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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 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.
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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?
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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.