 + 10 - 9
= 1
NB:

this requires 9 calculations.

Nested or Synthetic Method
Outcome 1


Higher

This involves using the coefficients and requires fewer
calculations so is more efficient.
It can also be carried out quite easily using a calculator.
g(x) = 2x3 - 4x2 + 5x - 9
Coefficients are 2, -4,
5, -9
2
2

-4

5

-9

4

g(2) =

0
5

10

0

1

This requires only 6 calculations so is 1/3 more efficient.

Nested or Synthetic Method
Outcome 1


Higher

Example
If

f(x) = 2x3 - 8x

then the coefficients are

and

2

0

f(2) = 2 2

0

-8

0

8
0

0

2

4
4

0

-8

0

Factor Theorem
Outcome 1


Higher

If

(x – a) is a factor of the polynomial f(x)
Then

f(a) = 0.

Reason
Say f(x) = a3x3 + a2x2 + a1x + a0 = (x – a)(x – b)(x – c)
polynomial form
factorised form
Since (x – a), (x – b) and (x – c) are factors
then f(a) = f(b) = f(c ) = 0
Check
f(b) = (b – a)(b – b)(b – c) = (b – a) X 0 X (b – c) = 0

Factor Theorem
Outcome 1


Higher

Now consider the polynomial
f(x) = x3 – 6x2 – x + 30 = (x – 5)(x – 3)(x + 2)
So

f(5) = f(3) = f(-2) = 0

The polynomial can be expressed in 3 other factorised forms

A f(x) = (x – 5)(x2 – x – 6)
B f(x) = (x – 3)(x2 – 3x – 10)
C f(x) = (x + 2)(x2 – 8x + 15)

These can be
checked by
multiplying out
the brackets !

Keeping coefficients in mind an interesting thing occurs when
we calculate f(5) , f(3) and f(-2) by the nested method.

Factor Theorem
Outcome 1


Higher

A f(5) = 5

1
1

-6
5
-1

Other factor is x2 – x - 6
= (x – 3)(x + 2)

-1
-5
-6

30
-30
0

f(5) = 0 so (x – 5) a factor

Factor Theorem
Outcome 1


Higher

B

f(3) = 3

1
1

-6
3
-3

-1
-9
-10

30
-30
0

Other factor is x2 – 3x - 10 f(3) = 0 so (x – 3) a factor
= (x – 5)(x + 2)

Factor Theorem
Outcome 1


Higher

C

f(-2) = -2

1
1

-6
-2
-8

-1
16
15

30
-30
0

Other factor is x2 – 8x + 15 f(-2) = 0 so (x +2) a factor
= (x – 3)(x - 5)
This connection gives us a method of factorising
polynomials that are more complicated then quadratics
ie cubics, quartics and others.

Factor Theorem

Higher

Outcome 1

Example
Factorise

x3 + 3x2 – 10x - 24 We need some trial &
error with factors of –24
ie
+/-1, +/-2, +/-3 etc

f(-1) = -1 1
1
f(1) = 1

1
1

3
-1
2

-10
-2
-12

-24
12
-12

No good

3
1
4

-10
4
-6

-24
-6
-30

No good

Factor Theorem
Outcome 1


Higher

f(-2) = -2 1
1
Other factor is
So

3
-2
1

-10
-2
-12

x2 + x - 12

-24
24
0

f(-2) = 0
so (x + 2) a
factor

= (x + 4)(x – 3)

x3 + 3x2 – 10x – 24 = (x + 4)(x + 2)(x – 3)

Roots/Zeros
The roots or zeros of a polynomial tell us where
it cuts the X-axis. ie where f(x) = 0.
If a cubic polynomial has zeros a, b & c then it has
factors (x – a), (x – b) and (x – c).

We need some
trial & error with
factors of –9 ie

Factorising Higher Orders
Outcome 1

Higher

+/-1, +/-3 etc


Example
Solve
f(-1) = -1

x4 + 2x3 - 8x2 – 18x – 9 = 0
1

2
-1

-8
-1

1

1

-9

-18
9
-9

-9
9
0

f(-1) = 0
so
(x + 1) a
factor

Other factor is x3 + x2 – 9x - 9 which we can call g(x)

test

+/-1, +/-3 etc

Outcome 1


Higher

g(-1) = -1

1

1
-1

-9
0

-9
9

1

0

-9

0

g(-1) = 0
so (x
+ 1) a factor

Other factor is x2 – 9 = (x + 3)(x – 3)
if

x4 + 2x3 - 8x2 – 18x – 9 = 0

then

(x + 3)(x + 1)(x + 1)(x – 3) = 0
So

x = -3 or x = -1 or x = 3


Higher

Summary

Outcome 1

A cubic polynomial ie

ax3 + bx2 + cx + d

could be factorised into either
(i) Three linear factors of the form (x + a) or (ax + b)

or

(ii) A linear factor of the form (x + a) or (ax + b) and
a quadratic factor (ax2 + bx + c) which doesn’t
factorise.
or
IT DIZNAE
(iii) It may be irreducible.
FACTORISE

Linear Factors in the form (ax + b)
Outcome 1


Higher

If

(ax + b) is a factor of the polynomial f(x)
then f(-b/a) = 0

Reason

Suppose f(x) = (ax + b)(………..)
If f(x) = 0 then (ax + b)(………..) = 0
So (ax + b) = 0 or (…….) = 0
so ax = -b
so x = -b/a

NB:

When using such factors we need to take care
with the other coefficients.



Higher

Outcome 1

Example

Show that (3x + 1) is a factor of g(x) = 3x3 + 4x2 – 59x – 20
and hence factorise the polynomial completely.

Since (3x + 1) is a factor then g(-1/3) should equal zero.
g(-1/3) =

-1

/3 3

3

4
-1
3

-59
-1
-60

-20
20
0

g(- 1/3) = 0
so (x + 1/3)
is a factor

3x2 + 3x - 60

Outcome 1


Higher

Other factor is
3x2 + 3x - 60 = 3(x2 + x – 20) = 3(x + 5)(x – 4)
NB: common factor

Hence g(x) = (x + 1/3) X 3(x + 5)(x – 4)
= (3x + 1)(x + 5)(x – 4)

Missing Coefficients
Higher

Example

Outcome 1


Given that (x + 4) is a factor of the polynomial
f(x) = 2x3 + x2 + ax – 16

find the value of a

and hence factorise f(x) .
Since
f(-4)

=

(x + 4) a factor

then

2

a
28
(a + 28)

-4

2

1
-8
-7

f(-4) = 0 .
-16
(-4a – 112)
(-4a – 128)

Outcome 1


Higher

Since -4a – 128 = 0
then 4a = -128
so
a = -32
If a = -32
then the other factor is

2x2 – 7x - 4
= (2x + 1)(x – 4)

So

f(x) = (2x + 1)(x + 4)(x – 4)

Higher

Outcome 1

Example


(x – 4) is a factor of f(x) = x3 + ax2 + bx – 48
while f(-2) = -12.
Find a and b and hence factorise f(x) completely.

(x – 4) a factor so f(4) = 0
f(4) = 4 1
1

a
4

b
(4a + 16)

(a + 4) (4a + b + 16)
16a + 4b + 16 = 0 (÷4)
4a + b + 4 = 0
4a + b = -4

-48
(16a + 4b + 64)
(16a + 4b + 16)

Outcome 1

Higher


f(-2) = -12 so
f(-2) = -2 1
1

a
-2

b
(-2a + 4)

-48
(4a - 2b - 8)

(a - 2) (-2a + b + 4)

(4a - 2b - 56)

4a - 2b - 56 = -12 (÷2)
2a - b - 28 = -6
2a - b = 22
We now use simultaneous equations ….

Outcome 1


Higher

4a + b = -4
2a - b = 22
add

6a = 18
a=3

Using 4a + b = -4
12 + b = -4
b = -16

When (x – 4) is a factor the quadratic factor is
x2 + (a + 4)x + (4a + b + 16) = x2 + 7x + 12 =(x + 4)(x + 3)
So f(x) = (x - 4)(x + 3)(x + 4)

Finding a Polynomial From Its Zeros


Higher

Caution
Suppose that
f(x) = x2 + 4x - 12

Outcome 1
and g(x) = -3x2 - 12x + 36

f(x) = 0

g(x) = 0

x2 + 4x – 12 = 0

-3x2 - 12x + 36 = 0

(x + 6)(x – 2) = 0

-3(x2 + 4x – 12) = 0

x = -6 or x = 2

-3(x + 6)(x – 2) = 0
x = -6 or x = 2

Although f(x) and g(x) have identical roots/zeros they are
clearly different functions and we need to keep this in mind
when working backwards from the roots.



Higher

Outcome 1

If a polynomial f(x) has roots/zeros at a, b and c
then it has factors (x – a), (x – b) and (x – c)
And can be written as f(x) = k(x – a)(x – b)(x – c).

NB: In the two previous examples
k = 1 and k = -3 respectively.

Outcome 1


Higher

Example
y = f(x)
30

-2

1

5

Outcome 1


Higher

f(x) has zeros at x = -2, x = 1 and x = 5,
so it has factors (x +2), (x – 1) and (x – 5)
so

f(x) = k (x +2)(x – 1)(x – 5)

f(x) also passes through (0,30) so replacing x by 0
and f(x) by 30 the equation becomes
30 = k X 2 X (-1) X (-5)
ie

10k = 30

ie

k=3

Outcome 1


Higher

Formula is

f(x) = 3(x + 2)(x – 1)(x – 5)
f(x) = (3x + 6)(x2 – 6x + 5)
f(x) = 3x3 – 12x2 – 21x + 30

Higher Maths

Quadratic Theory
Strategies

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Quadratic Theory

Higher

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Quadratic Theory
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Quadratic Theory

Higher

Show that the line with equation y = 2 x + 1
does not intersect the parabola

y = x 2 + 3x + 4
with equation
Put two equations equal
Use discriminant
Show discriminant < 0
No real roots
Hint

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Quadratic Theory

a) Write

Higher

f ( x) = x + 6 x + 11 in the form( x + a ) + b
2

2

b) Hence or otherwise sketch the graph of
a)

f ( x) = ( x + 3) 2 + 2

b)

This is graph of

y = f ( x)

y = x2

minimum t.p. at (-3, 2)

moved 3 places to left and 2 units up.
y-intercept at (0, 11)

Hint

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Quadratic Theory

Show that the equation

Higher

(1 − 2k ) x 2 − 5kx − 2k = 0

has real roots for all integer values of k
Use discriminant

a = (1 − 2k )

b = −5k

c = −2k

b 2 − 4ac = 25k 2 − 4 × (1 − 2k ) × ( −2k )

→ 9 k 2 + 8k

= 25k 2 + 8k − 16k 2

Consider when this is greater than or equal to zero
Sketch graph

cuts x axis at

k =0

and

k =−

8
9

Hence equation has real roots for all integer k
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Quadratic Theory

Higher

The diagram shows a sketch of a parabola
passing through (–1, 0), (0, p) and (p, 0).
a) Show that the equation of the parabola is
y =p + p − x − 2
(
1)
x

y =x +p a tangent to this curve?
be

b)For what value of p will the line

a)

y = k ( x + 1)( x − p )

p = − pk

Use point (0, p) to find k

k = −1

y = −( x + 1)( x − p )
→ y = p + ( p − 1) x − x 2

→ y = − x 2 + px − x + p
b)

Simultaneous equations

0 = ( p − 2) x − x2
Previous

p = k (0 + 1)(0 − p )

x + p = p + ( p − 1) x − x 2
Discriminant = 0 for tangency
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p=2

Hint

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Quadratic Theory

Given

Higher

f ( x) = x + 2 x −, 8
2

express f ( x) in the form

( x + a)

2

−b

f ( x) = ( x + 1) 2 − 10

Hint

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Quadratic Theory

Higher

For what value of k does the equation
Discriminant

x 2 − 5 x + (k + 6) = 0 have equal roots?

a = 1 b = −5 c = k + 6
b 2 − 4ac = 25 − 4(k + 6)

For equal roots
discriminant = 0

0 = 25 − 4k − 24
4k = 1
1
k=
4
Hint

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Quadratic Theory

Higher

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Polynomials

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