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Core Mathematics 2 Algebra 1

Core Mathematics 2

Algebra

Edited by: K V Kumaran

Email: [email protected]

mailto:[email protected]


Algebra and functions

Simple algebraic division; use of the Factor Theorem and the

Remainder Theorem.

Only division by (x + a) or (x – a) will be required.

Students should know that if

f(x) = 0 when x = a, then (x – a) is a factor of f(x).

Students may be required to factorise cubic expressions

such as

x3 + 3x2 – 4 and 6x3 + 11x2 – x – 6.

Students should be familiar with the terms ‘quotient’ and

‘remainder’ and be able to determine the remainder when

the polynomial f(x) is divided by (ax + b).


Simplifying Algebraic Fractions

Algebraic fractions can be simplified by division.

Example 1: Simplify 3x

3+ 5x

2– 2x

x =

3x3

x+

5x2

x–

2x

x

= 3x2

+ 5x – 2


3+ 5x

2– 2x

x

= 3x

3

x+

5x2

x–

2x

x

= 3x2

+ 5x – 2


4– 5x

3+ 2x

2

– 2x =

6x4

– 2x–

5x3

– 2x+

2x2

– 2x = – 3x

3+

5x2

2– x


4– 5x

3+ 2x

2

– 2x

= 6x

4

– 2x–

5x3

– 2x+

2x2

– 2x

= – 3x3

+5x

2

2– x

Factorising to Cancel

Sometimes you need to factorise before you can simplify

Example 1: Simplify x

2+ 7x + 12

x + 3

= (x + 3)(x + 4)

x + 3

= x + 4

Example 1: Simplify x

2+ 7x + 12

x + 3

= (x + 3)(x + 4)

x + 3

= x + 4


2+ 5x – 12

2x2

– 7x + 6

= (2x – 3)(x + 4)

(x – 2)(2x – 3) =

x + 4

x – 2


2+ 5x – 12

2x2

– 7x + 6

= (2x – 3)(x + 4)

(x – 2)(2x – 3)

= x + 4

x – 2

________


Dividing a Polynomial by (x±p)

You divide polynomials in the same way as you perform long division.

Example 1: Divide 6x3

+ 28x2

– 7x + 15 by (x + 5) 6x2

– 2x + 3 = (x + 5) 6x3

+ 28x2

– 7x + 15 – 6x3

+ 30x2 – 2x

2– 7x – – 2x

2– 10x 3x + 15 – 3x + 15 0

6x

3+ 28x

2– 7x + 15

x + 5 = 6x

2– 2x + 3


+ 28x2

– 7x + 15 by (x + 5)

6x2

– 2x + 3

= (x + 5) 6x3

+ 28x2

– 7x + 15

– 6x3

+ 30x2

– 2x2

– 7x

– – 2x2

– 10x

3x + 15

– 3x + 15

0

6x

3+ 28x

2– 7x + 15

x + 5 = 6x

2– 2x + 3

Example 2: Divide – 5x3

– 27x2

+ 23x + 30 by (x + 6) – 5x2 + 3x + 5 = (x + 6) – 5x

3– 27x

2+ 23x + 30 – – 5x

3– 30x

2 3x

2+ 23x – 3x

2+ 18x 5x + 30 – 5x + 30 0

– 5x

3– 27x

2+ 23x + 30

x + 6 = – 5x

2+ 3x + 5

Example 2: Divide – 5x3

– 27x2

+ 23x + 30 by (x + 6)

– 5x2 + 3x + 5

= (x + 6) – 5x3

– 27x2

+ 23x + 30

– – 5x3

– 30x2

3x2

+ 23x

– 3x2

+ 18x

5x + 30

– 5x + 30

0

– 5x

3– 27x

2+ 23x + 30

x + 6 = – 5x

2+ 3x + 5

____________

____________

___________

___________

__________

____________


Finding the Remainder

After a division of a polynomial, if there is anything left then it is called the remainder. The answer

is called the quotient. If there is no remainder then what you are dividing by is said to be a factor.


– 5x2

– 16x + 10 by (x – 4) and state the remainder and quotient 2x2

+ 3x – 4 = (x – 4) 2x3

– 5x2

– 16x + 10 – 2x3

– 8x2 3x

2– 16x – 3x

2– 12x – 4x + 10 – – 4x + 16 – 6 The remainder is – 6 and the quotient is 2x

2+ 3x – 4


– 5x2

– 16x + 10 by (x – 4) and state the remainder and quotient

2x2

+ 3x – 4

= (x – 4) 2x3

– 5x2

– 16x + 10

– 2x3

– 8x2

3x2

– 16x

– 3x2

– 12x

– 4x + 10

– – 4x + 16

– 6

The remainder is – 6 and the quotient is 2x2

+ 3x – 4


+ 9x2

+ 25 by (x + 5) and state the remainder and quotient 2x2 – x + 5 = (x + 5) 2x

3+ 9x

2+ 0x + 25 – 2x

3+ 10x

2 – x

2+ 0x – – x

2– 5x 5x + 25 – 5x + 25 0 The remainder is 0 and the quotient is 2x

2– x + 5


+ 9x2

+ 25 by (x + 5) and state the remainder and quotient

2x2 – x + 5

= (x + 5) 2x3

+ 9x2

+ 0x + 25

– 2x3

+ 10x2

– x2

+ 0x

– – x2

– 5x

5x + 25

– 5x + 25

0

The remainder is 0 and the quotient is 2x2

– x + 5

____________

____________

____________

________

________

________

Remember to use 0x

as there is no x term.


Factor Theorem

Notes:

Factor Theorem: (x – a) is a factor of a polynomial f(x) if f(a) = 0.

Remainder Theorem: The remainder when a polynomial f(x) is divided by (x – a) is f(a).

Extended version of the factor theorem:

(ax + b) is a factor of a polynomial f(x) if 0b

fa

.

To find out the factors of a polynomial you can quickly substitute in values of x to see which give

you a value of zero.

Example 1: Given f(x) = x3

+ x2

– 4x – 4. Use the factor theorem to find a factor f(x) = x3

+ x2

– 4x – 4 f(1) = (1)3

+ (1)2

– (4 1) – 4 = – 6 (x – 1) is not a factor f(2) = (2)3

+ (2)2

– (4 2) – 4 = 0 (x – 2) is a factor

Example 1: Given f(x) = x3

+ x2

– 4x – 4. Use the factor theorem to find a factor

f(x) = x3

+ x2

– 4x – 4

f(1) = (1)3

+ (1)2

– (4 1) – 4

= – 6 (x – 1) is not a factor

f(2) = (2)3

+ (2)2

– (4 2) – 4

= 0 (x – 2) is a factor

To go from the x value to the factor simply put it into a bracket and change the sign.

Remember this:-

(x + 3)(x – 2) = 0 x = – 3 and x = 2

(x + 3)(x – 2) = 0

x = – 3 and x = 2

We are now going in reverse by putting the x

value back into the brackets


Example 2: Given f(x) = 3x2

+ 8x2 + 3x – 2. Factorise fully the given function f(x) = 3x

3+ 8x

2+ 3x – 2 f(1) = 3 (1)

3+ 8 (1)

2+ (3 1) – 2 = 12 (x – 1) is not a factor f(2) = 3 (2)

3+ 8 (2)

2+ (3 2) – 2 = 60 (x – 2) is not a factor f( – 1) = 3 ( – 1)

3+ 8 ( – 1)

2+ (3 – 1) – 2 = 0 (x + 1) is a factor

Example 2: Given f(x) = 3x2

+ 8x2 + 3x – 2. Factorise fully the given function

f(x) = 3x3

+ 8x2

+ 3x – 2

f(1) = 3 (1)3

+ 8 (1)2

+ (3 1) – 2

= 12 (x – 1) is not a factor

f(2) = 3 (2)3

+ 8 (2)2

+ (3 2) – 2

= 60 (x – 2) is not a factor

f( – 1) = 3 ( – 1)3

+ 8 ( – 1)2

+ (3 – 1) – 2

= 0 (x + 1) is a factor

To fully factorise it we now need to find the quotient. We do this by dividing by the factor we have

just found.

3x2

+ 5x – 2 = (x + 1) 3x3

+ 8x2

+ 3x – 2 – 3x3

+ 3x2 5x2

+ 3x – 5x2

+ 5x – 2x – 2 – – 2x – 2 0 (x + 1)(3x2

+ 5x – 2) = (x + 1)(3x – 1)(x + 2)

3x2

+ 5x – 2

= (x + 1) 3x3

+ 8x2

+ 3x – 2

– 3x3

+ 3x2

5x2

+ 3x

– 5x2

+ 5x

– 2x – 2

– – 2x – 2

0

(x + 1)(3x2

+ 5x – 2) = (x + 1)(3x – 1)(x + 2)

___________

___________

__________

_________

_________

_________

We have factorised the

quotient further to get the

fully factorised answer


Using the Factor Theorem to find the Remainder

If the polynomial f(x) is divided by (ax – b) then the remainder is f

b

a

If the polynomial f(x) is divided by (ax – b) then the remainder is f

b

a

Example 1: Find the remainder when f(x) = 4x4

– 4x2

+ 8x – 1 is divided by (2x – 1)

Using (ax – b) then a = 2, b = 1 so f

1

2

f(x) = 4x4

– 4x2

+ 8x – 1

f

1

2

= 4

1

2

4– 4

1

2

2+

8

1

2

– 1

=4

16 –

4

4 +

8

2 – 1 =

1

4 – 1 + 4 – 1 = 2 1

4 Remainder is 2

1

4 Example 2: Find the remainder if f(x) = x

3– 20x + 3 is divided by (x – 4) Using (ax – b) then a = 1, b = 4 so f(4) f(x) = x

3– 20x + 3 f(4) = (4)

3– (20 4) + 3 = 64 – 80 + 3 = – 13 Remainder is – 13

Example 1: Find the remainder when f(x) = 4x4

– 4x2

+ 8x – 1 is divided by (2x – 1)

Using (ax – b) then a = 2, b = 1 so f

1

2

f(x) = 4x4

– 4x2

+ 8x – 1

f

1

2

= 4

1

2

4– 4

1

2

2+

8

1

2

– 1

=4

16 –

4

4 +

8

2 – 1

=1

4 – 1 + 4 – 1

= 2 1

4

Remainder is 2 1

4

Example 2: Find the remainder if f(x) = x3

– 20x + 3 is divided by (x – 4)

Using (ax – b) then a = 1, b = 4 so f(4)

f(x) = x3

– 20x + 3

f(4) = (4)3

– (20 4) + 3

= 64 – 80 + 3

= – 13

Remainder is – 13


(C2-1.1) Name:

Homework Questions 1 – Simplifying Algebraic Fractions

Simplify the following by division

a)

3x3

+ 2x2

+ x

x

3x3

+ 2x2

+ x

x

b)

5x3

+ 4x2

– 2x

– 2x

5x3

+ 4x2

– 2x

– 2x

c)

4x3

+ 3x2

+ 7x

x

4x3

+ 3x2

+ 7x

x

d)

– 8x4

+ 6x3

– 6x2

– 2x2

– 8x4

+ 6x3

– 6x2

– 2x2

e)

5x2

+ 6x – 3

x2

5x2

+ 6x – 3

x2

f)

7x3

– 3x2

+ 5x

x2

7x3

– 3x2

+ 5x

x2

g)

3x5

– 6x3

2x2

3x5

– 6x3

2x2

h)

7x5

+ 3x2

+ 6

3x2

7x5

+ 3x2

+ 6

3x2

3

i)

4x2

+ 6x – 5

2x

4x2

+ 6x – 5

2x

j)

12x3

– 9x2

+ 3x

3x2

12x3

– 9x2

+ 3x

3x2


(C2-1.2) Name:

Homework Questions 2 – Factorising to Cancel

Simplify the following by factorizing first

a)

x2

+ 5x + 6

x + 2

x2

+ 5x + 6

x + 2

b)

x2

+ 6x – 7

x – 1

x2

+ 6x – 7

x – 1

c)

x2

+ 2x – 15

x – 3

x2

+ 2x – 15

x – 3

d)

x2

+ 10x + 24

x + 6

x2

+ 10x + 24

x + 6

e)

2x2

– 10x + 12

x – 3

2x2

– 10x + 12

x – 3

f)

x2

– 9x + 20

x – 4

x2

– 9x + 20

x – 4

g)

x2

– 49

x – 7

x2

– 49

x – 7

h)

x2

+ 10x + 9

x + 1

x2

+ 10x + 9

x + 1

i)

4x2

+ 12x + 9

2x + 3

4x2

+ 12x + 9

2x + 3

j)

7x3y – 28xy

3

x – 2y

7x3y – 28xy

3

x – 2y


(C2-1.3) Name:

Homework Questions 3 – Dividing a Polynomial

Divide the following to find the quotient

a)

2x3

+ 3x2

– 3x – 2 by (x + 2)

2x3

+ 3x2

– 3x – 2 by (x + 2)

b)

3x3

– 2x2

– 19x – 6 by (x – 3)

3x3

– 2x2

– 19x – 6 by (x – 3)

c)

x3

+ 7x2

– 6x – 72 by (x – 3)

x3

+ 7x2

– 6x – 72 by (x – 3)

e)

4x3

+ 3x2

– 25x – 6 by (x – 2)

4x3

+ 3x2

– 25x – 6 by (x – 2)


(C2-1.4) Name:

Homework Questions 4 – Finding the Remainder

Find the remainder by division when the following polynomials are divided by:-

a)

x2

– 4x + 5 by (x – 1)

x2

– 4x + 5 by (x – 1)

b)

2x3

+ 3x2

– 3x + 2 by (x + 2)

2x3

+ 3x2

– 3x + 2 by (x + 2)

c)

2x2

+ 7x – 3 by (2x – 1)

2x2

+ 7x – 3 by (2x – 1)

d)

x3

– 2x2

– 7x – 1 by (x + 1)

x3

– 2x2

– 7x – 1 by (x + 1)


(C2-1.5) Name:

Homework Questions 5 – Factor Theorem

Use the factor theorem to show that the following are factors, and hence fully factorise.

a)

(x – 3) is a factor of x3

– 3x2

– 4x + 12


– 3x2

– 4x + 12

b)

(x + 2) is a factor of x3

– 3x2

– 6x + 8


– 3x2

– 6x + 8

c)


+ 8x2

+ 19x + 12


+ 8x2

+ 19x + 12

d)


– 4x2

– 11x + 30


– 4x2

– 11x + 30

e)


+ 2x2

– 5x – 6


+ 2x2

– 5x – 6


(C2-1.6) Name:

Homework Questions 6 – Finding Remainder Using Factor Theorem

Find the remainder when the following polynomial is divided by:-

a)

3x3

– 2x2

+ 5x – 12 by (x – 1)

3x3

– 2x2

+ 5x – 12 by (x – 1)

b)

7x3

– 5x2

+ 12x + 16 by (x + 2)

7x3

– 5x2

+ 12x + 16 by (x + 2)

c)

6x3

+ 4x2

– 9x + 10 by (x + 1)

6x3

+ 4x2

– 9x + 10 by (x + 1)

d)

5x5

– 4x3

+ 2x2

– 8x + 11 by (x + 3)

5x5

– 4x3

+ 2x2

– 8x + 11 by (x + 3)

e)

7x4

+ 7x3

– 6x2

+ 9x + 2 by (x – 2)

7x4

+ 7x3

– 6x2

+ 9x + 2 by (x – 2)

f)

5x3

– 5x2

+ 9x + 12 by (x + 1)

5x3

– 5x2

+ 9x + 12 by (x + 1)

g)

7x4

– 4x3

+ 5x2

– 6x + 15 by (x + 2)

7x4

– 4x3

+ 5x2

– 6x + 15 by (x + 2)

h)

4x3

+ 6x2

+ 3x + 4 by (x – 3)

4x3

+ 6x2

+ 3x + 4 by (x – 3)

i)

7x4

– 9x3

– 8x2

– 7x + 5 by (x – 1)

7x4

– 9x3

– 8x2

– 7x + 5 by (x – 1)

j)

6x5

+ 4x3

+ 2x2

– x – 4 by (x – 3)

6x5

+ 4x3

+ 2x2

– x – 4 by (x – 3)


Past examination questions.

1. f(x) = x3 – 2x2 + ax + b, where a and b are constants.

When f(x) is divided by (x – 2), the remainder is 1.

When f(x) is divided by (x + 1), the remainder is 28.

(a) Find the value of a and the value of b. (6)

(b) Show that (x – 3) is a factor of f(x). (2)

(C2, Jan 2005 Q5)

2. (a) Use the factor theorem to show that (x + 4) is a factor of 2x3 + x2 – 25x + 12.

(2)

(b) Factorise 2x3 + x2 – 25x + 12 completely.

(4)

(C2, June 2005 Q3)

3. f(x) = 2x3 + x2 – 5x + c, where c is a constant.

Given that f(1) = 0,

(a) find the value of c,

(2)

(b) factorise f(x) completely,

(4)

(c) find the remainder when f(x) is divided by (2x – 3).

(2)

(C2, Jan 2006 Q1)

4. f(x) = 2x3 + 3x2 – 29x – 60.

(a) Find the remainder when f(x) is divided by (x + 2).

(2)

(b) Use the factor theorem to show that (x + 3) is a factor of f(x).

(2)

(c) Factorise f(x) completely.

(4)

(C2, May 2006 Q4)

5. f(x) = x3 + 4x2 + x – 6.

(a) Use the factor theorem to show that (x + 2) is a factor of f(x). (2)

(b) Factorise f(x) completely.


(4)

(c) Write down all the solutions to the equation

x3 + 4x2 + x – 6 = 0. (1)

(C2, Jan 2007 Q5)

6. f(x) = 3x3 – 5x2 – 16x + 12.

(a) Find the remainder when f(x) is divided by (x – 2).

(2)

Given that (x + 2) is a factor of f(x),

(b) factorise f(x) completely.

(4)

(C2, May 2007 Q2)

7. (a) Find the remainder when

x3 – 2x2 – 4x + 8

is divided by

(i) x – 3,

(ii) x + 2.

(3)

(b) Hence, or otherwise, find all the solutions to the equation

x3 – 2x2 – 4x + 8 = 0.

(4)

(C2, Jan 2008 Q1)

8. f(x) = 2x3 – 3x2 – 39x + 20

(a) Use the factor theorem to show that (x + 4) is a factor of f (x).

(2)

(b) Factorise f (x) completely.

(4)

(C2, June 2008 Q1)

9. f (x) = x4 + 5x3 + ax + b,

where a and b are constants.

The remainder when f(x) is divided by (x – 2) is equal to the remainder when f(x) is divided by (x +

1).

(a) Find the value of a. (5)


Given that (x + 3) is a factor of f(x),

(b) find the value of b. (3)

(C2, Jan 2009 Q6)

10. f(x) = (3x − 2)(x − k) − 8

where k is a constant.

(a) Write down the value of f (k).

(1)

When f(x) is divided by (x − 2) the remainder is 4.

(b) Find the value of k.

(2)

(c) Factorise f (x) completely.

(3)

(C2, June 2009 Q3)

11. f(x) = 2x3 + ax2 + bx – 6,


When f(x) is divided by (2x – 1) the remainder is –5.

When f(x) is divided by (x + 2) there is no remainder.

(a) Find the value of a and the value of b.

(6)


(3)

(C2, Jan 2010 Q3)

12. f(x) = 3x3 − 5x2 − 58x + 40.

(a) Find the remainder when f (x) is divided by (x − 3) .

(2)

Given that (x − 5) is a factor of f(x) ,

(b) find all the solutions of f(x) = 0 .

(5)

(C2, June 2010 Q2)


13. f(x) = x4 + x3 + 2x2 + ax + b,


When f(x) is divided by (x – 1), the remainder is 7.

(a) Show that a + b = 3.

(2)

When f(x) is divided by (x + 2), the remainder is 8.

(b) Find the value of a and the value of b.

(5)

(C2, Jan 2011 Q1)

14. f(x) = 2x3 − 7x2 − 5x + 4

(a) Find the remainder when f(x) is divided by (x – 1).

(2)

(b) Use the factor theorem to show that (x + 1) is a factor of f(x).

(2)

(c) Factorise f(x) completely.

(4)

(C2, May 2011 Q1)

15. f(x) = x3 + ax2 + bx + 3, where a and b are constants.

Given that when f (x) is divided by (x + 2) the remainder is 7,

(a) show that 2a − b = 6.

(2)

Given also that when f(x) is divided by (x −1) the remainder is 4,

(b) find the value of a and the value of b.

(4)

(C2, Jan 2012 Q5)

16. f(x) = 2x3 – 7x2 – 10x + 24.

(a) Use the factor theorem to show that (x + 2) is a factor of f(x).

(2)


(4)

(C2, May 2012 Q4)


17. f(x) = ax3 + bx2 − 4x − 3, where a and b are constants.

Given that (x – 1) is a factor of f(x),

(a) show that a + b = 7.

(2)

Given also that, when f(x) is divided by (x + 2), the remainder is 9,

(b) find the value of a and the value of b, showing each step in your working.

(4)

(C2, Jan 2013 Q2)

18. f(x) = 2x3 – 5x2 + ax + 18

where a is a constant.


(a) show that a = –9,

(2)

(b) factorise f(x) completely.

(4)

Given that

g(y) = 2(33y) – 5(32y) – 9(3y) + 18,

(c) find the values of y that satisfy g(y) = 0, giving your answers to 2 decimal places where

appropriate.

(3)

(C2, May 2013 Q3)

19. f(x) = –4x3 + ax2 + 9x – 18, where a is a constant.


(a) find the value of a,

(2)

(b) factorise f(x) completely,

(3)

(c) find the remainder when f(x) is divided by (2x – 1).

(2)

(C2, May 2014 Q4)


20. f(x) = 2x3 – 7x2 + 4x + 4.

(a) Use the factor theorem to show that (x – 2) is a factor of f(x).

(2)


(4)

(C2, May 2014 Q2)

21. f(x) = 6x3 + 3x2 + Ax + B, where A and B are constants.

Given that when f(x) is divided by (x + 1) the remainder is 45,

(a) show that B – A = 48.

(2)

Given also that (2x + 1) is a factor of f(x),

(b) find the value of A and the value of B.

(4)

(c) Factorise f(x) fully.

(3)

(C2, May 2015 Q3)

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