factoring trinomials

Desan65

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Mar 12, 2012
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Can you help me factor the trinomials step by step in this problem?
12q ² + 34 q - 28
 
Can you help me factor the trinomials step by step in this problem?
12q ² + 34 q - 28

Here's another method that you can use to factor some quadratic trinomials of the form

ax2 + bx + c

Note that after removing a common factor of 2, as JeffM did in his approach, you have 2(6q2 + 17q - 14)

What is inside the parentheses is a quadratic trinomial of this form:

ax2 + bx + c
6q2 + 17q - 14

Step 1: multiply a*c. 6*(-14) is -84

Step 2: look for two factors of this product (-84) which add up to "b" in the original expression, which in this case is +17

-4*21 = -84, and -4 + 21 = +17

Step 3: Use the numbers you found in step 2 to rewrite the middle term. We'll rewrite +17q as -4q + 21q:

6q2 - 4q + 21q - 14

Step 4: Factor by grouping the first two terms together and the last two terms together, and then remove a common factor from each group:

(6q2 - 4q) + (21q - 14)

2q (3q - 2) + 7 (3q - 2)

Now, (3q - 2) is a common factor. Remove it, and you have

(3q - 2)(2q + 7)

remember that you originally removed a common factor of 2; this needs to be included in the "final answer":

2 (3q - 2)(2q + 7)
 
Thanks! I think I'm starting to understand it. How about this problem? Factor each polynomial.

12h ² - 60h + 75
 
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