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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 x, squared, plus, b, x, plus, c as a product of two binomials.

Review: Multiplying binomials

Let's consider the expression left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis.
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\begin{aligned} \tealD{(x+2)}(x+4)&=\tealD{(x+2)}(x)+\tealD{(x+2)}(4)\\\\ &=x^2+2x+4x+8\\\\ &=x^2+6x+8 \end{aligned}
So we have that left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, equals, x, squared, plus, 6, x, plus, 8.
From this, we see that x, plus, 2 and x, plus, 4 are factors of x, squared, plus, 6, x, plus, 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 x, squared, plus, 6, x, plus, 8, we can use factoring to write it as a product of two binomials, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis.
Let's take a look at a few examples to see how this is done.

Example 1: Factoring x, squared, plus, 5, x, plus, 6

To factor x, squared, plus, start color #e07d10, 5, end color #e07d10, x, plus, start color #aa87ff, 6, end color #aa87ff, we first need to find two numbers that multiply to start color #aa87ff, 6, end color #aa87ff (the constant number) and add up to start color #e07d10, 5, end color #e07d10 (the x-coefficient).
These two numbers are start color #11accd, 2, end color #11accd and start color #1fab54, 3, end color #1fab54 since start color #11accd, 2, end color #11accd, dot, start color #1fab54, 3, end color #1fab54, equals, 6 and start color #11accd, 2, end color #11accd, plus, start color #1fab54, 3, end color #1fab54, equals, 5.
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color #11accd, 2, end color #11accd, right parenthesis and left parenthesis, x, plus, start color #1fab54, 3, end color #1fab54, right parenthesis.
In conclusion, we factored the trinomial as follows:
x, squared, plus, 5, x, plus, 6, equals, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis
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\begin{aligned}(x+2)(x+3)&=(x+2)(x)+(x+2)(3)\\ \\ &=x^2+2x+3x+6\\ \\ &=x^2+5x+6 \end{aligned}
The product of x, plus, 2 and x, plus, 3 is indeed x, squared, plus, 5, x, plus, 6. Our factorization is correct!

Check your understanding

1) Factor x, squared, plus, 7, x, plus, 10.
Choose 1 answer:
Choose 1 answer:

2) Factor x, squared, plus, 9, x, plus, 20.

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

Example 2: Factoring x, squared, minus, 5, x, plus, 6

To factor x, squared, start color #e07d10, minus, 5, end color #e07d10, x, plus, start color #aa87ff, 6, end color #aa87ff, let's first find two numbers that multiply to start color #aa87ff, 6, end color #aa87ff and add up to start color #e07d10, minus, 5, end color #e07d10.
These two numbers are start color #11accd, minus, 2, end color #11accd and start color #1fab54, minus, 3, end color #1fab54 since left parenthesis, start color #11accd, minus, 2, end color #11accd, right parenthesis, dot, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, equals, 6 and left parenthesis, start color #11accd, minus, 2, end color #11accd, right parenthesis, plus, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, equals, minus, 5.
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, left parenthesis, start color #11accd, minus, 2, end color #11accd, right parenthesis, right parenthesis and left parenthesis, x, plus, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, right parenthesis.
The factorization is given below:
x25x+6=(x+(2))(x+(3))=(x2)(x3)\begin{aligned}x^2-5x+6&=(x+(\blueD{-2}))(x+(\greenD{-3}))\\ \\ &=(x\blueD{-2})(x\greenD{-3}) \end{aligned}
Factoring pattern: Notice that the numbers needed to factor x, squared, minus, 5, x, plus, 6 are both negative left parenthesis, minus, 2 and minus, 3, right parenthesis. This is because their product needs to be positive left parenthesis, 6, right parenthesis and their sum negative left parenthesis, minus, 5, right parenthesis.
In general, when factoring x, squared, plus, b, x, plus, c, if c is positive and b is negative, then both factors will be negative!

Example 3: Factoring x, squared, minus, x, minus, 6

We can write x, squared, minus, x, minus, 6 as x, squared, minus, 1, x, minus, 6.
To factor x, squared, start color #e07d10, minus, 1, end color #e07d10, x, start color #aa87ff, minus, 6, end color #aa87ff, let's first find two numbers that multiply to start color #aa87ff, minus, 6, end color #aa87ff and add up to start color #e07d10, minus, 1, end color #e07d10.
These two numbers are start color #11accd, 2, end color #11accd and start color #1fab54, minus, 3, end color #1fab54 since left parenthesis, start color #11accd, 2, end color #11accd, right parenthesis, dot, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, equals, minus, 6 and start color #11accd, 2, end color #11accd, plus, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, equals, minus, 1.
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color #11accd, 2, end color #11accd, right parenthesis and left parenthesis, x, plus, left parenthesis, start color #1fab54, minus, 3, end color #1fab54, right parenthesis, right parenthesis.
The factorization is given below:
x2x6=(x+2)(x+(3))=(x+2)(x3)\begin{aligned}x^2-x-6&=(x+\blueD2)(x+(\greenD{-3}))\\ \\ &=(x+\blueD2)(x\greenD{-3}) \end{aligned}
Factoring patterns: Notice that to factor x, squared, minus, x, minus, 6, we need one positive number left parenthesis, 2, right parenthesis and one negative number left parenthesis, minus, 3, right parenthesis. This is because their product needs to be negative left parenthesis, minus, 6, right parenthesis.
In general, when factoring x, squared, plus, b, x, plus, 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 x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff, we need to find factors of start color #aa87ff, c, end color #aa87ff that add up to start color #e07d10, b, end color #e07d10.
Suppose these two numbers are m and n so that c, equals, m, n and b, equals, m, plus, n, then x, squared, plus, b, x, plus, c, equals, left parenthesis, x, plus, m, right parenthesis, left parenthesis, x, plus, n, right parenthesis.

Check your understanding

3) Factor x, squared, minus, 8, x, minus, 9.

4) Factor x, squared, minus, 10, x, plus, 24.

5) Factor x, squared, plus, 7, x, minus, 30.

Why does this work?

To understand why this factorization method works, let's return to the original example in which we factored x, squared, plus, 5, x, plus, 6 as left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis.
If we go back and multiply the two binomial factors, we can see the effect that the start color #11accd, 2, end color #11accd and the start color #1fab54, 3, end color #1fab54 have on forming the product x, squared, plus, 5, x, plus, 6.
(x+2)(x+3)=(x+2)(x)+(x+2)(3)=x2+2x+3x+23=x2+(2+3)x+23\begin{aligned}(x+\blueD 2)(x+\greenD3)&={(x+\blueD2)}(x)+(x+\blueD 2)(\greenD{3})\\ \\ &=x^2+\blueD2x+\greenD3x+\blueD2\cdot \greenD3\\ \\ &=x^2+(\blueD 2+\greenD 3)x+\blueD2\cdot \greenD3 \end{aligned}
We see that the coefficient of the x-term is the sum of start color #11accd, 2, end color #11accd and start color #1fab54, 3, end color #1fab54, and the constant term is the product of start color #11accd, 2, end color #11accd and start color #1fab54, 3, end color #1fab54.

The sum-product pattern

Let's repeat what we just did with left parenthesis, x, plus, start color #11accd, 2, end color #11accd, right parenthesis, left parenthesis, x, plus, start color #1fab54, 3, end color #1fab54, right parenthesis for left parenthesis, x, plus, start color #11accd, m, end color #11accd, right parenthesis, left parenthesis, x, plus, start color #1fab54, n, end color #1fab54, right parenthesis:
(x+m)(x+n)=(x+m)(x)+(x+m)(n)=x2+mx+nx+mn=x2+(m+n)x+mn\begin{aligned}(x+\blueD m)(x+\greenD n)&={(x+\blueD m)}(x)+(x+\blueD m)(\greenD{n})\\ \\ &=x^2+\blueD mx+\greenD nx+\blueD m\cdot \greenD n\\ \\ &=x^2+(\blueD m+\greenD n)x+\blueD m\cdot \greenD n \end{aligned}
To summarize this process, we get the following equation:
left parenthesis, x, plus, start color #11accd, m, end color #11accd, right parenthesis, left parenthesis, x, plus, start color #1fab54, n, end color #1fab54, right parenthesis, equals, x, squared, plus, left parenthesis, start color #11accd, m, end color #11accd, plus, start color #1fab54, n, end color #1fab54, right parenthesis, x, plus, start color #11accd, m, end color #11accd, dot, start color #1fab54, n, end color #1fab54
This is called the sum-product pattern.
It shows why, once we express a trinomial x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff as x, squared, plus, left parenthesis, start color #11accd, m, end color #11accd, plus, start color #1fab54, n, end color #1fab54, right parenthesis, x, plus, start color #11accd, m, end color #11accd, dot, start color #1fab54, n, end color #1fab54 (by finding two numbers start color #11accd, m, end color #11accd and start color #1fab54, n, end color #1fab54 so start color #e07d10, b, end color #e07d10, equals, start color #11accd, m, end color #11accd, plus, start color #1fab54, n, end color #1fab54 and start color #aa87ff, c, end color #aa87ff, equals, start color #11accd, m, end color #11accd, dot, start color #1fab54, n, end color #1fab54), we can factor that trinomial as left parenthesis, x, plus, start color #11accd, m, end color #11accd, right parenthesis, left parenthesis, x, plus, start color #1fab54, n, end color #1fab54, right parenthesis.

Reflection question

6) Can this factorization method be used to factor 2, x, squared, plus, 3, x, plus, 1?
Choose 1 answer:
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 left parenthesis, x, plus, m, right parenthesis, left parenthesis, x, plus, n, right parenthesis for some integers m and n.
This means that the leading term of the trinomial must be x, squared (and not, for instance, 2, x, squared) in order to even consider this method. This is because the product of left parenthesis, x, plus, m, right parenthesis and left parenthesis, x, plus, n, right parenthesis will always be a polynomial with a leading term of x, squared.
However, not all trinomials with x, squared as a leading term can be factored. For example, x, squared, plus, 2, x, plus, 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 x, squared, plus, 5, x, y, plus, 6, y, squared.

8*) Factor x, start superscript, 4, end superscript, minus, 5, x, squared, plus, 6.

Want to join the conversation?

  • 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) .
    (7 votes)
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  • 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.
    (6 votes)
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  • male robot johnny style avatar for user Tony
    For question 8 there can be another answer= (x^2-1)(-x^2-6)
    (2 votes)
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    • 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.
      (12 votes)
  • blobby green style avatar for user Yarilis842
    can there be more than one variable
    (3 votes)
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    • 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.
      (7 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)?
    (3 votes)
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    • 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.
      (6 votes)
  • starky seedling style avatar for user Tyler Istre
    does it matter if i factor (x^2-5x^2+6) as (x-6)(x+1) instead of (x-3)(x-2)?
    (3 votes)
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  • starky tree style avatar for user Courtney472
    What if the two numbers have nothing in common,
    like for instance 51 and 13?
    (2 votes)
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  • duskpin sapling style avatar for user wongchris616
    For question 5, why can't (x-10)(x+3) work?
    (2 votes)
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  • spunky sam blue style avatar for user 💎Chυcκ Lørrε💎
    What should I do if I don't know what kind of form will the factor form be?
    Like what kind of form will 56x^5-72x^4 be after it has been factored?
    (3 votes)
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  • starky tree style avatar for user Megan Puth
    The reflection question for number 6 confused me a little. When you x-factor a*c and b (ax^2+bx+c) you can still use the method and just divide by the GCF.
    EX: 2x^2+3x+1 2(a*c)=2*1; 3(b)=2+1
    ([2÷]2x+[2÷]2)(2x+1)
    2x^2+3x+1=(x+1)(2x+1)
    (1 vote)
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