Main content
Course: Algebra 1 > Unit 13
Lesson 5: Factoring quadratics intro- Factoring quadratics as (x+a)(x+b)
- Factoring quadratics: leading coefficient = 1
- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Factoring quadratics with a common factor
- Factoring completely with a common factor
- Factoring quadratics with a common factor
- Factoring simple quadratics review
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Factoring simple quadratics review
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.
Our goal is to rewrite the expression in the form:
Expanding
So
After playing around with different possibilities for
Plugging these in, we get:
And we can multiply the binomials to check our solution if we'd like:
Yep, we get our original expression back, so we know we factored correctly to get our answer:
Want to see another example? Check out this video.
Want to join the conversation?
What are you supposed to do when the exponent that is squared has an integer before it?(37 votes)
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)
Hope you find this helpful!(72 votes)
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...(13 votes)
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.(31 votes)
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(11 votes)
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.(11 votes)- 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.(12 votes)
- how do i do this if an expression only has two terms(4 votes)
- If you have 2 terms, then you have a binomial. Assuming you have a quadratic (highest exponent is 2 on the variable), then your options to factor are:
1) A greatest common factor that includes the variable. For example: 3x^2-15x factors into 3x(x-5)
2) The binomial is a difference of 2 squares like: x^2-9 which factors into (x-3)(x+3)
Hope this helps.(25 votes)
- how do i factorise the
expression 18pq-7+4pq-4(7 votes)- First, you should simplify the polynomial because in its current form, it is not factorable. Combine like terms.
22pq-11
Now, you have a binomial with a common factor. Factor out the common factor. Can you identify the factor and do the factoring from here?
Comment back if you have questions.(6 votes)
- is there a way to make negatives easier(7 votes)
- 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.(2 votes)
- why do we do this stuff(5 votes)
Algebra is used in many scientific jobs and in programming.(6 votes)
- is there a way to make negatives easier(7 votes)
no. just do a lot of problems and get used to it(2 votes)
Is a quadratic simply the product of two or more polynomials?(4 votes)- A quadratic is any expression that can be written as: Ax^2+Bx+C where A, B, C are real numbers and A does not =0. Basically is has an x^2 term as its highest degree term.(6 votes)
