Unit 11 · Lesson 11.2

Factoring trinomials x²+bx+c

This is the lesson the unit builds toward, so give it the most time. It's the exact reverse of multiplying two binomials in Unit 10, and it's the move that makes solving quadratics in Unit 12 possible. Once you can turn x²+5x+6 into (x+2)(x+3), setting it equal to zero and reading off the answers is one short step away.

Picture the area box from Unit 10 again, but backward. When you multiplied (x+2)(x+3), the box filled in: the corner cells came out x² and 6, and the two middle cells, 2x and 3x, added to 5x. Factoring runs that in reverse. Now you're handed the inside of the box, with the x², the middle, and the corner, and you have to find the edges:

$$\begin{array}{c|c|c} & x & \,? \\ \hline x & x^2 & ?x \\ \hline ? & ?x & 6 \end{array}$$

The two unknown edges have to multiply to 6 (that's the corner cell), and their x-terms have to add to 5x (that's the middle). Which is exactly one question, and it's the question behind every trinomial: what two numbers multiply to c and add to b?

So the method is a search for two numbers. List the pairs that multiply to c, and pick the pair that adds to b. For x²+5x+6, the pairs multiplying to 6 are 1·6 and 2·3; only 2 + 3 gives 5. So the factors are (x+2)(x+3).

The part that takes real care is the signs, and they aren't a guess: the signs of c and b tell you the signs of the two numbers. Here's how to read them. Look first at c.

When c is positive, the two numbers have the same sign, because their product came out positive. Then b breaks the tie: if b is positive too, both numbers are positive, and if b is negative, both are negative.

When c is negative, the two numbers have opposite signs, because their product is negative. Then the number with the larger size carries the sign of b.

Take the signs slowly, in that order: c first, then b. And if one slips, the multiply-back check catches it, since the middle term comes out wrong.

When nothing fits: prime trinomials

Not every trinomial factors over the integers, and saying so plainly matters, because the alternative is hunting forever for an answer that isn't there. The two-number search has a finite list to check, just the factor pairs of c, so it can genuinely run out. When it does, the trinomial is prime, and that's the correct, complete answer, not a sign you failed.

Here's the rule that tells you when to stop. To factor x²+bx+c, list every integer pair that multiplies to c, then check each pair's sum against b. If no pair sums to b, stop: the trinomial is prime (irreducible over the integers). Working through the whole list is the proof. It's the same "I actually checked" spirit as multiplying back, turned on the search itself. The key is that the list has to be exhausted: you've earned the word "prime" only after every pair has been tried, including the negative pairs when the signs call for them.

A prime one (c>0, b>0): 11.2.w7 $$x^2+2x+5 \;\Rightarrow\; \text{prime (irreducible over the integers)}$$ Since c = 5 is positive, the two numbers would share a sign, and since b is positive, both would be positive. The only positive pair multiplying to 5 is 1·5, and that sums to 6, not 2. There's no other pair. The search is exhausted, so it's prime. There's nothing to multiply back here, and that's fine, because the finished list is the proof.

Another prime one: 11.2.w8 $$x^2+x+1 \;\Rightarrow\; \text{prime (irreducible over the integers)}$$ Here c = 1 and b = 1. The only integer pairs multiplying to 1 are 1·1 (sum 2) and (−1)(−1) (sum −2); neither sums to 1. List exhausted, so it's prime.

Two honest cautions about "prime." First, "prime" doesn't mean "I gave up." It means you finished a finite search and it came up empty, so always list the pairs and read off the sums, the way the examples above do.

Second, don't reach for non-integers to rescue a near-miss. You might notice that x²+5x+9 almost works and feel tempted to try fractions or decimals. Hold the line: over the integers it's prime, full stop. Other tools for these come in Unit 12.

And one trap that hides a prime answer: a GCF can disguise things. Take 2x²+4x+10. As written it looks unfactorable, but it isn't prime. Pull the 2 out first to get 2(x²+2x+5), and then the inside, x²+2x+5, is prime. So "factor completely" here means 2(x²+2x+5): the GCF out front, the prime piece left intact.

This is the reason for a habit worth adopting now: always try the GCF before the two-number search. Pulling the shared factor leaves a smaller, friendlier trinomial inside, and it keeps you from wrongly calling something prime.

A couple more places people stumble, both caught by the same multiply-back habit. One is declaring "prime" too soon: quitting after one or two pairs instead of all of them, or forgetting the negative pairs when c is positive and b is negative. Only an exhausted list earns the word.

The other is skipping the check entirely, accepting, say, (x+2)(x+3) for x²+6x+6 without expanding. Always multiply back, because the middle term is where errors surface. (And as it happens, x²+6x+6 is itself prime: no integer pair multiplies to 6 and adds to 6.)

If a problem still isn't landing after you've walked the signs and the pairs, it's fine to leave it and come back tomorrow. A break genuinely helps, and you haven't lost anything. When you return, re-read the area-box picture at the top of the lesson first; that's the version of this idea everything else is built on.