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Factoring_Trinomials30
Factoring Trinomials
Trinomials of the Form x^{2} +b x +c
If a polynomial of the form x^{2} + b x + c is factorable
then it can be written as (x + u) (x + v).
Note: (x + u) (x + v) = x^{2} + v x + u x + u ٠ v = x^{2} + (u + v) x + u ٠ v
Looking at this carefully we see that if a trinomial of the form x^{2} + b x + c
is factorable then b = u + v and c = u٠v, for some
numbers u and v . So to factor this form of a polynomial we need to find two
numbers that have product c and that have sum b.
Examples
Example 1: Factor x ^{2} + 7 x + 12
Solution : We need to find two numbers that multiply to be
12, but add to be 7. These two numbers are 3 and 4. This then gives
the factorization
x^{2} + 7 x + 12 = (x + 3) (x + 4)
Example 2: Factor x^{2}  12 x + 35
Solution : We need to find two numbers that multiply to be
35, but add to be 12. These two numbers are 7 and  5. This then
gives the factorization
x^{2}  12 x + 35 = (x  7) (x  5)
Example 3: Factor x^{2} + 5 x  24
Solution: We need to find two numbers that multiply to be
24, but add to be 5. These two numbers are 3 and 8. This then
gives the factorization
x^{2}+ 5 x  24 = (x  3) (x + 8)
Trinomials of the Form ax^{2} +b x +c
If a polynomial of the form a x^{2} + b x + c is factorable
then it can be written as (m x + u) (n x + v).
Note:
Looking at this carefully we see that if a trinomial of
the form a x ^{2} + b x + c is factorable then b = m٠ v + n ٠ u and
a٠ c = m٠ n ٠u ٠v = （m٠v）（n ٠u）, for some numbers m , n, u and v. So to
factor this form of a polynomial we need to find two
numbers that have product a ÿ c and that have sum b . We can then rewrite the
polynomial as four terms (similar to the first step
above) and then factor by grouping .
Examples
Example 1: Factor 6 x^{2} + 7 x + 2
Solution: First we find the product of a and c. In this
case it's 6 * 2 = 12. So we need two numbers that multiply to be 12, but add
to be 7. These numbers are 3 and 4. We use these numbers to rewrite the middle
term as 3 x + 4 x and then factor by grouping as
follows:
Example 2: Factor 10 x^{2}  23 x + 12
Solution: First we find the product 10 * 12 = 120. So we
need two numbers that multiply to be 120 and add to be 23. These two
numbers are 15 and  8. So the middle term can be rewritten as 15 x  8 x. Now
factor by grouping.
Example 3: Factor 24 x^{2}  14 x  3
Solution: First we find the product 24 *3 = 72. So we
need two numbers that multiply to be 72 and add to be 14. These
two numbers are 18 and 4. So the middle term can be rewritten as 18 x + 4 x.
Now factor by grouping.
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