1 1 review on factoring

Review on Factoring
Frank Ma © 2011

Review on Factoring
To factormeans to rewrite an expression as a product in a nontrivial way.

Review on Factoring
To factormeans to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression.

Review on Factoring
I. Always pull out the greatest common factor first.

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II. If the expression is a trinomial (three-term) ax2 + bx + c,
use the reverse-FOIL methodor the ac-method.

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III. Use the following factoring formulas if possible
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+

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x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
A common factor of two or more quantities is a factor that
belongs to all the quantities.

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
Example A.
a. Since 6 = (2)(3) and 15 = (3)(5),

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
Example A.
a. Since 6 = (2)(3) and 15 = (3)(5), so 3 is a common factor.

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
Example A.
b. The common factor of 4ab, 6a are 2, a, 2a.

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
Example A.
c. The common factors of 6xy2, 15x2y2 are 3, x, y2, xy2, ..

Review on Factoring
x2 – y2 = (x + y)(x–y).
x3 y3 = ( x y )( x2xy + y2 ).
–
±
±
+
Pulling out GCF
Example A.
c. The common factors of 6xy2, 15x2y2 are 3, x, y2, xy2, ..
d. The common factor of a(x+y), b(x+y) is (x+y).

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The greatestcommon factor(GCF)is the common factor that has the largest coefficient and highest degree of each factor among all common factors.

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Example B. Find the GCF of the given quantities.
a. GCF{24, 36}

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a. GCF{24, 36} = 12.

Review on Factoring
a. GCF{24, 36} = 12.
b. GCF{4ab, 6a}

Review on Factoring
a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.

Review on Factoring
a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2}

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a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.

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a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} =

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a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} = x3y4.

Review on Factoring
a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} = x3y4.
The Extraction Law
Distributive law interpreted backward gives the Extraction Law, that is, common factors may be extracted from sums or differences.

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a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} = x3y4.
The Extraction Law
AB ± AC A(B±C)

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a. GCF{24, 36} = 12.
b. GCF{4ab, 6a} = 2a.
c. GCF {6xy2, 15 x2y2} = 3xy2.
d. GCF{x3y5, x4y6, x5y4} = x3y4.
The Extraction Law
AB ± AC A(B±C)
This procedure is also called “factoring out common factor”. To factor, the first step always is to factor out the GCF.

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Example C. Factor out the GCF.
a. 12x2y3 + 6x2y2

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a. 12x2y3 + 6x2y2
(the GCF is 6x2y2)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1)
(the GCF is 6x2y2)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
Pull out the common factor (2x – 3),
(2x – 3)3x – 2(2x – 3)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
c. 3x – 3y + ax – ay

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
c. 3x – 3y + ax – ay Group them into two groups.
= (3x – 3y) + (ax – ay)

Review on Factoring
a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3x – 3y) + (ax – ay) Factor out the GCF of each group.
= 3(x – y) + a(x – y)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= 3(x – y) + a(x – y) Pull the factor (x – y) again.
= (3 + a)(x – y)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
Reverse-FOIL (Guessing Method)

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
To factor a trinomial (three-term) of the form ax2 + bx + c,
try an answer of the form (#x ± #) (#x ± #).

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a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have(#)(#) = a,

Review on Factoring
a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have(#)(#) = a, (#)(#) = c

Review on Factoring
a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have(#)(#) = a, (#)(#) = c and
that the inner and outer products combined to be bx.

Review on Factoring
a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1)
(the GCF is 6x2y2)
b. (2x – 3)3x – 2(2x – 3)
(2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2)
= (3 + a)(x – y)
try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have(#)(#) = a, (#)(#) = c and
that the inner and outer products combined to be bx.
That is (#x ± #) (#x ± #).
combined to be bx

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Example D. Factor x2 – 4x – 12
We are to fill in ( x ± ) ( x ± ).

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It must be (1x ± ) (1x ± ) to get the 1x2.

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It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore,
it must be (x + ) (x – ) to get the –12. Let's factor

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12 = (1)(12) = (2)(6) = (3)(4)

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12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x,

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we see that x2– 4x – 12 = (x + 2)(x – 6).

Review on Factoring
we see that x2– 4x – 12 = (x + 2)(x – 6).
If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c,
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)

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we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
Example E. Factor 20x2 – 4x – 3

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we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3.

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we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
20 = (1)(20) = (2)(10) = (4)(5),

Review on Factoring
we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
20 = (1)(20) = (2)(10) = (4)(5), observe that the difference
between (2)(3) and (10)(1) is ±4

Review on Factoring
we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
between (2)(3) and (10)(1) is ±4 which matches the coefficient of the middle term –4x.

Review on Factoring
we see that x2– 4x – 12 = (x + 2)(x – 6).
i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
between (2)(3) and (10)(1) is ±4 which matches the coefficient of the middle term –4x. Use 2xand 10xfor the
factors, we get 20x2 – 4x – 3 = (10x + 3)(2x – 1 ).

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The ac-Method

Review on Factoring
The ac-Method
The product of two binomials has four terms.

Review on Factoring
The ac-Method
The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials.

Review on Factoring
The ac-Method
Example F. Factor x2 – x – 6 by grouping.

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The ac-Method
x2– x – 6
= x2– 3x + 2x – 6

Review on Factoring
The ac-Method
x2 – x – 6
= x2 – 3x + 2x – 6 Put them into two groups
= (x2 – 3x) + (2x – 6)

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The ac-Method
x2 – x – 6
= (x2 – 3x) + (2x – 6) Pull out the common factor
= x(x – 3) + 2(x – 3)

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The ac-Method
x2 – x – 6
= x(x – 3) + 2(x – 3)Pull out the common (x – 3)
= (x – 3)(x + 2)

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The ac-Method
x2 – x – 6
= x(x – 3) + 2(x – 3) Pull out the common (x – 3)
= (x – 3)(x + 2)
But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place?

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The ac-Method
x2 – x – 6
= (x – 3)(x + 2)
But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method.

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The ac-Method
x2 – x – 6
= (x – 3)(x + 2)
But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we can use the grouping method.

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The ac-Method
x2 – x – 6
= (x – 3)(x + 2)
But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we can use the grouping method. If it isn't possible to rewrite the trinomial according to the ac-method, then the trinomial is prime.

Review on Factoring
ac-Method:

Review on Factoring
ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c.

Review on Factoring
ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c.
1. Calculate ac,

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ac-Method: We assume that there is no common factor for the trinomial ax2 + bx+ c.
1. Calculate ac, and find two numbers u and v such that
uv = ac,

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ac-Method: We assume that there is no common factor for the trinomial ax2+ bx + c.
uv = ac, and u + v = b.

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2. Write ax2 + bx + c as ax2 + ux + vx +c.

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2. Write ax2 + bx + c as ax2 + ux + vx +c
3. Use the grouping method to factor (ax2 + ux) + (vx + c)

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If step 1 can’t be done, then the expression is prime.

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Example G. Factor 3x2 – 4x – 20 using the ac-method.

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a = 3, c = –20.

Review on Factoring
a = 3, c = –20. Hence ac = 3(–20) = –60.

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Example G. Factor 3x2– 4x – 20 using the ac-method.
a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4.

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By searching, we get that 6(–10) = –60 and 6 + (–10) = –4.
or u, v are 6 and –10.

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or u, v are 6 and –10. Replace –4x by ux and vx

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or u, v are 6 and –10. Replace –4x by ux and vx we get
3x2– 4x – 20 = 3x2 + 6x –10x – 20

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3x2– 4x – 20 = 3x2 + 6x –10x – 20 Put into two groups
= (3x2 + 6x ) + (–10x – 20)

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= (3x2 + 6x ) + (–10x – 20) Pull out common factor
= 3x(x + 2) – 10 (x + 2)

Review on Factoring
= (3x2 + 6x ) + (–10x – 20) Pull out common factor
= 3x(x + 2) – 10 (x + 2) Pull out common factor
= (3x – 10)(x + 2)

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Factoring Formula
If it fits, use the Difference of Squares Formula
x2 – y2 = (x + y)(x–y).

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Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2

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Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2

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Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)

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Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)
c. (89)(91)
=

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)
c. (89)(91)
= (90 – 1)(90 + 1)

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)
c. (89)(91)
= (90 – 1)(90 + 1)
= 902 – 12

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)
c. (89)(91)
= (90 – 1)(90 + 1)
= 902 – 12
= 8,100 – 1 = 7,099

Review on Factoring
Factoring Formula
x2 – y2 = (x + y)(x–y).
Example H.
a. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
b. 32x3– 2x
= 2x(16x2 – 1)
= 2x(4x + 1)(4x – 1)
c. (89)(91)
= (90 – 1)(90 + 1)
= 902 – 12
= 8,100 – 1 = 7,099
The terms (x + y) and (x – y) are said to be conjugates of each other.

Review on Factoring
The Difference of Powers Formula (Optional)

Review on Factoring
The Difference of Squares Formula is just a special case of a general Difference of Powers Formula.

Review on Factoring
The Difference of Squares Formula is just a special case of a general Difference of Powers Formula.
xn – yn= ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1)

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The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines.

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In particular that, if y = 1
xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1).

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1
29 + 28 + 27 + …+ 21 + 1
=

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1
29 + 28 + 27 + …+ 21 + 1
=
By inspection, we get x = 2, n = 10.

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1
29 + 28 + 27 + …+ 21 + 1
= (2 – 1)(29 + 28 + 27 + …+ 21 + 1)
this happens to be “1”

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1
29 + 28 + 27 + …+ 21 + 1
= (2 – 1)(29 + 28 + 27 + …+ 21 + 1)
= 210 – 1 = 1023
Apply the formula

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xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1)
Example I. Sum 29 + 28 + 27 + …+ 21 + 1
29 + 28 + 27 + …+ 21 + 1
= (2 – 1)(29 + 28 + 27 + …+ 21 + 1)
= 210 – 1 = 1023
Apply the formula
The cubic version is listed below for the sake of completeness
–
±
±
x3y3 = ( x y )( x2xy + y2 ).
+

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Ex. A. Factor the following trinomials. use any method.
If it’s prime, state so.
1. 3x2 – x – 2
2. 3x2 + x – 2
3. 3x2 – 2x – 1
4. 3x2 + 2x – 1
5. 2x2 – 3x + 1
6. 2x2 + 3x – 1
8. 2x2 – 3x – 2
7. 2x2 + 3x – 2
9. 5x2 – 3x – 2
12. 3x2 – 5x + 2
11. 3x2 + 5x + 2
10. 5x2 + 9x – 2
15. 6x2 + 5x – 6
14. 6x2 – 5x – 6
13. 3x2 – 5x + 2
16. 6x2 – x – 2
17. 6x2 – 13x + 2
18. 6x2 – 13x + 2
19. 6x2 + 7x + 2
20. 6x2 – 7x + 2
21. 6x2 – 13x + 6
23. 6x2 – 5x – 4
24. 6x2 – 13x + 8
22. 6x2 + 13x + 6
26. 4x2 – 49
27. 25x2 – 4
25. 6x2 – 13x – 8
29. 25x2 + 9
28. 4x2 + 9
Ex. B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first.
32. –6x3 – x2 + 2x
30. – 6x2 – 5xy + 6y2
31. – 3x2 + 2x3– 2x
33. –15x2 – 25x2 – 10x
34. 12x2y2 –14x2y2 + 4xy2

1 1 review on factoring

by math123b on Aug 28, 2011

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1 1 review on factoring — Presentation Transcript