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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:

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:

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.