When the Leading Coefficient Is Not 1
A.SSE.2
Read the standard text
A.SSE.2 Use the structure of an expression to identify ways to rewrite it. For example, to factor 3x(x − 5) + 2(x − 5), students should recognize that the "x − 5" is common to both expressions being added, so it simplifies to (3x + 2)(x − 5); or see x^4 − y^4 as (x^2)^2 − (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 − y^2)(x^2 + y^2).
Section 1 · LearnLearn the idea
In this lesson you will: Factor out -1 or a common factor so the leading coefficient becomes 1 whenever you can; Factor a quadratic with leading coefficient +/-2 or +/-3 by splitting the middle term; Keep the common factor in the final answer.
First ask whether it has to be hard
Two of the three quadratics that arrive with a leading coefficient other than 1 can be made easy in one move. If every term shares a factor, take it out: 2x^2+8x+6=2(x^2+4x+3)=2(x+1)(x+3). If the leading term is negative, take out -1: -x^2-5x-6=-(x^2+5x+6)=-(x+2)(x+3).
Ohio's content limits keep the leading coefficients small on purpose. Away from the special forms of Lesson 15.4, you will see 1, -1, +/-2 and +/-3, and nothing worse. That is the whole range this lesson covers.
Splitting the middle term
When there is no common factor — 2x^2+7x+3 is the standard example — use the product ac. Here a=2 and c=3, so ac=6. Find two numbers with product 6 and sum 7: they are 1 and 6.
Use them to split the middle term, then factor the two halves separately: 2x^2+x+6x+3=x(2x+1)+3(2x+1). The binomial (2x+1) is now a common factor, exactly as in Lesson 15.1, and the answer is (2x+1)(x+3).
The area model still works
Nothing about the rectangle picture changes when a is 2 or 3. The four pieces are still the four products, and the side lengths are still the factors.
For 2x^2+11x+12 the pieces are 2x^2, 8x, 3x and 12. The two middle pieces add to 11x, and the sides read off as 2x+3 and x+4.
Common mistake — Losing the common factor on the way to the answer
Factoring 2x^2+8x+6, a student divides by 2, factors x^2+4x+3 into (x+1)(x+3), and writes that down as the answer.
The middle of the work was correct, and the 2 was set aside for what felt like a moment. Then the page turned.
(x+1)(x+3)=x^2+4x+3, which is half of the expression that was asked about. Dividing by 2 is a legal move on an equation, where both sides change together, but an expression has no other side to balance it. Here the 2 was factored out, so it must stay in the answer: 2(x+1)(x+3).
How to catch it: Compare constants. The original ends in 6; your answer multiplies out to a constant of 3. Any mismatch in the first or last term means a factor went missing.
Section 2 · DoPractice this standard
These published practice sets are scored on the student site and cover A.SSE.2 directly.