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Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2
The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term.
Example: Factor 18x3 + 60x.
GCF = 6x18x3 + 60x = 6x (3x2) + 6x (10)
18x3 = 2 · 3 · 3 · x · x · x
Apply the distributive law
to factor the polynomial.
6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x
Check the answer by multiplication.
Factor each term.
Find the GCF.
60x = 2 · 2 · 3 · 5 · x
= 6x (3x2 + 10)
= (2 · 3 · x) · 3 · x · x
= (2 · 3 · x) · 2 · 5
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Example: Factor 4x2 – 12x + 20.
GCF = 4.
4(x2 – 3x + 5) = 4x2 – 12x + 20Check the answer.
A common binomial factor can be factored out of certain expressions.
Example: Factor the expression 5(x + 1) – y(x + 1).
5(x + 1) – y(x + 1) = (x + 1) (5 – y)
(x + 1) (5 – y) = 5(x + 1) – y(x + 1)Check.
= 4(x2 – 3x + 5)
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A difference of two squares can be factored using the formula
Example: Factor x2 – 9y2.
= (x)2 – (3y)2
= (x + 3y)(x – 3y)
Write terms as perfect squares.
Use the formula.
The same method can be used to factor any expression which can be written as a difference of squares.
Example: Factor (x + 1)2 – 25y 4.
= (x + 1)2 – (5y2)2
= [(x + 1) + (5y2)][(x + 1) – (5y2)]
= (x + 1 + 5y2)(x + 1 – 5y2)
a2 – b2 = (a + b)(a – b).
x2 – 9y2
(x + 1)2 – 25y 4
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2. Factor 2a2 + 3bc – 2ab – 3ac.
Some polynomials can be factored by grouping terms to produce a common binomial factor.
= 2a2 – 2ab + 3bc – 3ac
= y (2x + 3) – 2(2x + 3)
= (2a2 – 2ab) + (3bc – 3ac)
= 2a(a – b) + 3c(b – a)
= (2xy + 3y) – (4x + 6) Group terms.
Examples: 1. Factor 2xy + 3y – 4x – 6.
Factor each pair of terms.
= (2x + 3) ( y – 2) Factor out the common binomial.
Rearrange terms.
Group terms.
Factor.
= 2a(a – b) – 3c(a – b) b – a = – (a – b).
= (a – b) (2a – 3c) Factor.
Examples: Factor
2xy + 3y – 4x – 6
2a2 + 3bc – 2ab – 3ac
Notice the sign!
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Factoring these trinomials is based on reversing the distributive property process.
To factor a simple trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example,
x2 + 10x + 24 = (x + 4)(x + 6).
Example: Factor x2 + 3x + 2. = (x + a)(x + b)
Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b.
= x2 Multiply the binomials.
= x2 + (b + a) x + ba Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2.
= x2 + (1 + 2) x + 1 · 2
Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
+ bx + ax + ba
x2 + 3x + 2
(Product-sum method)
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Example: Factor x2 – 8x + 15.
= (x + a)(x + b)
(x – 3)(x – 5) = x2 – 5x – 3x + 15
x2 – 8x + 15 = (x – 3)(x – 5).
Therefore a + b = -8
Check:
= x2 + (a + b)x + ab
It follows that both a and b are negative.
= x2 – 8x + 15.
SumNegative Factors of 15
-3, - 5 - 8
- 1, - 15 -15
and ab = 15.
x2 – 8x + 15
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Example: Factor x2 + 13x + 36.
= (x + a)(x + b)
Check: (x + 4)(x + 9)
Therefore a and b are:
x2 + 13x + 36
= x2 + 9x + 4x + 36 = x2 + 13x + 36.
= (x + 4)(x + 9)
= x2 + (a + b) x + ab
SumPositive Factors of 36
1, 36 37
153, 12
4, 9 13
6, 6 12
2, 18 20
x2 + 13x + 36
two positive factors of 36
whose sum is 13.
36 13
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Example: Factor 4x3 – 40x2 + 100x.
A polynomial is factored completely when it is written as a product of factors that can not be factored further.
The GCF is 4x.
= 4x(x2 – 10x + 25) Use distributive property to factor out the GCF.
= 4x(x – 5)(x – 5) Factor the trinomial.
4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25)
= 4x(x2 – 10x + 25)
= 4x3 – 40x2 + 100x
4x3 – 40x2 + 100x
Check:
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Factoring complex trinomials of the form ax2 + bx + c, (a 1) can be done by decomposition or cross-check method.
Example: Factor 3x2 + 8x + 4.
1. Find the product of
first and last terms
3 4 =
12Decomposition Method
2. We need to find factors of 12whose sum is 8
1, 122, 63, 4
3. Rewrite the middle
term decomposed
into the two numbers
3x2 + 2x + 6x + 4
= x(3x + 2) + 2(3x + 2)
= (3x2 + 2x) + (6x + 4)4. Factor by grouping
in pairs = (3x + 2) (x + 2)
3x2 + 8x + 4 = (3x + 2) (x + 2)
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Example: Factor 4x2 + 8x – 5.
4x2 + 8x – 5 = (2x –1)(2x – 5)
4 5 =
20 We need to find factors of 20
whose difference is 81, 202, 104, 5
Rewrite the middle term
decomposed into the
two numbers
4x2 – 2x + 10x – 5
= 2x(2x – 1) + 5(2x – 1)
= (4x2 – 2x) + (10x – 5)
= (2x – 1) (2x + 5)
Factor by grouping
in pairs
Solve by Factoring
• To solve a quadratic by factoring you first get the equation equal to zero and then factor.
• Once factored you set each expression equal to zero and solve for the variable.
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Creating a Perfect Square Trinomial
In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____
Find the constant term by squaring half the coefficient of the linear term.
(14/2)2
X2 + 14x + 49
Perfect Square Trinomials
Create perfect square trinomials.
x2 + 20x + ___ x2 - 4x + ___ x2 + 5x + ___
100
4
25/4
Solving Quadratic Equations by Completing
the SquareSolve the following
equation by completing the square:
Step 1: Move quadratic term, and linear term to left side of the equation
2 8 20 0x x
2 8 20x x
Solving Quadratic Equations by Completing
the SquareStep 2: Find the term
that completes the square on the left side of the equation. Add that term to both sides.
2 8 =20 + x x 21
( ) 4 then square it, 4 162
8
2 8 2016 16x x
Solving Quadratic Equations by Completing the Square
Step 3: Factor
the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.
2 8 2016 16x x
2
( 4)( 4) 36
( 4) 36
x x
x
Solving Quadratic Equations by Completing the Square
Step 4: Take the square root of each side
2( 4) 36x
( 4) 6x
Solving Quadratic Equations by Completing the Square
Step 5: Set up the two possibilities and solve
4 6
4 6 an
d 4 6
10 and 2 x=
x
x x
x
Completing the Square-Example #2
Solve the following equation by completing the square:
Step 1: Move quadratic term, and linear term to left side of the equation, the constant to the right side of the equation.
22 7 12 0x x
22 7 12x x
Solving Quadratic Equations by Completing the Square
2
2
2
2
2
1. 2 63 0
2. 8 84 0
3. 5 24 0
4. 7 13 0
5. 3 5 6 0
x x
x x
x x
x x
x x
Try the following examples. Do your work on your paper and then check your answers.
1. 9,7
2.(6, 14)
3. 3,8
7 34.
2
5 475.
6
i
i
What Does The Formula Do ?
The Quadratic formula allows you to find the roots of a quadratic
equation (if they exist) even if the quadratic equation does not
factorise.
The formula states that for a quadratic equation of the form :
ax2 + bx + c = 0
The roots of the quadratic equation are given by :
a
acbbx
2
42
Example 1
Use the quadratic formula to solve the equation :
x 2 + 5x + 6= 0
Solution:
x 2 + 5x + 6= 0
a = 1 b = 5 c = 6
a
acbbx
2
42
12
)614(55 2
x
2
)24(255 x
2
15 x
2
15
2
15
xorx
x = - 2 or x = - 3
These are the roots of the equation.
Example 2
Use the quadratic formula to solve the equation :
8x 2 + 2x - 3= 0
Solution:
8x 2 + 2x - 3= 0
a = 8 b = 2 c = -3
a
acbbx
2
42
82
)384(22 2
x
16
)96(42 x
16
1002 x
16
102
16
102
xorx
x = ½ or x = - ¾
These are the roots of the equation.
Example 3
Use the quadratic formula to solve the equation :
8x 2 - 22x + 15= 0
Solution:
8x 2 - 22x + 15= 0
a = 8 b = -22 c = 15
a
acbbx
2
42
82
)1584()22()22( 2
x
16
)480(484(22 x
16
422 x
16
222
16
222
xorx
x = 3/2 or x = 5/4
These are the roots of the equation.
Use the quadratic formula to solve the equation :
8x 2 - 22x + 15= 0
Solution:
8x 2 - 22x + 15= 0
a = 8 b = -22 c = 15
x = 3/2 or x = 5/4
These are the roots of the equation.