© Boardworks Ltd 2010 1 of 8 The Factor Theorem. © Boardworks Ltd 2010 2 of 8 The Factor Theorem...

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© Boardworks Ltd 2010 1 of 8 The Factor Theorem The Factor Theorem

Transcript of © Boardworks Ltd 2010 1 of 8 The Factor Theorem. © Boardworks Ltd 2010 2 of 8 The Factor Theorem...

Page 1: © Boardworks Ltd 2010 1 of 8 The Factor Theorem. © Boardworks Ltd 2010 2 of 8 The Factor Theorem If ( x – a ) is a factor of a polynomial then substituting.

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The Factor TheoremThe Factor Theorem

Page 2: © Boardworks Ltd 2010 1 of 8 The Factor Theorem. © Boardworks Ltd 2010 2 of 8 The Factor Theorem If ( x – a ) is a factor of a polynomial then substituting.

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The Factor Theorem

If (x – a) is a factor of a polynomial then substituting a in x will give an answer of zero.If (x – a) is a factor of a polynomial then substituting a in x will give an answer of zero.

The converse is also true:

If substituting a in for x gives an answer of zero then (x – a) is a factor of a polynomial.If substituting a in for x gives an answer of zero then (x – a) is a factor of a polynomial.

Suppose that a polynomial is divided by an expression of the form (x – a) the remainder is 0.

What can you conclude about (x – a)?

If the remainder is 0 then (x – a) is a factor of the polynomial.

This is the Factor Theorem:

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The Factor Theorem

(x - 3) is a factor of 2x2 - 7x + 3 if we substitute 3 in for x and get zero.

2x2 - 7x + 3 = 2(3)2 - 7(3) + 3= 18 – 21 + 3

= 0 as required.

Use the Factor Theorem to show that (x - 3) is a factor of 2x2 - 7x + 3.

And so 2x2 - 7x + 3 = (x - 3)(2x – 1)

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The Factor Theorem

(x + 2) is a factor of 3x2 + 5x – 2 if we substitute –2 in for x and get zero.

3x2 + 5x – 2 = 3(–2)2 + 5(–2) – 2= 12 – 10 – 2

= 0 as required.

Use the Factor Theorem to show that (x + 2) is a factor of 3x2 + 5x – 2.

And so 3x2 + 5x – 2 = (x + 2)(3x – 1)

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Factoring polynomials

The Factor Theorem can be used to factor polynomials by systematically looking for values of x that will make the polynomial equal to 0. For example:

Factor the cubic polynomial x3 – 3x2 – 6x + 8.

Let x3 – 3x2 – 6x + 8.

Start by testing 1, then -1, then 2, etc.

x3 – 3x2 – 6x + 8 = 1 – 3 – 6 + 8 = 0

(x – 1) is a factor of x3 – 3x2 – 6x + 8.

Now long divide

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Long Divide

8631 23 xxxx

2x

23 1xx 23 1xx 22x x6

x2

xx 22 2 xx 22 2 x8 8

8

88 x 88 x

0

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Factor Fully

)2)(4)(1(

)82)(1(

8632

23

xxx

xxx

xxx

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Your turn

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