# Findthethreeintegers 1stintegerx2ndintegerx2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lternating signs; that is, one is negative and one is positive. If b is positive, then the larger factor is positive. If b is negative, then the larger factor is negative. In either case, you're looking for factors that are b units apart. To factor a "hard" quadratic, we have to handle all three coefficients, not just the two we handled in the "easy" case, because the leading coefficient (the number on the x2 term) is not 1. The first step in factoring will be to multiply "a" and "c"; then we'll need to find factors of the product "ac" that add up to "b". Factor 2x2 + x – 6. Looking at this quadratic, a = 2, b = 1, and c = –6, so ac = (2)(–6) = –12. Need to find factors of –12 that add up to +1. The pairs of factors for 12 are 1 and 12, 2 and 6, and 3 and 4. Factor 2x2 + x – 6. Since –12 is negative, need one factor to be positive and the other to be negative (because positive times negative is negative). This means that we want to use the pair "3 and 4", and we want the 3 to be negative, because –3 + 4 = +1. We need an inner product of ­3 and an outer product of 4. Since a = 2 and 2∙2 = 4 we will use a 2 for our outer product. 2x2 + x – 6 = (2x – 3)(x + 2). Factor 6x2 + 23x + 20 Multiply the leading coefficient a and the constant. 6 ∙20 = 120 Find a pair of factors , p and q, so tha...
View Full Document

## This document was uploaded on 02/18/2014.

Ask a homework question - tutors are online