Remainder Theorem

Remainder theorem

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Remainder theorem

We can denote polynomials in x by using the notation of a function.

Example: ƒ(x) = 2x² + x – 1

g(x) = 3x³ + 9x² – 4x – 7

Q(x) = 9x³ – 3

For the polynomial ƒ(x) = 2x² + x – 1, when x = 3,

ƒ(3) = 2(3)² + (3) – 1

The value of the polynomial ƒ(x) is 20 when x = 3. = 20

2

Remainder theorem

ƒ(x) (x – a)

Remainder

(by long division)

ƒ(a)

(x³ + 2x² + 7x – 4) ÷ (x – 2)

(x³ + 2x² + 7x – 4) ÷ (x + 2)

26

ƒ(2) = 26

–18 ƒ(–2) = –18

Can you describe the relationship between

the remainder and ƒ(a)?

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Remainder theorem

When ƒ(x) ÷ (x – a) , remainder = ƒ(a).

x³ + 2x² + 7x – 4 = (x – 2)(x² + 4x + 15) + 26

Put x = 2 into (*). Then we have

ƒ(2) = (2 – 2)[(2)² + 4(2) + 15] + 26

= 26

0

ƒ(x) = (x – 2)(x² + 4x + 15) + 26 ......(*)

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Remainder theorem

When a polynomial ƒ(x) is divided by x – a and the degree of

the divisor is 1, the remainder must be of degree 0, i.e. the

remainder is a constant. Let Q(x) be the quotient and R be the remainder. Then we have:

ƒ(x) = (x – a) × Q(x) + R ............(**)

Put x = a into (**). Then we have

ƒ(a) = (a – a) × Q(a) + R

= 0 × Q(a) + R

= 0 + R

= R

When a polynomial ƒ(x) is divided by x – a,

the remainder is equal to ƒ(a).

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Remainder theorem

When a polynomial ƒ(x) is divided by mx – n, we have

ƒ(x) ≡ (mx – n) × Q(x) + R ............(***)

Put x = into (***). Then we have

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m

n

ƒ( ) = [m( ) – n] × Q( ) + R

= (n – n) × Q( ) + R

= 0 × Q( ) + R

= R

m

n

m

n

m

n

m

n

m

n

When a polynomial ƒ(x) is divided by mx – n,

the remainder is equal to ƒ( ).

m

n

Remainder theorem

When a polynomial ƒ(x) divided by

Using the remainder theorem, we can find the remainder

without performing division of polynomials.

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(i) x – 3, then the remainder =

(ii) x + 9 , then the remainder =

(iii) 3x – 2 , then the remainder =

(iv) 4x + 1 , then the remainder =

ƒ(3)

ƒ(–9)

3

2ƒ( )

4

1ƒ( )

Remainder theorem

Find the remainder when 2x³ + 5x² – 8x + 4 is divided by

(a) x – 2, (b) x + 4.

Let ƒ(x) = 2x³ + 5x² – 8x + 4.

By the remainder theorem,

remainder = ƒ(2)

= 2(2)³ + 5(2)² – 8(2) + 4

= 24

Let 2x³ + 5x² – 8x + 4 = (x – 2)Q(x) + R,

where Q(x) is the quotient and R is the remainder.

16 + 20 – 16 + 4 = R

R = 24

∴ The remainder is 24.

Alternative

method

When x = 2,

2(2)³ + 5(2)² – 8(2) + 4 = (2 – 2)Q(2) + R

(a)

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Example 1

Remainder theorem

By the remainder theorem,

remainder = ƒ(–4)

= 2(–4)³ + 5(–4)² – 8(–4) + 4

= –12

Let 2x³ + 5x² – 8x + 4 = (x + 4)Q(x) + R.

When x = –4,

2(–4)³ + 5(–4)² – 8(–4) + 4 = (–4 + 4)Q(–4) + R

(b)

–128 + 80 + 32 + 4 = R

R = –12

∴ The remainder is –12.

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Find the remainder when 2x³ + 5x² – 8x + 4 is divided by

(a) x – 2, (b) x + 4.

Example 1

Alternative

method

Remainder theorem

Find the remainder when x³ + 3x² – 2x is divided by

(a) 3x + 1, (b) 2x – 1.

Let ƒ(x) = x³ + 3x² – 2x.

=

10

(a) By the remainder theorem,

remainder = ƒ( )

= ( )³ + 3( )² – 2( )

3

1

3

1

3

1

3

1

3

2

3

1

27

1

27

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Example 2

Remainder theorem

11

By the remainder theorem,

remainder = ƒ( )

= ( )³ + 3( )² – 2( )

=

=

(b)

2

1

2

1

2

1

2

1

14

3

8

1

8

1

Find the remainder when x³ + 3x² – 2x is divided by

(a) 3x + 1, (b) 2x – 1.

Example 2

Remainder theorem

When the polynomial 3x² – 7x + 2 is divided by x – k

where k is a constant, the remainder is 2 – k. Find the value

of k.

Let ƒ(x) = 3x² – 7x + 2.

By the remainder theorem,

remainder = ƒ(k)

2 – k = 3k² – 7k + 2

0 = 3k² – 6k

3k(k – 2) = 0

3k = 0 or k – 2 = 0

k = 0 or k = 2

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Example 3

Remainder theorem

When the polynomial 9x² + 3x – k is divided by 3x – 1

where k is a constant, the remainder is –1. Find the value

of k.

Let ƒ(x) = 9x² + 3x – k, by the remainder theorem,

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–1 = 9( )³ + 3( ) – k

–1 = 1 + 1 – k

k = 3

remainder = ƒ( ) 3

1

3

1

3

1

Example 4

Remainder theorem

Consider the polynomial 4x³ + 2x² + mx + n, where m and n

are constants. When the polynomial is divided by x + 2, the

remainder is 7. When the polynomial is divided by 1 – x, the

remainder is 4. Find the values of m and n.

Let ƒ(x) = 4x³ + 2x² + mx + n.

When ƒ(x) is divided by x + 2, the remainder is 7.

By the remainder theorem,

7 = 4(–2)³ + 2(–2)² + m(–2) + n

remainder = ƒ(–2)

7 = –24 – 2m + n

31 = –2m + n

–2m + n = 31 ............ (1)

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Example 5

Remainder theorem

When ƒ(x) is divided by 1 – x, the remainder is 4.

By the remainder theorem,

4 = 4(1)³ + 2(1)² + m(1) + n

4 = 6 + m + n

–2 = m + n

m + n = –2 ............(2)

remainder = ƒ(1)

15

Consider the polynomial 4x³ + 2x² + mx + n, where m and n

are constants. When the polynomial is divided by x + 2, the

remainder is 7. When the polynomial is divided by 1 – x, the

remainder is 4. Find the values of m and n.

Example 5

Remainder theorem

(1) – (2) : –2m + n – (m + n) = 31 – (–2)

–2m + n – m – n = 33

–3m = 33

m = –11

Put m = –11 into (2).

m + n = –2

–11 + n = –2

n = 9

16

Consider the polynomial 4x³ + 2x² + mx + n, where m and n

are constants. When the polynomial is divided by x + 2, the

remainder is 7. When the polynomial is divided by 1 – x, the

remainder is 4. Find the values of m and n.

Example 5