P.5Factoring Polynomials

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P.5 Factoring Polynomials
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Factoring
Factoring is the process of writing a polynomial
as the product of two or more polynomials. The
factors of 6x2 x 2 are 2x 1 and 3x 2. In
this section, we will be factoring over the
integers. Polynomials that cannot be factored
using integer coefficients are called irreducible
over the integers or prime.
The goal in factoring a polynomial is to use one
or more factoring techniques until each of the
polynomials factors is prime or irreducible. In
this situation, the polynomial is said to be
factored completely.
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Common Factors
In any factoring problem, the first step is to
look for the greatest common factor. The greatest
common factor is an expression of the highest
degree that divides each term of the polynomial.
The distributive property in the reverse
direction ab ac a(b c) can be used to
factor out the greatest common factor. The GCF
will be the largest integer (only negatively
signed if the leading coefficient is negatively
signed) that can divide without remainder into
each term, and the lowest power for any one
variable that appears in each term. (Examples to
follow.)
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Example

• Factor a. 18x3 27x2 b. x2(x 3) 5(x 3)

• Solution
• We begin by determining the greatest common
  factor. 9 is the greatest integer that divides 18
  and 27. Furthermore, x2 is the greatest
  expression (SMALLEST exponent for the variable x
  that appears in each term) that divides x3 and
  x2. Thus, the greatest common factor of the two
  terms in the polynomial is 9x2.
• 18x3 27x2
• 9x2(2x) 9x2(3) Express each term with the
  greatest common factor as a factor.
• 9x2(2 x 3) Factor out the greatest
  common factor.

b. In this situation, the greatest common factor
is the common polynomial factor (x 3). We
factor out this common factor as follows.
x2(x 3) 5(x 3) (x 3)(x2 5) Factor
out the greatest common factor polynomial (GCFP).
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Factoring using the sum/product method.

• EX Factor a. x2 6x 8 b. x 4 3x 2y 18 y 2

Solution a. The factors of the first term are x
and x (x )( x ) To find the second
term of each factor, we must find two numbers
whose product is 8 and whose sum is 6. (Always
start with finding the integers for the product,
then select the ones that add to the sum.)
SUM PRODUCT 6 8 9 81 6 42
From the table above, we see that 4 and 2 are the
required integers. Thus, x2 6x 8 (x 4)( x
2) or (x 2)( x 4).
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• Ex., cont. Factor b. x 4 3x 2y 18 y 2

Solution b. We begin with x 4 3x 2y 18 y 2
( )(
). To find the second term of each factor, we
must find two numbers whose sum is __________,
and whose product is __________.
SUM PRODUCT
From the table above, we see that _______ and
________ are the required integers. Thus, (ans)
x 4 3x 2y 18 y 2 or
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A Strategy for Factoring ax2 bx c
If no such combinations exist, the polynomial is
prime.
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Example

• Factor 8x2 10x 3.

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(using text method) Text Example cont.
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Factoring by the ac method

• For a polynomial in the form
• ax2 bx c
• 1. Multiply ac (product)
• 2. Find factors of the product whose sum is b
• 3. Arrange the factors of a and c so that they
  will yield those factors when the binomials are
  multiplied.
• 4. Check by multiplying the factors.
• Ex Factor 4x2 16 x 15 (on board) S
  P

Will do if asked, otherwise do your homework.
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Factoring Perfect Square Trinomials

• Let A and B be real numbers, variables, or
  algebraic expressions,
• 1. A2 2AB B2 (A B)2
• 2. A2 2AB B2 (A B)2
• If we recognize that we have a perfect square
  trinomial (look for perfect squares at the
  beginning and end), we can use this first before
  doing too much work.

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Text Example

• Factor x2 6x 9.

Solution
x2 6x 9 x2 2 x 3 32 (x 3)2

• Factor 81x2 - 90x 25
• (ans

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The Difference of Two Squares

• If A and B are real numbers, variables, or
  algebraic expressions, then
• A2 B2 (A B)(A B).
• In words The difference of the squares of two
  terms factors as the product of a sum and the
  difference of those terms. (That is, the
  difference of square can be factored as the
  product of CONJUGATES.)

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Text Example

• Factor 81x2 - 49

Solution
81x2 49 (9x)2 72 (9x 7)(9x
7). (You will notice when we check this by
multiplying it out, the middle terms are
additive inverses (add to zero) because they have
opposite signs.)
Q Would it be possible to factor 81x2 49? A
NO. There is no real number factorization for
the SUM of squares.
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Factoring the Sum and Difference of 2 Cubes
Note that there is NO FACTORIZATION for the SUM
of SQUARES!
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Factoring by Grouping (by way of Example)

• Factor x3 5x2 4x 20 (DO NOT LOOK AT YOUR
  NOTES, no need to write.)

Solution x3 5x2 4x 20 (Note that there is
no gcf among all 4 terms). (x3 5x2) (-4x
20) Group the terms with common factors.
x2(x 5) 4(x 5) Factor from each group.
(Note that the negative leading coefficient
must be factored out.) (x 5)(x2
4) Factor out the common binomial factor,
(x 5). (x 5)(x 2)(x 2) Factor
completely by factoring x2 4 as the
difference of two squares.
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A Strategy for Factoring a Polynomial

• Factor out the GCF, if there is one.
• Determine the number of terms in the polynomial
  and try factoring as follows
• 2 terms can the binomial be factored by one of
  the special forms including difference of two
  squares, sum of two cubes, or difference of two
  cubes?
• 3 terms is the trinomial a perfect square
  trinomial? If the trinomial is not a perfects
  square trinomial, try factoring by trial and
  error (or the ac method).
• 4 or more terms try factoring by grouping.
• Check to see if any factors with more than one
  term in the factored polynomial can be factored
  further. (Look at exponents.) If so, factor
  completely.