Download 1 Section R.4 – Review of Factoring Objective #1: Factoring Out the

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Section R.4 – Review of Factoring
Objective #1:
Factoring Out the Greatest Common Factor
The Greatest Common Factor (GCF) is the largest factor that can divide
into the terms of an expression evenly with no remainder. To find the GCF
of variables, list the variables that appear in all the terms and then choose
the lowest power of each respective variable as the power for the GCF.
Factor the following completely:
Ex. 1a 3p3 – 15p2
Ex. 1b 42x3y2z + 14xy2z2 – 35x4y4
Ex. 1c – 12p3t + 2p2t3 + 6pt2 – 2pt Ex. 1d 9x(2x + 3) – 6(2x + 3)
Solution:
a) Since the GCF = 3p2, then 3p3 – 15p2 = 3p2(p) – 3p2(5)
= 3p2(p – 5)
b) Since the GCF = 7xy2, then 42x3y2z + 14xy2z2 – 35x4y4
= 7xy2(6x2z) + 7xy2(2z2) – 7xy2(5x3y2) = 7xy2(6x2z + 2z2 – 5x3y2)
c) Since the GCF = – 2pt, then – 12p3t + 2p2t3 + 6pt2 – 2pt
= – 2pt(6p2) – 2pt(– pt2) – 2pt(3t) – 2pt(1)
= – 2pt(6p2 – pt2 – 3t + 1)
d) Since the GCF = 3(2x + 3), then 9x(2x + 3) – 6(2x + 3)
= 3(2x + 3)•3x – 3(2x + 3)•2
= 3(2x + 3)(3x – 2)
Objective #2:
Factoring by Grouping
Whenever a factoring problem has four or more terms, we put the terms
into groups and factor out the G.C.F. of each group. We then see if the
”groupings” have any pieces in common. If so, we then factor out the
common pieces.
Factor the following completely:
Ex. 2a my + 6y – 7m – 42
Solution:
my + 6y – 7m – 42
= my + 6y + – 7m – 42
= (my + 6y ) + (– 7m – 42)
= y(m + 6) – 7(m + 6)
= (m + 6)(y – 7)
Ex. 2b w2 + 5w + 20z + 4wz
Solution:
w2 + 5w + 20z + 4wz
= (w2 + 5w) + (20z + 4wz)
= w(w + 5) + 4z(5 + w)
= w(w + 5) + 4z(w + 5)
= (w + 5)(w + 4z)
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Ex. 2c 10uv3 – 5v3 – 30uv2 + 15v2
Solution:
Factor out GCF first.
10uv3 – 5v3 – 30uv2 + 15v2
= 5v2[2uv – v – 6u + 3]
= 5v2[(2uv – v) + (– 6u + 3)]
= 5v2[v(2u – 1) – 3(2u – 1)]
= 5v2(2u – 1)(v – 3)
Objective #3:
Factoring Trinomials in the form x2 + bx + c
To factor trinomials where the coefficient of the square term is 1, we will
look at the factors of c to see which pair of factors add up to b. We will then
use that result to rewrite the problem as a product of two binomials.
Factoring x2 + bx + c
1)
List all pairs of factors of c.
2)
Determine which pair of factors e1 and e2 found in part 1 you can add
to get b. If you get the opposite of b for the sum, change the signs of
e1 and e2. If a pair factors cannot be found, then the trinomial is said
to be "prime."
3)
Rewrite the trinomial as (x + e1)(x + e2). If the original trinomial was
in the form x2 + bxy + cy2, then the result will be (x + e1y)(x + e2y).
4)
Verify that the product gives you the original trinomial.
Factor the following completely:
Ex. 3a
25x3 – 275x2y + 600xy2
Solution:
The G.C.F. = 25x, so 25x3 – 275x2 + 600x = 25x(x2 – 11xy + 24y2).
We want to find a pair of factors of 24 that add up to – 11.
Pairs of factors of 24
b = Sum of factors of c = – 11
1•24
1 + 24 = 25 No
2•12
2 + 12 = 14 No
3•8
3 + 8 = 11 Almost, change the signs
4•6
4 + 6 = 10 No
So, – 3 and – 8 are what we want to use. Since 24 had a y2 next to it, our
answer will be 25x(x – 3y)(x – 8y).
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Ex. 3b
– 4a3b2 + 40a2b3 + 224ab4
Solution:
G.C.F. = – 4ab2, so – 4a3b2 + 40a2b3 + 224ab4 = – 4ab2(a2 – 10ab – 56b2)
Pairs of factors of – 56
b = Sum of factors of c = – 10
– 1•56
– 1 + 56 = 55 No
– 2•28
– 2 + 28 = 26 No
– 4•14
– 4 + 14 = 10 Almost, change the signs
– 7•8
– 7 + 8 = 1 No
So, 4 and – 14 are what we want to use. Since – 56 had a b2 next to it, our
answer will be – 4ab2(a – 14b)(a + 4b).
Objective #4:
Factoring Trinomials Using AC Method.
To factor trinomials, we will use a method called the AC method. With this
method, you “uncombine” the middle term and then factor the problem by
grouping. Here is the procedure:
AC Method
1)
Multiply the coefficient of the first and last terms (a•c).
2)
List all pairs of factors of result from step #1 until you find a pair
whose sum is the coefficient of the middle term.
3)
Rewrite the middle term as the sum of two terms using the factors
found in step #2 as the coefficients.
4)
Factor by grouping.
Factor the following completely:
Ex. 4a
3x2 – 5x – 12
Solution:
The coefficients of the first and last terms are 3 and – 12. Their
product is – 36.
Rewrite – 5x as 4x – 9x
factors of – 36 sum = – 5
3x2 – 5x – 12
– 1•36
– 1 + 36 = 35 No
= 3x2 + 4x – 9x – 12
– 2•18
– 2 + 18 = 16 No
= (3x2 + 4x) + (– 9x – 12)
– 3•12
– 3 + 12 = 9 No
= x(3x + 4) – 3(3x + 4)
– 4•9
–4+9=5
Yes, change the signs = (3x + 4)(x – 3)
– 6•6
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Ex. 4b
– 135x2 + 9xy + 54y2
Solution:
The G.C.F. = – 9, so – 135x2 + 9xy + 54y2 = – 9(15x2 – xy – 6y2).
The coefficients of the first and last terms are 15 and – 6. Their
product is – 90.
factors of – 90
sum = – 1
Rewrite – xy = 9xy – 10xy,
– 1•90
– 1 + 90 = 89 No
– 9(15x2 – xy – 6y2)
– 2•45
– 2 + 45 = 43 No
= – 9(15x2 + 9xy – 10xy – 6y2)
– 3•30
– 3 + 30 = 27 No
= – 9([15x2 + 9xy] + [– 10xy – 6y2])
– 5•18
– 5 + 18 = 13 No
= – 9(3x[5x + 3y] – 2y[5x + 3y])
– 6•15
– 6 + 15 = 9 No
= – 9(5x + 3y)(3x – 2y)
– 9•10
– 9 + 10 = 1 Almost,
change the signs
Ex. 4c
– 4a3b – 25a2b2 – 30ab3
Solution:
G.C.F. = – ab, so – 4a3b – 25a2b2 – 30ab3 = – ab(4a2 + 25ab + 30b2)
The coefficients of the first and last terms are 4 and 30. Their product
is 120.
factors of 120
Sum = 25
1•120
1 + 120 = 121 No
2•60
2 + 60 = 62 No
3•40
3 + 40 = 43 No
4•30
4 + 30 = 34 No
5•24
5 + 24 = 29 No
•
6 20
6 + 20 = 26 No
8•15
8 + 15 = 23 No
10•12
10 + 12 = 22 No
Since there are no pairs that yield a sum of 25, the trinomial is prime.
Thus, our answer is – ab(4a2 + 25ab + 30b2).
Objective #5:
Factoring the Difference of Squares
To factor a binomial that is a difference of squares, we can use the
following special product:
F2 – L2 = (F – L)(F + L)
In order for a binomial to fit the pattern, the first term has to be a perfect
square, the last term has to be a perfect square, and the operation has to
be subtraction.
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Factor the following completely:
Ex. 5a 9x2 – 16y2
Solution:
9x2 – 16y2
= (3x)2 – (4y)2
F = 3x, L = 4y and operation is –.
9x2 – 16y2 = (3x – 4y)(3x + 4y)
Ex. 5c 81x2 + 25z2
Solution:
The operation is + so this cannot
be factored. 81x2 + 25z2 is prime*.
* - Note that 81x2 + 25z2 does
factor in the complex numbers. If
we let i = −1 , so i 2 = 1, then
81x2 + 25z2 = (9x – 5i z)(9x + 5i z).
Objective #6:
€
Ex. 5b 49y3 – 36y
Solution:
49y3 – 36y
= y(49y2 – 36) = y([7y]2 – [6]2)
F = 7y, L = 6 and operation is –.
y(49y2 – 36) = y(7y – 6)(7y + 6)
Ex. 5d 16x4 – 81
Solution:
16x4 – 81 = (4x2)2 – (9)2
F = 4x2, L = 9 and operation is –.
16x4 – 81 = (4x2 – 9)(4x2 + 9)
= ([2x]2 – [3]2)(4x2 + 9)
F = 2x, L = 3 and operation is –.
= (2x – 3)(2x + 3)(4x2 + 9)
Factoring a Perfect Square Trinomial
To factor a trinomial that is a perfect, we can use the following special
products:
1)
F2 + 2FL + L2 = (F + L)2
2)
F2 – 2FL + L2 = (F – L)2
For a trinomial to fit the pattern, the first term has to be a perfect square,
the last term has to be a perfect square, and the middle term has to match.
Factor the following completely:
Ex. 6a 49x2 + 56xy + 16y2
Solution:
F = 7x, L = 4y and 2FL = 56xy.
= (7x)2 + 2(7x)(4y) + (4y)2
49x2 + 56xy + 16y2 = (7x + 4y)2
Ex. 6c 100x2 + 360xz + 324z2
Solution:
= 4(25x2 + 90xz + 81z2)
= 4([5x]2 + 2[5x][9z] + [9z]2)
F = 5x, L = 9z and 2FL = 90xz
4(25x2 + 90xz + 81z2) = 4(5x + 9z)2
Ex. 6b 20a3 – 60a2 + 45a
Solution:
20a3 – 60a2 + 9a = 5a(4a2 – 12a + 9)
F = 2a, L = 3 and – 2FL = – 12a.
= 5a([2a]2 – 2[2a][3] + [3]2)
5a(4a2 – 12a + 9) = 5a(2a – 3)2
Ex. 6d 4w2 – 25w + 36
Solution:
F = 2w, L = 6, but – 2FL = – 24w.
So, the pattern does not work. We
will have to factor using the AC
Method to get (4w – 9)(w – 4).
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Objective #7:
Factoring the Sum or Difference of Cubes
To factor a binomial that is a sum or difference of cubes, we can use the
following special products:
F3 + L3 = (F + L)(F2 – FL + L2)
* - if the degree of F2 – FL + L2 and
3
3
2
2
F – L = (F – L)(F + FL + L )
F2 + FL + L2 is 2, then they are
prime in the real numbers.
In order for a binomial to fit the pattern, the first term has to be a perfect
cube and the last term has to be a perfect cube.
Factor the following completely:
Ex. 7a 27x3 – 64y3
Solution:
27x3 – 64y3 = (3x)3 – (4y)3
F = 3x, L = 4y and operation is –.
2
= (3x – 4y)([3x] + [3x][4y] + [4y]2)= (3x – 4y)(9x2 + 12xy + 16y2)
Ex. 7b 125a3 + 8b3
Solution:
125a3 + 8b3 = (5a)3 + (2b)3 F = 5a, L = 2b and operation is +.
= (5a + 2b)([5a]2 – [5a][2b] + [2b]2) = (5a + 2b)(25a2 – 10ab + 4b2)
Objective #8:
Factoring Using Substitution
Sometimes is difficult to factor a problem directly. What we can do is to
notice a pattern, make a substitution into the problem, which gives us an
easier problem to factoring. After factoring the easier problem, we
substitute back in the original expression.
Factor the following completely:
Ex. 8 4x4 – 63x2 – 16
Solution:
Let u = x2. Our problem becomes:
4x4 – 63x2 – 16 = 4u2 – 63u – 16
The coefficients of the first and last terms are 4 and – 16.
Their product is – 64. Using the AC Method, we get:
Rewrite – 63u = u – 64u,
factors of – 64
sum = – 63
4u2 – 63u – 16
– 1•64
– 1 + 64 = 63 Almost,
= 4u2 + u – 64u – 16
change the signs
= u[4u + 1] – 16[4u – 1]
Now, replace u = x2 and simplify:
2
2
= (4u + 1)(u – 16)
(4u + 1)(u – 16) = (4x + 1)(x – 16)
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But, x – 16 = (x – 4)(x + 4), so,
(4x2 + 1)(x2 – 16) = (4x2 + 1)(x – 4)(x + 4)