Factoring ax² + bx + c, part 2

By Tutor GuyNo Comments


In my last post, I showed you a procedure for how to factor a quadratic when the leading coefficient is not equal to 1. Here I refine that method slightly with a short cut that makes your work even quicker. However, this method is a bit more difficult—if you are not really good at mental calculations, you will need to practice this a few times to get it down. To compare methods, I’ll use the same quadratic we factored in the last post.

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

  1. Our first step is the same as in the previous method—we need to find two numbers that multiply together to give 60 (that’s 12*-5) and add together to give +4. Of course, those numbers are the same as they were in the last post: +10 and -6.
  2. Now here’s the shortcut. Divide both of these numbers by the leading coefficient, 12. That gives two fractions:
    ….. \dfrac{10}{12} \;\; \text{and } \dfrac{-6}{12}
  3. Next, simplify the fractions if possible. In our example, we get
    ….. \dfrac{5}{6} \;\; \text{and } \dfrac{-1}{2}
  4. In each fraction, the denominator is the coefficient of the x term in our factor and the numerator is the constant term. So our two factors must be 6x+5 and 2x-1. See the diagram below:



Algebra 1, Algebra 2
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