We will now work in the opposite direction—factoring. First we will factor quadratic polynomials, expressions of the form ax2 + bx + c (where a is not 0). For example x2 + 5x + 6 is factored as (x + 2) (x + 3). Quadratic polynomials whose first factors are x2 are the easiest to factor. Their factorization always begins as (x ± __) (x ± __). This forces the first factor to be x2 when the FOIL method is used All you need to do is fill in the two blanks and decide when to use plus and minus signs. All quadratic polynomials factor though some do not factor ‘‘nicely.’’ We will only concern ourselves with ‘‘nicely’’ factorable polynomials in this chapter.
If the second sign is minus, then the signs in the factors
will be different (one
plus and one minus). If the second sign is plus then both of the signs will be the same. In this case, if the
first sign in the trinomial is a plus sign, both signs in the factors will be plus; and if the first
sign in the trinomial is a minus sign, both signs in the factors will be minus.
Practice
1. x2 – 5x – 6 =
2.
x2 + 3x – 10 =
3. x2 + 4x – 21 =
Solutions
1. x2 – 5x – 6 = (x – __) (x + __)
2. x2 + 3x – 10 = (x – __) (x + __)
3. x2 + 4x – 21 = (x – __) (x
+ __)
Once the signs are determined all that remains is to fill in
the two blanks. Look
at all of the pairs of factors of the constant term. These pairs will be the candidates for the blanks. For
example, if the constant term is 12, you will need to consider 1 and 12, 2 and 6, and 3 and 4. If both
signs in the factors are the
same, these will be the only ones you need to try. If the signs are different, you will need to
reverse the order: 1 and 12 as well as 12 and 1; 2 and 6 as well as 6 and 2; 3 and 4 as well as 4
and 3. Try the FOIL method on
these pairs. (Not every trinomial can be factored in this
way).
Practice
1.
x2
– 5x – 6 =
2.
x2 + 3x – 10 =
3.
x2 + 4x – 21
=
Solutions
1. x2 – 5x – 6 = (x – 6) (x + 1)
2. x2 + 3x – 10 = (x
+ 5) (x – 2)
3. x2 + 4x – 21 = (x + 7) (x
– 3)
There is a factoring shortcut when the first term is x2.
If the second sign is plus,
choose the factors whose sum is the coefficient of the second term. For example the factors of 6 we need for x2 – 7x + 6 need to sum to 7: x2 – 7x + 6 = (x – 1)(x – 6). The factors of 6 we need for x2 + 5x + 6 need to sum to 5: x2 + 5x + 6 = (x
+ 2)(x + 3).
If the second sign is minus, the difference of the
factors needs to be the coefficient
of the middle term. If the first sign is plus, the bigger factor will have the plus sign. If the first sign
is minus, the bigger factor will have the minus sign.
Practice
1.
x2 – 6x + 9 =
2.
x2 + x – 30 =
3.
x2 – 49 =
Solutions
1.
x2 – 6x + 9 = (x – 3) (x – 3) = (x – 3)2
2.
x2 + x – 30 = (x + 5) (x – 4)
3.
x2 – 49 = (x
– 7) (x + 7)
Sumber
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