If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Factoring quadratics: leading coefficient = 1

Learn how to factor quadratic expressions as the product of two linear binomials. For example, x²+5x+6=(x+2)(x+3).

What you need to know for this lesson

Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. For more on this, check out our previous article on taking common factors.

What you will learn in this lesson

In this lesson, you will learn how to factor a polynomial of the form x2+bx+c as a product of two binomials.

Review: Multiplying binomials

Let's consider the expression (x+2)(x+4).
We can find the product by applying the distributive property multiple times.
(x+2)(x+4)=(x+2)(x)+(x+2)(4)=x2+2x+4x+8=x2+6x+8
So we have that (x+2)(x+4)=x2+6x+8.
From this, we see that x+2 and x+4 are factors of x2+6x+8, but how would we find these factors if we didn't start with them?

Factoring trinomials

We can reverse the process of binomial multiplication shown above in order to factor a trinomial (which is a polynomial with 3 terms).
In other words, if we start with the polynomial x2+6x+8, we can use factoring to write it as a product of two binomials, (x+2)(x+4).
Let's take a look at a few examples to see how this is done.

Example 1: Factoring x2+5x+6

To factor x2+5x+6, we first need to find two numbers that multiply to 6 (the constant number) and add up to 5 (the x-coefficient).
These two numbers are 2 and 3 since 23=6 and 2+3=5.
We can then add each of these numbers to x to form the two binomial factors: (x+2) and (x+3).
In conclusion, we factored the trinomial as follows:
x2+5x+6=(x+2)(x+3)
To check the factorization, we can multiply the two binomials:
(x+2)(x+3)=(x+2)(x)+(x+2)(3)=x2+2x+3x+6=x2+5x+6
The product of x+2 and x+3 is indeed x2+5x+6. Our factorization is correct!

Check your understanding

1) Factor x2+7x+10.
Choose 1 answer:

2) Factor x2+9x+20.

Let's take a look at a few more examples and see what we can learn from them.

Example 2: Factoring x25x+6

To factor x25x+6, let's first find two numbers that multiply to 6 and add up to 5.
These two numbers are 2 and 3 since (2)(3)=6 and (2)+(3)=5.
We can then add each of these numbers to x to form the two binomial factors: (x+(2)) and (x+(3)).
The factorization is given below:
x25x+6=(x+(2))(x+(3))=(x2)(x3)
Factoring pattern: Notice that the numbers needed to factor x25x+6 are both negative (2 and 3). This is because their product needs to be positive (6) and their sum negative (5).
In general, when factoring x2+bx+c, if c is positive and b is negative, then both factors will be negative!

Example 3: Factoring x2x6

We can write x2x6 as x21x6.
To factor x21x6, let's first find two numbers that multiply to 6 and add up to 1.
These two numbers are 2 and 3 since (2)(3)=6 and 2+(3)=1.
We can then add each of these numbers to x to form the two binomial factors: (x+2) and (x+(3)).
The factorization is given below:
x2x6=(x+2)(x+(3))=(x+2)(x3)
Factoring patterns: Notice that to factor x2x6, we need one positive number (2) and one negative number (3). This is because their product needs to be negative (6).
In general, when factoring x2+bx+c, if c is negative, then one factor will be positive and one factor will be negative.

Summary

In general, to factor a trinomial of the form x2+bx+c, we need to find factors of c that add up to b.
Suppose these two numbers are m and n so that c=mn and b=m+n, then x2+bx+c=(x+m)(x+n).

Check your understanding

3) Factor x28x9.

4) Factor x210x+24.

5) Factor x2+7x30.

Why does this work?

To understand why this factorization method works, let's return to the original example in which we factored x2+5x+6 as (x+2)(x+3).
If we go back and multiply the two binomial factors, we can see the effect that the 2 and the 3 have on forming the product x2+5x+6.
(x+2)(x+3)=(x+2)(x)+(x+2)(3)=x2+2x+3x+23=x2+(2+3)x+23
We see that the coefficient of the x-term is the sum of 2 and 3, and the constant term is the product of 2 and 3.

The sum-product pattern

Let's repeat what we just did with (x+2)(x+3) for (x+m)(x+n):
(x+m)(x+n)=(x+m)(x)+(x+m)(n)=x2+mx+nx+mn=x2+(m+n)x+mn
To summarize this process, we get the following equation:
(x+m)(x+n)=x2+(m+n)x+mn
This is called the sum-product pattern.
It shows why, once we express a trinomial x2+bx+c as x2+(m+n)x+mn (by finding two numbers m and n so b=m+n and c=mn), we can factor that trinomial as (x+m)(x+n).

Reflection question

6) Can this factorization method be used to factor 2x2+3x+1?
Choose 1 answer:

When can we use this method to factor?

In general, the sum-product method is only applicable when we can actually write a trinomial as (x+m)(x+n) for some integers m and n.
This means that the leading term of the trinomial must be x2 (and not, for instance, 2x2) in order to even consider this method. This is because the product of (x+m) and (x+n) will always be a polynomial with a leading term of x2.
However, not all trinomials with x2 as a leading term can be factored. For example, x2+2x+2 cannot be factored because there are no two integers whose sum is 2 and whose product is 2.
In future lessons we will learn more ways of factoring more types of polynomials.

Challenge problems

7*) Factor x2+5xy+6y2.

8*) Factor x45x2+6.

Want to join the conversation?

  • duskpin sapling style avatar for user wongchris616
    For question 5, why can't (x-10)(x+3) work?
    (26 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user ✞☢♛Clatus♛☢✞
      Here's the question: Factor x^2+7x-30x
      The way Khan wants you to solve is this to find two numbers that add up to positive 7 and multiply to negative 30. If you add those two numbers together, they add up to negative 7. You were close, you just have to check that the numbers you get actually add and multiply correctly.

      So the answer is (x+10)(x-3)
      10+(-3) equals 7 and 10*(-3) equals 30.
      (74 votes)
  • blobby green style avatar for user jasmina.yorkulova
    I can't be the only one who finds this the most confusing thing I've ever learned
    (27 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user OnePlusThreeEqualsB
    On the "reflection question" there are two things wrong. The 2x^2 messes things up and also there are no two numbers that give you a sum of 3 and a product of 1.
    (10 votes)
    Default Khan Academy avatar avatar for user
  • leaf blue style avatar for user ltn7ntl
    Reflection Q6: 2x^2+3x+1 = (2x+1)(x+1) and is = 2x(x+1) + 1(x+1) .
    (10 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user zehor048.student
    help me please
    (3 votes)
    Default Khan Academy avatar avatar for user
    • sneak peak green style avatar for user somethingishere
      I had a lot of trouble with this too, so don't worry. I'm going to use an expression as an example. Say we have the expression x^2+5x+6. We need to figure out what two integers will add up to 5, and when multiplied, will make 6. To figure this out, you have to guess and check. The best way is to find out what numbers make 6 when they are multiplied. 3 and 2 work here. When 2 and 3 are added together, they make five. Therefore, to correctly factor x^2+5x+6, you would say (x+3)(x+2), or you could do (x+3)(x+2). There are also times when the middle term or the third number is negative. For example, x^2+x-6. The first step would be to find what two numbers make 6 when they are multiplied. 2 and 3 do. And to make positive one with these two numbers, 2 has to be negative, so you would factor x^2+x-6 as (x-2)(x+3). Sometimes the middle term will be negative. Let's take another example. x^2-8x+16. First, what we would figure out is what to numbers make 8. Now, since the two numbers are both negative, because the middle term is negative and the constant is positive, we can disregard the negative in -8. Two numbers that would work are 4 and 4. Now we have to un disregard the -8. So now, the numbers would be -4 and -4. -4*-4 is 16, so the factored form of x^2-8x+16 would be (x-4)(x-4). If both the middle term and the constant are negative, that means the greater number that makes the product is negative, and the smaller number is positive. For example, x^2-5x-14. The factored from would be (x-7)(x+2). I hope this helped, or at least made things a little bit more clear!
      (23 votes)
  • starky seedling style avatar for user crozier.james
    The challenge problems are. Challenging.
    (14 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user Tony
    For question 8 there can be another answer= (x^2-1)(-x^2-6)
    (3 votes)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      Sorry to tell you... your factors will not work.
      Let's check them by multiplying:
      (x^2-1)(-x^2-6) = -x^4 -6x^2 +x^2 + 6 = -x^4 -5x^2 + 6
      Notice: You have the wrong sign on the x^4 term. So, your factors will not work.
      Tip: If the leading term (the x^4) is positive, don't make one of them negative. A negative time a positive = a negative, not a positive.

      Hope this helps.
      (21 votes)
  • duskpin ultimate style avatar for user Black Fox
    In practice question #3 would not (x-1)(x+9) work the same way as (x+1)(x-9)?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      No... your 2 examples are not the same. You can see this if you multiply the binomials.
      (x-1)(x+9) = x^2 + 9x - x - 9 = x^2 + 8x - 9
      (x+1)(x-9) = x^2 - 9x + x - 9 = x^2 - 8x - 9
      Notice: the placement of the signs in your factors does matter. Practice question #3 wanted factors of x^2 - 8x -9, so the correct factors are (x+1)(x-9) because these would recreate the original polynomial.
      (10 votes)
  • blobby green style avatar for user Yarilis842
    can there be more than one variable
    (4 votes)
    Default Khan Academy avatar avatar for user
    • duskpin tree style avatar for user gser0616
      Yes! There can be more than one variable. We see this in challenge problem #7. When there are two variables, I prefer to solve by grouping as it makes more sense (at least to me). You can check out articles and videos under "Factoring Quadratics by Grouping". Assuming you took a look at those videos -
      I solved number 7 by first setting up equations for which two numbers the middle term (5xy) will be split into: (f and g represent my two numbers - I randomly chose these variables)

      f*g = 1*6 = 6
      (the 1 comes from x^2 and the 6 comes from 6y^2 because our two numbers need to multiply to the coefficients of the two outer terms.)

      f + g = 5 (our two numbers need to equal the coefficient of the middle term).

      So, what two numbers multiply to 6 and add to 5? That's 3 and 2.

      Now, we need to split up the 5xy term into two terms with coefficients 3 and 2. (3xy) and (2xy) will replace 5xy in the polynomial we are factoring. So now we have-

      x^2 + 2xy + 3xy + 6y^2

      The order of these terms (2xy) and (3xy) does not matter in this particular problem, but it does in some (watch the videos on Factoring by Grouping for more details).

      Now, we will separate the polynomial as shown and think of each part separately.

      x^2 + 2xy + 3xy + 6y^2

      Now we'll factor each separate part.

      x(x+2y) + 3y(x+2y)

      We can then factor out the (x+2y) so that we are left with our finished product:

      (x+2y) (3y+x)


      Hope this helps! Don't forget to check out those videos if you are confused on any part of this explanation.
      (9 votes)
  • sneak peak green style avatar for user Robert Zak
    By definition, what's a quadratic?
    (6 votes)
    Default Khan Academy avatar avatar for user