Algebra 1
Solving by factoring (zero-product property)
Start with something you already trust about multiplying. If you multiply two numbers and the result is 0, one of them had to be 0. There's no other way to reach zero. Two non-zero numbers never multiply to zero. That small, obvious fact is the engine of this lesson.
Turn it on an equation. Suppose (x − 2)(x − 3) = 0. You've got two things multiplied together giving 0, so one of them must be 0: either x − 2 = 0 or x − 3 = 0. Each of those is a little one-step equation you can solve: x = 2 from the first, x = 3 from the second. So the quadratic has two solutions, 2 and 3, and you found them by reading off the factors 12.3.f1. This is the zero-product property, and it's the everyday method for the many quadratics that factor.
The one thing to hold onto: this trick works only against zero. Zero is special. It's the single value a product can reach only by one of its parts being zero. A product equal to 8 tells you nothing useful about the factors, because lots of pairs multiply to 8. So the first move, always, is to get the equation into the form (something) = 0, with everything on one side.
Each factor-equals-zero gives a genuine solution, not just a candidate. Here's why: the property runs both ways. Forward: if the product is 0, some factor is 0, and that's what finds your x-values. Backward: if a factor is 0, then the whole product is 0, so that x truly makes the equation true. The substitution check at the end is exactly that backward direction confirming the root is real.
And the factoring step itself is the reverse-distribute, area-box move from Unit 11, run to find the edges; once you've got the factors, each one set to zero is its own tiny equation.
So the procedure has four beats: get one side to 0 in the form ax² + bx + c = 0, factor the quadratic (Unit 11), set each factor equal to 0, and solve each little equation.
New words
- 12.3.d1 Zero-product property: if a product equals 0, then at least one factor is 0. If A·B=0, then A=0 or B=0. (No other number forces a product to zero; this is special to 0.)
- 12.3.d2 Root / solution: a value of x that makes the equation true. These are the same "roots" that are x-intercepts in 12.6.
Worked example
- 12.3.w1 (x-2)(x-3)=0. Already factored and equal to 0. Set each: x-2=0 ⇒ x=2; x-3=0 ⇒ x=3. x=2 or x=3. Check: (2-2)(2-3)=0·(-1)=0. The first factor went to zero, which is enough to make the whole product zero.
- 12.3.w2 x²-5x+6=0. Factor first (what multiplies to +6, adds to -5? -2,-3): (x-2)(x-3)=0 ⇒ x=2 or 3. Same two roots as #1, reached by factoring before reading them off.
- 12.3.w3 x²+x-6=0. (Multiply to -6, add to +1: +3,-2.) (x+3)(x-2)=0 ⇒ x=-3 or 2. Notice x+3=0 gives x=-3, not +3: the factor and its root carry opposite signs.
- 12.3.w4 x²-9=0. Difference of squares (Unit 11.3): (x-3)(x+3)=0 ⇒ x=±3. (Same answer 12.2 gives: two routes, one truth.)
- 12.3.w5 Must equal 0 first: x²-5x=-6 is not ready. Move the -6 over: x²-5x+6=0, then factor as in #2 → x=2,3. Setting factors of x²-5x=-6 to numbers is meaningless, because the zero-product property only works against 0.
Here's a clean case to settle the method before the set mixes things up: x² − x = 0. Both terms share an x, so factor it out: x(x − 1) = 0. Then x = 0 or x − 1 = 0, giving x = 0 or x = 1. Check: 0² − 0 = 0, and 1² − 1 = 0. Two roots, and one of them is zero itself, which is perfectly allowed.
Now that several are solved, two slips are worth a word. The first is setting factors equal to the wrong number, factoring x² − 5x + 6 = 8 and writing x − 2 = 8. The property needs a zero on the other side; an 8 gives you nothing, so move everything over first to make that side 0.
The second is stopping after one factor, reporting x = 2 and forgetting x = 3. Each factor is its own equation and (usually) its own root, so finish both: two factors, two answers. (If your factoring ever feels shaky, you can always check it by expanding the factors back out, or by substituting a root into the original; both catch a sign error at the factoring step.)
Check yourself
- 12.3.c1 Solve (x + 5)(x − 1) = 0 without expanding. What lets you split it into two easy equations? (The zero-product property: the product is 0, so a factor is 0. That means x + 5 = 0 or x − 1 = 0, giving x = −5 or x = 1.)
- 12.3.c2 A solution factors x² − 2x − 8 = 0 into (x − 4)(x + 2) and stops at x = 4. What's missing? (The second factor: x + 2 = 0 gives x = −2, so the full answer is x = 4 or x = −2. Each factor yields a root.)
- 12.3.c3 Why must x² − 3x = 10 be rewritten as x² − 3x − 10 = 0 before factoring to solve? (Because the zero-product property only works when the product equals 0. Against 10 the factors tell you nothing; moving the 10 over makes the right side 0 so the property applies.)
You can now solve a quadratic that factors by getting one side to 0, factoring it, setting each factor to 0, and solving. Then check by putting a root back in.
Mixed practice feels harder, and that's what makes it last. Every problem's answer is at the end of the lesson; if one stalls you, flip back to the worked example it's based on.
Solve (already factored):
Reveal answerHide to problem 1
x=1,4Reveal answerHide to problem 2
x=-5,2Factor, then solve:
Reveal answerHide to problem 3
(x-3)(x-4), x=3,4Reveal answerHide to problem 4
(x+2)(x+3), x=-2,-3Reveal answerHide to problem 5
(x-4)(x+2), x=-2,4Reveal answerHide to problem 6
(x+2)(x+5), x=-2,-5Reveal answerHide to problem 7
(x-4)(x+3), x=-3,4Reveal answerHide to problem 8
(x-3)(x-5), x=3,5Reveal answerHide to problem 9
x(x+3), x=0,-3Special patterns: