Algebra 1
Greatest common factor (GCF)
Start with a picture you can hold. Imagine a row of gift bags, and every bag has a chocolate inside. You can reach into each bag, take out one chocolate, and set them all in a single pile out front. That pile is the chocolates every bag was carrying. What's left stays in the bags.
That's exactly what pulling out a greatest common factor does. Take 6x + 9. Both terms are divisible by 3, so both bags are carrying a 3. The 3 comes out front, and what's left of each term goes inside the parentheses:
$$6x + 9 = 3(2x + 3)$$
This is the first thing to try on any factoring problem. Pulling out a shared factor often shrinks a scary-looking expression into something plain, and it's the exact reverse of the single-term distributing you did in Units 2 and 10. It also gives you the multiply-back check that the rest of the unit leans on.
Here's the same idea as a picture, using the area box from Unit 10 run backward. Think of 4x² + 8x as the inside of a one-row box whose left edge is 4x. The top edges are whatever you multiply 4x by to land on each piece:
$$\begin{array}{c|c|c} & x & 2 \\ \hline 4x & 4x^2 & 8x \end{array}\qquad\Rightarrow\qquad 4x^2+8x = 4x(x+2)$$
The left edge, 4x, is the common factor; the top edges, x and 2, are what's left inside.
Now the symbols. Finding the GCF comes in two parts. First the number: the biggest number that divides every coefficient evenly. For 4x² + 8x, that's the biggest number dividing both 4 and 8, which is 4.
Then the variables: the lowest power of the variable that shows up in every term. Here you have x² in the first term and x¹ in the second, so the most that every term carries is x¹. Multiply the two parts together and the GCF is 4x. Divide each term by 4x to find what stays inside.
One thing to name before it trips you up. When you divide 8x by 4x, you get 2, not 0. The term doesn't vanish; it just gets smaller by the factor you removed. Nothing disappears by magic. The piece is still there, shrunk down to what's left after the common factor came out.
New words
- 11.1.d1 Factor (noun): something multiplied. In 6=2·3, both 2 and 3 are factors of 6. In 3(2x+3), the factors are 3 and (2x+3).
- 11.1.d2 Greatest common factor (GCF): the largest factor (biggest number times the most variables) that divides evenly into every term.
- 11.1.d3 Factor (verb): to rewrite a sum as a product, i.e. to undo distributing.
Worked example
(each pulls the GCF, then checks by multiplying back):
Numbers only: 11.1.w1 $$6x+9 \;=\; 3(2x+3) \qquad \text{Check: } 3\cdot 2x + 3\cdot 3 = 6x+9$$
Number and a variable: 11.1.w2 $$4x^2+8x \;=\; 4x(x+2) \qquad \text{Check: } 4x\cdot x + 4x\cdot 2 = 4x^2+8x$$
Higher powers, take the lowest power of x: 11.1.w3 $$15x^3-10x^2 \;=\; 5x^2(3x-2) \qquad \text{Check: } 5x^2\cdot 3x - 5x^2\cdot 2 = 15x^3-10x^2$$ Both terms carry at least x², so x² comes out; the biggest number dividing 15 and 10 is 5, so 5 comes out. Together the GCF is 5x².
The "stopped too early" trap, done right: 11.1.w4 $$12x+18 \;=\; 6(2x+3) \qquad \text{Check: } 6\cdot 2x + 6\cdot 3 = 12x+18$$ Here's the slip worth knowing now that you've factored a few cleanly. It's easy to pull out a common factor and stop, to write 2(6x + 9) or 3(4x + 6) and call it done. Those are common factors, but not the greatest one: look inside the parentheses and the pieces still share a factor. Only pulling the full 6 leaves an inside, 2x + 3, with nothing left to take. A quick self-check: after you factor, glance inside the parentheses and ask whether those terms still have something in common. If they do, you didn't take it all.
Variable factor with a negative inside: 11.1.w5 $$10x^2-25x \;=\; 5x(2x-5) \qquad \text{Check: } 5x\cdot 2x - 5x\cdot 5 = 10x^2-25x$$ Notice the minus stayed put. When a term is subtracted, keep its sign as you pull the factor out, and the multiply-back check catches a dropped sign instantly.
Two more slips that the multiply-back check will always expose, so you're never relying on memory alone. One is forgetting the variable part: writing 4x² + 8x = 4(x² + 2x) and leaving an x behind. The fix is to ask whether every bag has at least one x, and if so, the x comes out too.
The other is taking too high a power: trying to pull x³ out of 15x³ − 10x², when the second term only has x². You can only take the lowest power that every term actually has.
Here's a clean one to get the move running before the practice mixes things up. Factor 8x² + 12x. The biggest number dividing 8 and 12 is 4, and both terms carry at least one x, so the GCF is 4x. That leaves 4x(2x + 3). Multiply back: 4x·2x + 4x·3 = 8x² + 12x. It matches. Just the one move, start to finish.
Check yourself
- 11.1.c1 Factor 8x²+12x, then prove it's right on your own. (The GCF is 4x, giving 4x(2x+3); multiply it back out. 4x·2x + 4x·3 = 8x²+12x, and it matches, so it's right.)
- 11.1.c2 Someone writes 6x+9=3(2x+9). Multiply it back. Where did it go wrong? (Multiplying back gives 3·2x + 3·9 = 6x+27, not 6x+9. The 9 inside should be 3, since 3·3 = 9; the right answer is 3(2x+3).)
- 11.1.c3 Why can't you pull x² out of 15x³−10x² as x² and still leave an x in the second term? (The second term is the x² piece. Once you take x² out of −10x², all that's left is the coefficient −10. There's no extra x hiding in it to leave behind.)
You can now find the greatest common factor of a polynomial, the biggest number times the most variables that every term shares, pull it out front, and confirm your work by multiplying back.
Mixed practice feels harder than repeating one kind of problem, and that's the point. It's what makes the skill last. Every problem below has its answer at the end of the lesson, and if one stalls you, flip back to the worked example it's based on. That's what it's there for.
Numbers only:
Reveal answerHide to problem 1
2(x+3)Reveal answerHide to problem 2
5(x+3)Reveal answerHide to problem 3
3(2x+3)Reveal answerHide to problem 4
4(x-3)Reveal answerHide to problem 5
4(2x+5)Reveal answerHide to problem 6
3(3x-2)With a variable:
Reveal answerHide to problem 7
x(x+7)Reveal answerHide to problem 8
3x(x+2)Reveal answerHide to problem 9
5x(2x-3)Reveal answerHide to problem 10
4x(x+2)Higher powers (take the lowest power):