5.6 Factoring Trinomials

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Transcript 5.6 Factoring Trinomials

5.6 Factoring Trinomials
ax  bx  c
2
How to Factor
• 1. Write the trinomial in descending powers of
one variable.
• 2. Factor out any GCF (including (-1) to make
the 1st term positive)
• 3. Test the trinomial for factorability.
2
– Check the value of b  4ac
• If it’s a perfect square, then it’s factorable into 2 different
binomials
• If it’s zero, then it’s factorable and the factors will be the
same
More steps
• 4. When the sign of the 1st term of the trinomial is
positive:
– And the sign of the 3rd term is positive, the signs in each
parentheses will be the same
x  6x  5   x  5 x 1
2
– And the sign of the 3rd term is negative, the signs in each
parentheses will be different
x  6x  7   x  7 x  1
2
More steps continued
• 5. Try different combos until you find the one
that works.
– If the leading coefficient is 1
• Pick the factorization where the sum of the factors of c is
equal to the coefficient of the middle term (b)
– If the leading coefficient is not 1
• Can use trial and error
• Can use the key method
Step 5 Cont’d (Key Method)
• i. Find product of ac (this is the key #)
• ii. Find 2 factors of the key # whose sum is b
• iii. Use the factors of the key # as coefficient of 2
2
terms to be placed between ax and c
• iv. Factor using grouping
Example of key method
15a  17 a  4
2
i. ac  15  4   60
ii. factors of  60 that add up to 17
20 and  3
iii. 15a  3a  20a  4
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iv. 3a  5a  1  4  5a  1
 5a  1 3a  4 
• 6. Check by multiplying out.
Another example – key method
6 x  17 x  12
2
i. ac  6 12  72
ii. factors of 72 that add up to 17 (b term)
9 and 8
iii. 6 x  9 x  8x  12
2
iv. 3x  2x  3  4  2x  3
 2x  33x  4
Using Substitution to Factor
 x  y
Let
2
 7  x  y   12
 x  y  z
z  7 z  12
2
 z  3 z  4
Resubtitute:
 x  y  3 x  y  4
Using Grouping to Factor Trinomials
10 x  13x  3
2
10 x  15x  2 x  3
2
10 x
2
 15 x   2 x  3
5x  2x  3   2x  3
 2x  35x 1