1 p.4 factoring objectives: greatest common factor factoring trinomial special factoring ...
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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!