Quadratic equations come up all the time in algebra, as do binomials. You might be hoping for some shortcuts for factoring quadratic binomials – luckily, there are some that you can use!

So, how do you factor a quadratic binomial? **For a quadratic binomial in standard form, if the constant term is zero (c = 0), the quadratic binomial factors as x(ax + b) with roots x = 0 and x = –b/a. When the linear term is zero (b = 0), the quadratic binomial factors as a(x + r)(x – r) with roots r and –r, where r is the square root of -c/a.**

Of course, you can also derive these shortcut formulas from the quadratic formula by plugging in either b = 0 or c = 0 as the case may be.

In this article, we’ll talk about how to factor a quadratic binomial and look at several examples of each of the two cases. We’ll also talk about the shortcut formulas and where they come from.

Let’s get going.

## How To Factor A Quadratic Binomial

Factoring a quadratic binomial is more straightforward than factoring a quadratic trinomial. For one thing, there are fewer terms to worry about.

There are also some shortcut formulas (derived from the quadratic formula) that can make your life much easier when working to factor or solve quadratic binomials.

The best way to start this discussion is a review of vocabulary so that we know exactly what we are looking at.

First, remember that the **standard form of a quadratic equation** is:

ax^{2} + bx + c = 0

where a is not equal to zero.

Here are some examples of quadratic equations in standard form:

**x**^{2}– 4x + 1**8x**^{2}+ 5x + 7**1.5x**(the coefficients are rational here, not integers)^{2}– 4.2x + 5.1**3x**(the coefficient of ‘x’ is zero here)^{2}+ 11**2x**(the constant term is zero here)^{2}– 13x

Also, remember that a **binomial** has two terms added together (sort of like how a bicycle has two wheels).

Here are some examples of binomials:

**2x + 5****x – 7****x**^{2}+ 5**3x**^{3}– 8**4x**^{2}– 9x

Putting these definitions together, a **quadratic binomial** is a quadratic with two terms. It must always have an x^{2} term (since a cannot equal zero in a quadratic) and one other term: either an x term (linear) or a constant term that has a nonzero coefficient.

This is what gives us our two cases for factoring a quadratic binomial: whether we have b = 0 (zero linear term) or c = 0 (zero constant term). Let’s take a closer look at each of these.

### Factoring Quadratic Binomials: Two Cases

There are two basic cases to consider when factoring a quadratic binomial of the form ax^{2} + bx + c = 0:

**Case 1: c = 0**– this case is fairly easy to factor, since both nonzero terms have an x that we can factor out. No complex numbers will be necessary here: one root is zero, and the other is –b/a.**Case 2: b = 0**– this case can be factored as a difference of squares. However, we might need to use complex numbers for some quadratic binomials in this case (we can check the discriminant to find out when this is the case!)

Note that in the second case, we might also need to factor out a GCF (greatest common factor) first in order to factor completely and to make our work easier. Also remember that quadratic binomials with b = 0 are even functions.

We’ll take a closer look at each case, a shortcut formula, and some examples so you will know exactly how to solve them.

#### Factoring Quadratic Binomials: Case 1 (c = 0):

In this case, we have c = 0 in the standard form of the quadratic equation, which gives us:

**ax**^{2}+ bx + c = 0**ax**[since c = 0]^{2}+ bx = 0**x(ax + b) = 0**[factor out x]**x = 0 or ax + b = 0**[solve 2 separate equations, one for each factor]**x = 0 or x = -b/a**

So, the two solutions are x = 0 and x = -b/a. You can also derive this shortcut formula from the quadratic formula with c = 0:

Either way, we can use this formula to easily solve quadratic binomials with c = 0. Here are some examples.

##### Example 1: Factoring A Quadratic Binomial With c = 0

Let’s factor 2x^{2} + 5x = 0 to solve.

For this quadratic, we have a = 2, b = 5, and c = 0.

We know the factors are x and ax + b. Plugging in a = 2 and b = 5, we get:

x and 2x + 5 are the factors, so 2x^{2} + 5x = x(2x + 5).

We also know that the roots are x = 0 and x = -b/a. Plugging in a = 2 and b = 5, we get:

x = 0 and x = -5/2 are the roots (solutions).

##### Example 2: Factoring A Quadratic Binomial With c = 0

Let’s factor 2.5x^{2} – 7.5x = 0 to solve.

For this quadratic, we have a = 2.5, b = -7.5, and c = 0.

We know the factors are x and ax + b. Plugging in a = 2.5 and b = -7.5, we get:

x and 2.5x – 7.5 are the factors, so 2.5x^{2} – 7.5x = x(2.5x – 7.5).

If we want to factor completely, we can factor out the GCF of 2.5:

2.5x^{2} – 7.5x = 2.5(x)(x – 3).

We also know that the roots are x = 0 and x = -b/a. Plugging in a = 2.5 and b = -7.5, we get:

x = 0 and x = -7.5/2.5 = -3 are the roots (solutions).

#### Factoring Quadratic Binomials: Case 2 (b = 0):

In this case, we have b = 0 in the standard form of the quadratic equation, which gives us:

**ax**^{2}+ bx + c = 0**ax**^{2}+ c = 0**ax**^{2}= -c**x**^{2}= -c/a

To solve we take the square root of both sides of the last equation (considering both the positive and negative roots).

So, the two solutions are x = r and x = -r, where r is the square root of –c/a.

You can also derive this shortcut formula from the quadratic formula with b = 0:

Since we are taking the square root of –c/a to find our solutions, we know we will get imaginary (complex) solutions in some cases – but when?

Well, -c/a will be negative when a and c have the same sign.

In other words, if a and c are both positive, we will get imaginary roots. Likewise, if a and c are both negative, we will get two distinct imaginary roots.

If a and c have different signs, then we will get two distinct real roots.

Either way, we can use this formula to easily solve quadratic binomials with c = 0. Here are some examples.

##### Example 1: Factoring A Quadratic Binomial With b = 0 (Real Roots)

Let’s factor x^{2} – 9 = 0 to solve.

For this quadratic, we have a = 1, b = 0, and c = -9.

We know that this will factor as a(x + r)(x – r), where r is the square root of –c/a. Plugging in a = 1 and c = -9, we get:

- -c/a
- =-(-9) / 1 [c = -9, a = 1]
- =9

Taking the square root, we get r = 3. So, this quadratic factors as:

- a(x + r)(x – r)
- =1(x + 3)(x – 3)
- =(x + 3)(x – 3)

Then x^{2} – 9 = (x + 3)(x – 3). Note that we can also recognize this as a difference of squares!

Now for the roots (solutions). Since a and c have opposite signs (a is positive, c is negative), we know we will get two distinct real roots.

We also know that the roots are x = r and x = -r. So, the roots are x = 3 and x = -3 (we can verify that these two values make the quadratic equation true).

##### Example 2: Factoring A Quadratic Binomial With b = 0 (Imaginary Roots)

Let’s factor 4x^{2} + 16 = 0 to solve.

For this quadratic, we have a = 1, b = 0, and c = 16.

We know that this will factor as a(x + r)(x – r), where r is the square root of –c/a. Plugging in a = 4 and c = 16, we get:

- -c/a
- =-(16) / 4 [c = 16, a = 4]
- =-4

Taking the square root, we get r = 2i. So, this quadratic factors as:

- a(x + r)(x – r)
- =4(x + 2i)(x – 2i)

Then 4x^{2} + 16 = 4(x + 2i)(x – 2i). Note that we can also recognize this as a difference of squares if we rewrite it as 4x^{2} + 16 = 4x^{2} – (-16) = (2x)^{2} – (2i)^{2}.

Now for the roots (solutions). Since a and c have the same signs (a is positive, c is positive), we know we will get two distinct complex roots.

We also know that the roots are x = r and x = -r. So, the roots are x = 2i and x = -2i (we can verify that these two values make the quadratic equation true).

## Conclusion

Now you know how to factor a quadratic binomial in the case of b = 0 or c = 0. You also know about the shortcut formulas you can use and where they come from.

You can learn more about factoring polynomials in my article on factoring by grouping.

You can find out about quadratic factored form and what it can tell you in my article here.

You might also want to read my article on whether parabolas are one-to-one functions and my article on quadratic functions (and the 3 forms to know).

I hope you found this article helpful. If so, please share it with someone who can use the information.

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~Jonathon