Remainder and Factor Theorem Polynomials Polynomials Polynomials Combining polynomials Combining polynomials Combining polynomials Combining polynomials

Remainder and Factor Remainder and Factor TheoremTheorem

•PolynomialsPolynomials•Combining polynomialsCombining polynomials

•Function notationFunction notation•Division of PolynomialDivision of Polynomial•Remainder TheoremRemainder Theorem

•Factor TheoremFactor Theorem

Polynomials

• An expression that can be written in the form

• a + bx + cx2 + dx3 +ex4 + ….

• Things with Surds (e.g. x + 4x +1 ) and reciprocals (e.g. 1/x + x) are not polynomials

• The degree is the highest index• e.g 4x5 + 13x3 + 27x is of degree 5

Polynomials can be combined to give new

polynomials

Set-up a multiplication table

3x2 -5x -32x4

+4x2

6x6 -10x5 -6x4

+12x4-20x3 -12x2

Gather like terms= 6x6 - 10x5 + 6x4 - 20x3 -12x2

They can be added:2x2 - 5x - 3 + 2x4 + 4x2 + 2x

= 2x4 + 6x2 -3x -3

(2x4 + 4x2)(3x2 - 5x - 3)Or multiplied:

6x6 -10x5 -6x4+12x4-20x3 -12x2

Function Notation

• An polynomials function can be written as

• f(x) = a + bx + cx2 + dx3 +ex4 + ….• f(x) means ‘function of x’• instead of y = ….

• e.g f(x) = 4x5 + 13x3 + 27x

• f(3) means ….– “the value of the function when x=3”– e.g. for f(x) = 4x5 + 13x3 + 27x – f(3) = 4 x 35 + 13 x 33 + 27 x 3– f(3) = 972 + 351 + 81 = 1404

x

f(x)

Combining Functions (1)

• Suppose– f(x) = x3 + 2x +1– g(x) = 3x2 - x - 2

• g(x) or p(x) or q(x) or ….. Can all be used to define different functions

We can define a new function by any linear or multiplicative combination of these…

• e.g. 2f(x) + 3g(x) = 2(x3 + 2x +1) + 3(3x2 - x - 2)• e.g. 3 f(x) g(x) = 3(x3 + 2x +1)(3x2 - x - 2)

Combining Functions (2)We can define a new function by any linear or multiplicative combination of these…

• e.g. 2f(x) + 3g(x) = 2(x3 + 2x +1) + 3(3x2 - x - 2)= 2x3 + 4x + 2 + 9x2 -3x -6GATHER LIKE TERMS

x3 x2 x

= 2x3 + 9x2 + x - 4

2 + 9 + - 4

Combining Functions (3)e.g. 3 f(x) g(x)

= 3(x3 + 2x +1)(3x2 - x - 2)

x3 +2x +13x2 -x -2

Do multiplication table; gather like terms and then multiply through by 3

Finding one bracket given the other

Fill in the empty bracket:

x2 - x - 20 = (x + 4)( )

To get x2 the x must be multiplied by another x

x2 - x - 20 = (x + 4)(x )

To get -20 the +4 must be multiplied by -5

x2 - x - 20 = (x + 4)(x - 5)

Expand it to check: (x + 4)(x - 5) = x2 + 4x - 5x -20 = x2 - x - 20

We can do division now

f(x) = (x2 - x - 20) (x + 4)

f(x) = (x2 - x - 20) (x + 4)

x (x+4) x (x+4)

(x + 4) x f(x) = (x2 - x - 20)

It is exactly the same question as:=Fill in the empty bracket:

x2 - x - 20 = (x + 4)( )

Finding one bracket given the other - cubics Fill in the empty bracket:

x3 + 3x2 - 12x + 4 = (x - 2)( )

To get x3 the x must be multiplied by x2

To get +4 the -2 must be multiplied by -2

x3 + 3x2 - 12x + 4 = (x - 2)(x2 )

x3 + 3x2 - 12x + 4 = (x - 2)(x2 -2)

These 2 give us -2x2,

but we need 3x2This x must be multiplied by 5x to give us another 5x2x3 + 3x2 - 12x + 4 = (x - 2)(x2 +5x -2)

Finding one bracket given the other - cubics (2.1)

Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)( )

Using a multiplication table:

+4-12x+3x2x3

x

-2

ax2 +bx +c

x3

+4

Can put x3 and +4 in. They can only come from 1 place.So a = 1, c = -2


Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)(x2 +bx -2 )


-12x+3x2

x

-2

x2 +bx -2

x3

+4

-2x

-2x2

Complete more of the table by multiplying known values

Complete the rest algebraically

+bx2

-2bx


Fill in the empty bracket:x3 + 3x2 - 12x + 4 = (x - 2)(x2 -2 )


-12x+3x2

x

-2

x2 +bx -2

x3

+4

-2x

-2x2

+bx2

-2bx

+3x2 = bx2 - 2x2

-12x = -2bx - 2x

+3 = b - 2

-12 = -2b - 2

Either way,b = 5

+5x

Gather like terms

Dealing with remainders

Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) + R


x

+2

ax2 +bx +c

x3

remainder

+x-x2

+15

x3

Can put x3

This can only come from 1 place.So a = 1




+x-x2

x

+2

x2 +bx +c

x3

remainder

+15

x3

x2

2x2

We can fill this bit in now

The rest of the x2 term must come from here

bx2

bx2 + 2x2 = -x2

b + 2 = -1b = -3




+x-x2

x

+2

x2 -3x +c

x3

remainder

+15

x3

x2 -3x

2x2

-3x2

-6x

We can fill this bit in nowThe rest of the x term must come from here

cx

cx - 6x = +xc - 6 = 1

c = 7


Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +R


+x-x2

x

+2

x2 -3x +7

x3

remainder

+15

x3

x2 – 3x +7

2x2

-3x2

-6x

7x

+14

We can fill this bit in now- and we’ve got our 2nd function


Fill in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +R


+x-x2

x

+2

x2 -3x +7

x3

remainder

+15

x3

x2 – 3x +7

2x2

-3x2

-6x

7x

+14Remainder

The numerical term (+15) comes from the +14 and the remainder R

+15 = +14 + R So, R = 1


Filled in the empty bracket:x3 - x2 + x + 15 = (x + 2)( ) +1


+x-x2

x

+2

x2 -3x +7

x3+15

x3

x2 – 3x +7

2x2

-3x2

-6x

7x

+14Remainder

The numerical term (+15) comes from the +14 and the remainder R

+15 = +14 + R So, R = 1

Fill in the empty bracket:

f(x) = x3 - x2 + x + 15 = (x + 2)( ) + R

If:f(x) = x3 - x2 + x + 15

Division with Remainders

What is f(x) divided by x+2 …. and what is the remainder

This is exactly the same as ……………

We found:x3 - x2 + x + 15 = (x + 2)( ) +1x2 – 3x +7

If:f(x) = x3 - x2 + x + 15

What is f(x) divided by x+2? …. and what is the remainder?

What is f(x) divided by x+2? (x2 – 3x +7)

…. and what is the remainder? 1

If:f(x) = x3 - x2 + x + 15 = (x + 2)( ) +1 What is f(x) divided by x+2? (x2 – 3x +7)


The Remainder Theorem – example 1

x2 – 3x +7

If we calculate f(-2) …..

f(-2) = (-2)3 – (-2)2 + -2 + 15 = -8 -4 - 2 +15 = 1

-2 from (x+2)=0Our remainder

If:p(x) = x3 + 2x2 - 9x + 10= (x - 2)( ) +8 What is p(x) divided by x-2? (x2 + 4x - 1)


The Remainder Theorem – example 2

x2 + 4x - 1

If we calculate p(2) …..

p(2) = (2)3 + 2(2)2 - 9(2) + 10 = 8 + 8 - 18 + 10 = 8

2 from (x-2)=0Our remainder

When p(x) is divided by (x-a) …. the remainder is p(a)

The Remainder Theorem

Given:p(x) = 2x3 - 5x2 + x - 12

What value of p(...)=0,hence will give no remainder?

The Factor Theorem – example

If we calculate p(0) = 2(0)3 – 5(0)2 + 0 - 12 = -12

p(1) = 2(1)3 – 5(1)2 + 1 - 12 = 2-5+1-12 = -14

p(2) = 2(2)3 – 5(2)2 + 2 - 12 = 16-20+2-12 = -16

p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0

By the Remainder Theorem :- the factor (x-3) gives no

remainder

For bigger values of ‘x’ the x3 term will dominate and make p(x) larger

Given:p(x) = 2x3 - 5x2 + x - 12

The Factor Theorem – example

p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0

By the Remainder Theorem :- the factor (x-3) gives no

remainderSo (x-3) divides exactly into p(x)

……… (x-3) is a factor

For a given polynomial p(x)

If p(a) = 0

… then (x-a) is a factor of p(x)

The Factor Theorem

If:p(x) = x3 + bx2 + bx + 5 When is p(x) divided by x+2 the remainder is 5

If we calculate p(-2) …..

p(-2) = (-2)3 + b(-2)2 + b(-2) + 5 = -8 + 4b - 2b + 5 = 2b - 3-2 from (x+2)

Which theorem?The Remainder Theorem

By the Remainder theorem: 2b - 3 = 5 Our remainder

2b = 8 b = 4

If:f(x) = x3 + 3x2 - 6x - 8 a) Find f(2) f(2) = (2)3 + 3(2)2 - 6(2) - 8

= 8 + 12 - 12 - 8 = 0

b) Use the Factor Theorem to write a factor of f(x)

For a given polynomial p(x)If p(a) = 0… then (x-a) is a factor of p(x)

f(2) = 0…. so (x-2) is a factor of x3 + 3x2 - 6x - 8

If:f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8c) Express f(x) as a product of 3 linear factors

.. means (x-a)(x-b)(x-c)=x3 + 3x2 - 6x - 8We know (x-2)(x-b)(x-c)=x3 + 3x2 - 6x - 8

…. consider (x-2)(ax2+bx+c)=x3 + 3x2 - 6x - 8

a=? a=1 : so x x ax2 = x3

(x-2)(x2+bx+c)=x3 + 3x2 - 6x - 8

c=? c=4 : so -2 x 4 = -8 (x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8


(x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8Expand : need only check the x2 or x terms

… + bx2 -2x2 + … = … + 3x2 + ….

Or … - 2bx + 4x + … = … - 6x + ….

b - 2 = 3 b=5

-2b + 4 = -6 b=5

EASIER

HARD

(x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8


(x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8

(x2+5x+4) = (x+4)(x+1)

So, (x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8

……. a product of 3 linear factors

(x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8

……. Sketch x3 + 3x2 - 6x - 8

y = x3 + 3x2 - 6x - 8y = (x-2)(x+4)(x+1)

Where does it cross the x-axis (y=0) ?

(x-2)(x+4)(x+1) = 0 Either (x-2) = 0 x=2 Or (x+4)= 0 x=-4Or (x+1) = 0 x=-1

Where does it cross the y-axis (x=0) ?

y = (0)3 + 3(0)2 - 6(0) - 8 = -8

Factor and Remainder Theorem

Where does it cross the x-axis (y=0) ? Either (x-2) = 0 x=2 Or (x+4)= 0 x=-4Or (x+1) = 0 x=-1

Where does it cross the y-axis (x=0) ? y = -8

x

y

Goes through these-sketch a nice curve

x-1

x-4

x2

x-8

Documents

Remainder and Factor Theorem Polynomials Polynomials Polynomials Combining polynomials Combining polynomials Combining polynomials Combining polynomials