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## Get ready for Algebra 2

Factoring quadratics is very similar to multiplying binomials, just going the other way. For example, x^2+3x+2 factors to (x+1)(x+2) because (x+1)(x+2) multiplies to x^2+3x+2. This article reviews the basics of how to factor quadratics into the product of two binomials.

### Example

Factor as the product of two binomials.
x, squared, plus, 3, x, plus, 2
Our goal is to rewrite the expression in the form:
left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis
Expanding left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis gives us a clue.
\begin{aligned} x^2+\goldD{3}x+\blueD{2}&=(x+a)(x+b) \\\\ &= x^2 +ax+bx + ab \\\\ &= x^2 +\goldD{(a+b)}x + \blueD{ab} \end{aligned}
So start color #e07d10, left parenthesis, a, plus, b, right parenthesis, equals, 3, end color #e07d10 and start color #11accd, a, b, equals, 2, end color #11accd.
After playing around with different possibilities for a and b, we discover that a, equals, start color #1fab54, 1, end color #1fab54, b, equals, start color #1fab54, 2, end color #1fab54 satisfies both conditions.
Plugging these in, we get:
left parenthesis, x, plus, start color #1fab54, 1, end color #1fab54, right parenthesis, left parenthesis, x, plus, start color #1fab54, 2, end color #1fab54, right parenthesis
And we can multiply the binomials to check our solution if we'd like:
\begin{aligned} &~(x+1)(x+2) \\\\ =&~x^2+2x+x+2 \\\\ =&~x^2+3x+2 \end{aligned}
Yep, we get our original expression back, so we know we factored correctly to get our answer:
left parenthesis, x, plus, start color #1fab54, 1, end color #1fab54, right parenthesis, left parenthesis, x, plus, start color #1fab54, 2, end color #1fab54, right parenthesis
Want to see another example? Check out this video.

## Practice

Factor the quadratic expression as the product of two binomials.
x, squared, minus, x, minus, 42, equals

Want more practice? Check out this exercise.

## Want to join the conversation?

• What are you supposed to do when the exponent that is squared has an integer before it?
• Here is an example: 3x^2 + 31x + 10

The way you solve this is pretty simple. Since the 3x is squared, you multiply 3x by x and you get 3x^2. The important thing to know about this is that 3x^2 does NOT equal 3x * 3x, because that would be 9x^2.

So when you factor this quadratic, you need to find a way to factor the 3x^2 into 3x * x. The answer for the quadratic is (3x + 1)(x + 10)

• I'm so confused... How do we find a and b? Do we just guess or something, because that's what it looks like to me.... I feel stupid because I don't understand this...
• It takes a while to get used to factoring polynomials. You need to be familiar with multiplying polynomials before you attempt factoring. Factoring reverses the processes used to multiply polynomials. You also need to be good at find all factor pairs for a number.

Consider what happens in this example...
Multiply (x + 3)(x + 5).
You would do this using FOIL or expanded distribution. As you multiply, here's what happens.
x(x) + 5x + 3x + 3(5)
x(x) + (5+3)x + 3(5)
x^2 + 8x + 15

I wrote the above in great detail because you need to see what the X's, the 3 and the 5 create (where they go in the final result). Factoring reverses this process.
Look at the 2nd row above: x(x) + (5+3)x + 3(5)
The 1st term: x(x) comes for the 1st term in both binomials. So, you know your binomials will look like: "(x + _ ) (x + _ )".
The 2nd and 3rd term are created using the 3 and the 5. In the 3rd term they are multiplied. In the 2nd term, they are added. So, we learn to factor polynomials by looking at the 3rd term (the 15 in the final result) and finding 2 factors of 15 that also add to the 2nd term (the 8)
Factors of 15 are:
1, 15
3, 5
Only 3 and 5 add to 8, thus we know the binomials need to contain the 3 and 5.
Thus, our factors become "(x+3)(x+5)"

2nd example: let's factor x^2 + 11x + 24
We know the lead terms in our binomials have to be "x" and "x".
So, we have "(x + _) (x + _)"
We start with the 24. We need to find 2 factors of 24 that add to 11 (the middle term)
This is where you need your factoring skills. Factors of 24 are:
1, 24
2, 12
3, 8
4, 6
Which pair adds to 11? The 3 and 8 add to 11, so that is the pair we want.
Thus, our factors become "(x + 3)(x + 8)"

Hope this helps.
• How would you do this if the leading coefficient isn't one? I can't seem to find an article for that on Khan. - Thanks
• My question was to factor -3x^2+6x+9 completely, and after I did all my work I ended up with the answer "-3(x+1)(x-3)" which I thought was correct, but then it told me I was wrong, so I looked at how they did the problem, and I saw we did the exact same steps, and we got the exact same answer. I don't know if it was just a glitch or something, but id love to know if I got it incorrect, or if it was Khan Academy that made a mistake.
• You can always check your factors - Do the multiplication and see if you get back to the original polynomial.
-3(x+1)(x-3) = -3(x^2-3x+x-3)
= -3x^2+9x-3x+9
= -3x^2+6x+9
Looks like your factors are correct.

Note: If you find something like this when you are doing a KA exercise, use the "report a problem" link within the exercise to report it to KA.
• is there a way to make negatives easier
• If all of the terms are negative, you could factor a negative. Ex: -x^2-4x-3 = -1(x^2+4x+3). If they aren't all negative, you just have to do it normally.
(1 vote)
• what do you do when you don't understand this even after you've watched the videos?
• is there a way to make negatives easier
• no. just do a lot of problems and get used to it
(1 vote)
• ok so how do I do the length and width ones? I couldn't find those in the videos
• Imagine the length and widths as just normal equations. Remember that the area is length x width, and once you can wrap your head it, it's easy to factor it the same way.