Factoring ax² + bx + c

By Tutor GuyNo Comments

 

You probably have a lot of experience factoring quadratics of the form x^2+bx+c. (That is, where a, the coefficient on the x^2 term is equal to 1.) But what do you do when a isn’t 1? For example, let’s factor 12x^2+4x-5. You may have been taught a trial-and-error technique where you look for all the numbers that multiply to give 12x^2 and combine them with all the numbers that multiply to give 5 and test each one to see which gives the correct middle term. But there’s a simple algorithm that takes all the guesswork out. I’ll show you the procedure with an example.

Problem: Factor 12x^2+4x-5

  1. Similar to the case where a=1, look for two numbers that multiply together to give a \cdot c (in our example, 12 \cdot -5=-60) and add together to give b (i.e., +4 in our example). [Actually, you can see that this is just like the a=1 case, because there you were looking for two numbers that multiply together to give c, but when a=1, a \cdot c=c…] For our example, those numbers are +10 and -6 \\.
    …..
  2. Bring down the first term without changing it:
    …..
    ….. 12x^2
    …..
  3. Write the linear term as a sum or difference using the two numbers you found in step A.:
    …..
    ….. 12x^2 +10x-6x
    …..
    (Don’t worry about which term comes first. As it turns out, you can write these two terms in either order.)
    …..
  4. Bring down the last term without changing it:
    …..
    …..12x^2 +10x-6x - \; 5
    …..
  5. Now comes the magic. Look at the first two terms only from step D (cover up the last two terms if that helps you) and factor them:
    …..
    ….. 2x \!\!(6x+5)
    …..
  6. Now look at the last two terms in step D (again, cover the first two terms if that helps you) and factor them. And this is very important: if the x term has a minus sign on it, always factor out the minus sign:
    …..
    ….. 2x \!\! (6x+5) - \; 1 \!\! (6x+5)
    …..
    (Note that if nothing factors out of the last two terms, write a 1 (you can always factor 1 out of any term).
    …..
  7. Are the two expressions in the parentheses the same? They have to be. If they’re not, you’ve done something wrong. That’s one of the factors in your answer. The other factor is made up of the numbers outside the parentheses:
    …..
    ….. (2x-1) \!\! (6x+5)

 

Tada! I’ll leave it to you to show that if you had written the linear terms in step C in the reverse order, you’d get the same answer.

 

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