Factoring

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Factoring

• Chapter 6

2
Chapter Sections

• 6.1 Factoring a Monomial from a Polynomial
• 6.2 Factoring by Grouping
• 6.3 Factoring Trinomials of the Form
• ax2 bx c, a 1
• 6.4 Factoring Trinomials of the Form
• ax2 bx c, a ? 1
• 6.5 Special Factoring Formulas and a General
  Review of Factoring
• 6.6 Solving Quadratic Equations Using Factoring
• 6.7 Applications of Quadratic Equations

3
6.1

• Factoring a Monomial from a Polynomial

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Factors

• To factor an expression means to write the
  expression as a product of its factors.

Recall that the greatest common factor (GCF) of
two or more numbers is the greatest number that
will divide (without remainder) into all the
numbers. Example The GCF of 27 and 45 is 9.
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Factors

• A prime number is an integer greater than 1 that
  has exactly two factors, 1 and itself.

A composite number is a positive integer that is
not prime.
Prime factorization is used to write a number as
a product of its primes. 24 2 2 2 3
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Determining the GCF

• Write each number as a product of prime factors.
• Determine the prime factors common to all the
  numbers.
• Multiply the common factors found in step 2. The
  product of these factors is the GCF.

Example Determine the GCF of 24 and 30. 24 2
2 2 3 30 2 3 5 A factor of 2
and a factor of 3 are common to both, therefore 2
3 6 is the GCF.
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Determining the GCF

• To determine the GCF of two or more terms, take
  each factor the largest number of times it
  appears in all of the terms.

b.) The GCF of 4x2y2, 3xy4, and 2xy2 is xy2.
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Factoring Monomials from Polynomials

• Determine the GCF of all the terms in the
  polynomial.
• Write each term as the product of the GCF and its
  other factors.
• Use the distributive property to factor out the
  GCF.

Example 24x 16x3 (GCF is 8x) 83x
82xx2 8x (3 2x2)
(To check, multiply the factors using the
distributive property. )
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6.2

• Factoring by Grouping

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Factoring by Grouping

• The process of factoring a polynomial containing
  four or more terms by removing common factors
  from groups of terms is called factoring by
  grouping.

Example Factor x2 7x 3x 21. x(x 7)
3(x 7) (x 7) (x 3)
(Use the FOIL method to check your answer.)
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Factoring by Grouping

• Determine whether there are any factors common to
  all four terms. If so, factor the GCF from each
  of the four terms.
• If necessary, arrange the four terms so that the
  first two terms have a common factor and the last
  tow terms have a common factor.
• Use the distributive property to factor each
  group of two terms.
• Factor the GCF from the results of step 3.

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Factoring by Grouping

• Example
• a.) Factor by grouping 4x2 6x 6x 9
• 2x(2x 3) 3(2x 3)
• (2x 3) (2x 3)
• b.) Factor by grouping 5x2 20x x 4
• 5x(x 4) ( 1) (x 4)
• (5x 1) (x 4)

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6.3

• Factoring Trinomials of the Form
• ax2 bx c, a 1

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Factoring Trinomials

• Recall that factoring is the reverse process of
  multiplication. Using the FOIL method, we can
  show that
• (x 3)(x 8) x2 11x 24.

Therefore x2 11x 24 (x 3)(x 8).
Note that this trinomial results in the product
of two binomials whose first term is x and second
term is a number (including its sign).
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Factoring Trinomials

• Factoring any polynomial of the form x2 bx c
• will result in a pair of binomials

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Factoring Trinomials

• Find two numbers whose product equals the
  constant, c, and whose sum equals the coefficient
  of the x-term, b.
• Use the two numbers found in step 1, including
  their signs, to write the trinomial in factored
  form. The trinomial in factored form will be
• (x one number) (x second number)

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Examples

• a.) Factor x2 8x 15.
• x2 8x 15 (x ?) (x ?)

Replace the ?s with two numbers that are the
product of 15 and the sum of 8.
x2 8x 15 (x 3) (x 5)
b.) Factor x2 10x 25. This is a prime
polynomial because it cannot be factored.
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Examples Continued

• c.) Factor 2a2 12a 32.

First, factor out a 2 from the polynomial.
2a2 12a 32 2(a2 6a 16)
Then factor the trinomial.
2(a2 6a 16) 2(a 8)(a 2)
d.) Factor 2b5 16b4 30b3. 2b5 16b4
30b3 2b3(b2 8b 15) 2b3(b 3)(b 5)
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6.4

• Factoring Trinomials of the Form
• ax2 bx c, a ? 1

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Trial and Error Method

• Determine whether there is any factor common to
  all three terms. If so, factor it out.
• Write all pairs of factors of the coefficient of
  the squared term, a.
• Write all pairs of factors of the constant term,
  c.
• Try various combinations of these factors until
  the correct middle term, bx, is found.

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Trial and Error Method

• Example Factor 2x2 9x 4.

There is no factor common to all three terms.
Since the first term is 2x2, one factor must
contain 2x and the other an x. (2x ?)(x ?)
The product of the last term in the factors must
be 4. Only the positive factors of 12 will be
considered.
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Trial and Error Method

• Factor 2x2 9x 4.

Since the product of (2x 1)(x 4) yields the
correct term, 9x, they are the correct
factors. 2x2 9x 4 (2x 1)(x 4)
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Factor by Grouping Method

• Determine whether there is any factor common to
  all three terms. If so, factor it out.
• Find two numbers whose product is equal to the
  product of a times c, and whose sum is equal to
  b.
• Rewrite the middle term, bx, as the sum or
  difference of two terms using the numbers found
  in step 2.
• Factor by grouping.

24
Factor by Grouping Method

• Example Factor 2x2 9x 4.

There is no factor common to all three terms.
a 2 b 9 c 4 Find two numbers whose
product is a c and whose sum is b.
Factors of 8 (1)(8) (2)(4)
Sum of Factors 9 6
?
Continued.
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Factor by Grouping Method
Example continued

• Factor 2x2 9x 4.

Use these factors to rewrite 9x.
Factor by grouping.
2x2 1x 8x 4 x (2x 1) 4 (2x 1) (2x
1) (x 4)
FOIL to check.
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Factor by Grouping Method

• Example Factor 18x3 21x2 9x.

Factor out 3x. 18x3 21x2 9x 3x (6x2 7x
3)
Rewrite the middle term. 3x (6x2 7x 3) 3x
(6x2 9x 2x 3)
Factor by grouping. 3x (6x2 9x 2x 3)
3x3x(2x 3) 1 (2x 3) 3x (3x
1) (2x 3)
FOIL to check.
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6.5

• Special Factoring Formulas and a General Review
  of Factoring

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Difference of Two Squares

• a2 b2 (a b) (a b)

Example a.) Factor x2 49. x2 49 x2 72
(x 7)(x 7) b.) Factor 4x4 25y8. 4x4
25y8 (2x2)2 (5y4)2 (2x2 5y4)(2x2 5y4)
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Sum of Two Cubes

• a3 b3 (a b) (a2 ab b2)

Example a.) Factor k3 8. k3 8 k3 23
(k 2)(k2 2k 4) b.) Factor 27c3 125d3.
(a 3c b 5d) 27c3 125d3 (3c
5d)(9c2 15cd 25d2)
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Difference of Two Cubes

• a3 b3 (a b) (a2 ab b2)

Example a.) Factor p3 1. p3 1 (p 1)(p2
p 1) b.) Factor 64x3 125y3. (a 4x
b 5y) 64x3 125y3 (4x 5y)(16x2 20xy
25y2)
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Helpful Hint for Factoring

• When factoring the sum or difference of two
  cubes, the sign between the terms in the binomial
  factor will be the same as the sign between the
  terms.

The sign of the ab term will be the opposite of
the sign between the terms of the binomial
factor.
The last term in the trinomial will always be
positive.
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6.6

• Solving Quadratic Equations Using Factoring

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Quadratic Equation

• A quadratic equation is an equation that
  contains a second-degree term and no term of a
  higher degree.

Quadratic equations have the form ax2 bx c
0 where a, b, and c are real numbers, a ? 0
Examples a.) z2 3z 7 0 b.) 4k2 5
0
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Zero-Factor Property

• To solve a quadratic equation by factoring, the
  zero-factor property is used.

If ab 0, then a 0 or b 0
Example Solve the equation 3x(x 4) 0 3x
0 or x 4 0 x 0 x -4
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Solving with Factoring

• Write the equation in standard form with the
  standard term having a positive coefficient.
  This will result in one side of the equation
  being 0.
• Factor the side of the equation that is not 0.
• Set each factor containing a variable equal to 0
  and solve each equation.
• Check each solution found in step 3 in the
  original equation.

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Solving with Factoring

• Example Solve the following equations.
• a.) 9x 20 -x2
• x2 9x 20 0
• (x 4) (x 5) 0
• x 4 0 or x 5 0
• x 4 or x 5

?
?
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Solving with Factoring

• b.) 2x2 50
• 2x2 50 0
• 2(x2 25) 0
• 2(x 5) (x 5) 0
• x 5 0 or x 5 0

x 5 or x 5
?
?
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6.7

• Applications of Quadratic Equations

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Solving Applications
Example

• The area of a rectangle is 84 square inches.
  Determine the length and width if the length is 2
  inches less than twice the width.

A 84 lw 84
Solve.
Continued.
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Solving Applications
Example continued
(2w 2)w 84 2w2 2w 84 2w2 2w 84
0 2(w2 w 42) 0 2(w 7) (w 6) 0 w 7
0 or w 6 0 w 7 or w -6
The width of the rectangle is 7 inches and the
length is 2(7) 2 14 inches.
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Pythagorean Theorem

• The square of the hypotenuse of a right triangle
  is equal to the sum of the squares of the two
  legs.

(leg)2 (leg)2 (hypotenuse)2
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Pythagorean Theorem
Example

• One leg of a right triangle is two inches more
  than twice the other leg. The hypotenuse is 13
  inches. Find the length of the three sides of
  the triangle.

Let a length of the first leg Let b length of
the second leg
b 2 2a
a2 b2 c2 a2 (2 2a)2 132
Solve.
Continued.
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Pythagorean Theorem
Example continued

• a2 b2 c2
• a2 (2 2a)2 132
• a2 4 8a 4a2 169
• 5a2 8a 4 169
• 5a2 8a 165 0
• 5a2 25a 33a 165 0
• 5a(a 5) 33(a 5) 0
• (5a 33)(a 5) 0
• 5a 33 0 or a 5 0
• a -33/5 or a 5

The lengths of the triangle are 5 in., 12 in.,
and 13 in.