Download Hybrid Elementary Algebra 5.3 – 5.5 Introduction Factoring

Hybrid Elementary Algebra
5.3 – 5.5 Introduction
2
Factoring trinomials of the form ax + bx + c, a ≠ 1 .
Steps to factor a polynomial of the form ax 2 + bx + c .
1. Find the GCF of all terms in the polynomial. If it is something other than 1, factor it out.
2. If the trinomial can be factored, it will be the product of two binomials (remember factoring is
“undoing” distributing).
3. The first terms have to multiply to ax 2 .
4. The second terms of each binomial must multiply to c . Furthermore the Outside and Inside
terms (remember FOIL) must multiply to bx . Factoring trinomials with a ≠ 1 is more
challenging than when a = 1 ; be persistent! Remember it is possible that the trinomial is
“prime” or does not factor.
5. Check your work by distributing.
Factor each trinomial, if possible.
1. 2 x 2 + 5 x − 3
No common factors. The only way to multiply to 2x 2 is if the first term of one binomial is 2x and the
other is x . Look at the signs: the second terms of the binomials need to multiply to a negative and
add to a positive. Thus one binomial will have a plus and the other a minus.
= ( 2 x + ? )( x − ? ) or ( 2 x − ? )( x + ? )
The only numbers that multiply to −3 are −1∗ 3 or −3 ∗1 .
If the trinomial factors, it must factor as one of the following four options:
( 2 x + 3)( x − 1)
( 2 x − 3)( x + 1)
( 2 x + 1)( x − 3)
( 2 x − 1)( x + 3)
Checking each possibility verifies the correct factoring is ( 2 x − 1)( x + 3) .
2. 2 x 2 − 5 x − 7
Hybrid Elementary Algebra
5.3 – 5.5 Introduction
Page 2 of 6
3. 3x 2 + 14 x + 15
No common factors. The only way to multiply to 3x 2 is if the first term of one binomial is 3x and the
other is x . Look at the signs: the second terms of the binomials need to multiply to a positive and
add to a positive. Thus both binomials will have a positive sign.
= ( 3 x + ? )( x + ? ) or ( 3 x − ? )( x − ? )
The numbers that multiply to 15 are 1 ∗15 or 3 ∗ 5 .
The possible factorings for the trinomial are:
( 3x + 1)( x + 15 )
( 3x + 15)( x + 1)
( 3 x + 3)( x + 5 )
( 3 x + 5 )( x + 3)
Some of these possibilities are unnecessary because they create binomials that have a common
factor. Since there was not a common factor in the trinomial, no binomial will have a common factor.
The unnecessary possibilities are crossed out. Multiply out the remaining possibilities to see which
one (if any) works.
The trinomial factors as ( 3 x + 5 )( x + 3) .
4. 3x 2 − 13x + 4
Hybrid Elementary Algebra
5.3 – 5.5 Introduction
Page 3 of 6
5. 2 x 2 + 3 x − 20
6. 6 x 2 − 13 x + 6
No common factors. There are two ways to multiply to 6x 2 : either 6x and x or 3x and 2x . Look at
the signs: the second terms of the binomials need to multiply to a positive and add to a negative.
Thus both binomials will have a minus sign.
= ( 6 x − ? )( x − ? ) or ( 3 x − ? )( 2 x − ? )
The numbers that multiply to 6 are 1∗ 6 or 3 ∗ 2 . Remember not to try any factorings that result in a
binomial having a common factor.
The possible factorings for the trinomial are:
( 6 x − 1)( x − 6 )
( 3 x − 2 )( 2 x − 3)
All other possible factorings result in at least one binomial with a common factor. Checking both
possibilities confirms the trinomial factors as ( 3 x − 2 )( 2 x − 3 ) .
Hybrid Elementary Algebra
5.3 – 5.5 Introduction
Page 4 of 6
7. 20 x 2 − 25 x + 5
Factoring Difference of Squares: This is a special case that is easy to recognize. First off, a
difference of squares only has two terms. Second, it is a difference, so one term is positive and the
other is negative. Finally each term is a perfect square.
Distribute
( x − 7 )( x + 7 ) = x 2 − 49
Factor
8. x 2 − 16
This is a binomial. There is no common factor. The first term is positive and the second is negative.
Further both x 2 and 16 are perfect squares, so this expression is of the form a 2 − b 2 .
Factors as ( x − 4 )( x + 4 ) .
9. x 2 − 64
Hybrid Elementary Algebra
5.3 – 5.5 Introduction
Page 5 of 6
10. 49 x 2 − 25
This is a binomial. There is no common factor. The first term is positive and the second is negative.
Further both 49x 2 and 25 are perfect squares, so this expression is of the form a 2 − b 2 .
Factors as ( 7 x + 5 )( 7 x − 5 ) .
11. −9 + 4z 2
12. 12 x 3 − 75 x
This is a binomial. Take out the common factor of 3x .
= 3x 4 x 2 − 25
(
)
The remaining binomial is a difference of squares.
= 3 x ( 2 x + 5 )( 2 x − 5 )
13. 48t 4 − 27t 2
Hybrid Elementary Algebra
5.3 – 5.5 Introduction
Page 6 of 6
General strategy to factor polynomials:
1. Find the GCF of all terms in the polynomial. If it is something other than 1, factor it out.
2. Count the terms in the polynomial or “leftovers” after you factor out the GCF.
3. Factor the “leftovers,” if possible. If a polynomial does not factor, it is called prime.
a. If two terms, the polynomial must be a difference of squares to factor.
b. If three terms, determine if a = 1 or if a ≠ 1 and follow the appropriate plan.
c. If four terms, try factor by grouping.
4. Repeat steps 2 and 3 until all factors are prime.
5. Use the distribution property to check your work.