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P.4 FACTORING

Objectives:

Greatest Common Factor

Factoring Trinomial

Special Factoring

Factoring by Grouping

General Factoring

التحلي)(ل

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Like prime factorization of a number

Only integer coefficients and constants allowed

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3 2a) 6 36t t

1. Factor Out the Greatest Common Factor( GCF) GCF = product of all prime factors raised to the smallest powers.

26 6t t factor out the GCF = 6t2

Def: Factoring means to write a polynomial as a product of

polynomials of lower degree.

Methods of Factoring:

Ex: Factor out the GCF.

write each term as product of its prime factors3 2 2 23 2 3 2t t

23 2GCF t

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2. Try to Factor A binomial by One of the Following Special Factoring Formulas:

a. Difference of Two Squares:

2

2 2

9

3

m

m

3 3m m

2 2x y x y x y

Ex.

No similar rule for a sum of squares

a b2 2Cannot be factored with real coefficients “prime”

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6 2x : 4E x y

3 3( 2 )( 2 )x y x y

2 23 2x y Perfect square – perfect square

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b. Sum/Difference of Two Cubes:

48 27x x

22 3 4 6 9x x x

3 3 2 2x y x y x xy y

Ex.

3 32 3x x

3 3 2 2x y x y x xy y

38 27x x GCF

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23. Try to Factor A trinomial , 0

where , ,and are integrs, as

ax bx c a

a b c

2ax bx c

Factorization Theorem:

The trinomial with integer coefficients a, b and c can be factored as the product of two binomial with integer coefficients if and only if

is a perfect square.2 4b ac

a. As a perfect –square trinomial

b. Using the trial method

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Ex: 25a2 – 90ac + 81c2

25a2 is a perfect square. 25a2 = 5a 5a

81c2 is a perfect square. 81c2 = (-9c) (-9c)

2(5a)(-9c) = -90ac

This is a perfect square trinomial.

25a2 – 90ac + 81c2 = (5a – 9c)2

Check 22 4 90 4 25 81 0( )b ac perfect square

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Ex: 2 8 12x x

Smile! The leading coefficient is 1! Easy! Set up for FOIL

( )( ) We know First term has to be x because only x*x =

x2

We know last term has to be factors of 12 … 12,1; or 4,3; or 6,2.

(x 12)(x 1) or (x 4)(x 3) or (x 6)(x 2)

Check 2 2 24 8 4 1 12 64 48 16 4b ac

Try Trail and Error Method

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X2 + 8x + 12

The “+” sign before the “12” lets us know we will be adding the two factors.

The sum of the 2 factors must = +8! Of the 3 pairs of factors only 6 and 2 have a sum of 8 The “+” sign before the 12 also lets us know both signs in the

solution will be the same.

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X2 + 8x + 12

Possibilities for solution: (x - 6)(x - 2) or… (x + 6)(x + 2)

(-6)(-2) = +12 and (6)(2) = +12 But -6 + -2 = -8; 6 + 2 = +8

Sooooo…. (x + 6)(x + 2) … check by FOIL

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X2 - 7x + 12

Here again our leading coefficient is 1… (x )(x )

The last terms must be factors of 12 … 6,2; or 12,1; or 3,4. The “+” before the 12 tells us we will be adding the 2 factors,

and that the signs will be the same! The sum of the factors must be -7!

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X2 - 7x + 12

Of the 3 pairs of factors of 12, only 4 & 3 sum to 7 Signs must be the same, so…

(4)(3) = +12; 4 + 3 = 7 (-4)(-3) = +12; -4 + -3 = -7 … these are the factors we are

looking for! (x - 4)(x - 3)

Check by FOIL

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X2 - 4x - 12

Leading coefficient is 1; we need two factors of 12 … whose difference is 4

The “-” sign in front of the 12 also tells us that the signs will be different in our solution.

Factors of 12 whose difference is 4 … 6 & 2 The “-” before the 4 lets us know the sign of the larger number

(6) must be negative … (x - 6)(x + 2)

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2x2 + 9x + 4

Perhaps by now we can recognize that both signs in the factors will be “+”

In this case we only have 3 possibilities (2x + 4)(x + 1) (2x + 1)(x + 4) (2x + 2)(x + 2)

Check by FOIL to see which is a solution.

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3. Factoring by Grouping2 2 2mx mx x

1 2 1mx x x

2 1mx x

Factor mx Factor 2

Group the first two terms and the last two terms

2 2 2mx mx x

watch

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Ex: Factor

2 24 4 1x x y

2 212 23 10ax axy ay 2 26 9 3x xy y x y

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You’reshining!