Lesson 3- Polynomials

21 April 2023 ML3 MH

Objectives : - Definition - Dividing Polynomials

Next Lesson

- Factor Theorem - Remainder Theorem

Polynomial


kjxcxbxax nnn ..................21

Real numbers called coefficients Constant

n is the Degree of the polynomial

Multiplying Polynomials

Expand all the terms


126263)63)(2( 2232 xxxxxxxx

Dividing Polynomials

This is trickier than multiplication

There are two main ways

─ Long Division─ By Inspection


This PowerPoint presentation demonstrates two different methods of polynomial division.

Click here to see algebraic long division

Click here to see dividing “in your head”


Divide 2x³ + 3x² - x + 1 by x + 2

3 22 2 3 1x x x x x + 2 is the divisor

The quotient will be here.

2x³ + 3x² - x + 1 is the dividend


First divide the first term of the dividend, 2x³, by x (the first term of the divisor).

3 22 2 3 1x x x x

22xThis gives 2x². This will be the first term of the quotient.


Now multiply 2x²by x + 2

3 22 2 3 1x x x x 3 22 4x x

22x

2xand subtract


Bring down the next term, -x.

3 22 2 3 1x x x x 3 22 4x x

22x

2x x


Now divide –x², the first term of –x² - x, by x, the first term of the divisor

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

which gives –x.


Multiply –x by x + 2

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

xand subtract


Bring down the next term, 1

x

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

1


Divide x, the first term of x + 1, by x, the first term of the divisor

13 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x 1which gives 1


Multiply x + 2 by 1

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x

1

12x 1and subtract


The remainder is –1.

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x

1

12x 1

The quotient is 2x² - x + 1


Click here to see this example of algebraic long division again

Click here to see dividing “in your head”

Click here to end the presentation


Divide 2x³ + 3x² - x + 1 by x + 2

When a cubic is divided by a linear expression, the quotient is a quadratic and the remainder, if any, is a constant.

Let the remainder be d.

Let the quotient by ax² + bx + c

2x³ + 3x² - x + 1 = (x + 2)(ax² + bx + c) + d


2x³ + 3x² - x + 1 = (x + 2)(ax² + bx + c) + d

The first terms in each bracket give the term in x³

x multiplied by ax² gives ax³

so a must be 2.


2x³ + 3x² - x + 1 = (x + 2)(2x² + bx + c) + d

The first terms in each bracket give the term in x³

x multiplied by ax² gives ax³

so a must be 2.


2x³ + 3x² - x + 1 = (x + 2)(2x² + bx + c) + d

Now look for pairs of terms that multiply to give terms in x²

x multiplied by bx gives bx²

bx² + 4x² must be 3x²

2 multiplied by 2x² gives 4x²

so b must be -1.


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

Now look for pairs of terms that multiply to give terms in x²

x multiplied by bx gives bx²

bx² + 4x² must be 3x²

2 multiplied by 2x² gives 4x²

so b must be -1.


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

Now look for pairs of terms that multiply to give terms in x

x multiplied by c gives cx

cx - 2x must be -x

2 multiplied by -x gives -2x

so c must be 1.


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

Now look for pairs of terms that multiply to give terms in x

x multiplied by c gives cx

cx - 2x must be -x

2 multiplied by -x gives -2x

so c must be 1.


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

Now look at the constant term

2 multiplied by 1 gives 2

2 + d must be 1

then add d

so d must be -1.


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

Now look at the constant term

2 multiplied by 1 gives 2

2 + d must be 1

then add d

so d must be -1.


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

The quotient is 2x² - x + 1 and the remainder is –1.


Click here to see algebraic long division

Click here to see this example of dividing “in your head” again

Click here to end the presentation


Do the following

1.

2.

3.

4.


)12()7136( 23 xxxx

)3()15171392( 234 xxxxx

)47()21643283( 2234 xxxxxx

)1()23( 23 xxx

Exercises C1/C2 Page 82 Ex 3A, Nos 3, 6, 9, 16 to 20

Documents

Lesson 3- Polynomials