If you want to solve a quadratic equation, factoring is one good option. Not only does factoring tell you the solutions, but it can make things easier if you want to graph the corresponding parabola.

So, how do you factor a quadratic? **To factor a quadratic, arrange the terms into standard form ax ^{2} + bx + c = 0. Next, try to identify a special case (such as a difference of squares). Then, apply a formula or the process for factoring a general quadratic. Finally, write an equation to show the original quadratic & its factored form.**

Of course, there are lots of shortcuts you can use when factoring. It’s just a matter of knowing the formulas and being able to recognize the shortcuts when you see them.

In this article, we’ll talk about how to factor a quadratic. We’ll look at the formulas and methods you can use to factor, along with examples of each to make the concepts clear.

Let’s get started.

## How To Factor A Quadratic

To factor a quadratic, take the following steps:

**First, rearrange the quadratic into standard form: ax**^{2}+ bx + c = 0.**Next, see if you can identify one of the special cases for factoring (these are listed below).****Then, apply the formula for the special case or the process for factoring a quadratic in the general case.****Finally, write an equation to show the original quadratic and the factored form.**

The special cases for factoring quadratics are as follows:

**Zero Constant Term:**in this case, we have c = 0, which gives us a quadratic binomial of the form ax^{2}+ bx, which factors easily as x(ax + b).**Difference of Squares:**in this case, we have b = 0 and a difference of terms d^{2}x^{2}– e^{2}, which is a quadratic binomial that factors as (dx + e)(dx – e).**Perfect Square Trinomial (positive middle term):**in this case, we have a quadratic of the form x^{2}+ 2dx + d^{2}, which factors as (x + d)(x + d) or (x + d)^{2}.**Perfect Square Trinomial (negative middle term):**in this case, we have a quadratic of the form x^{2}– 2dx + d^{2}, which factors as (x – d)(x – d) or (x – d)^{2}.**General Case (monic):**in this case, we have a = 1, and we try to find a pair of values d and e so that d + e = b and de = c. If we find such a pair, then we can factor x^{2}+ bx + c as (x + d)(x + e).**General Case (non-monic)”:**in this case, we have a not equal to 1, but we can divide all terms by a to get to the previous case (monic) and proceed as usual. Once we factor as (x + d)(x + e), we multiply by a to get the complete factorization a(x + d)(x + e).

Let’s take a look at some examples of each case to see how to factor them.

### Factor A Quadratic With Zero Constant Term (c = 0)

When a quadratic has a zero constant term (c = 0), we are left with a quadratic binomial where both terms have an x. We can factor out this x, which tells us that x = 0 is always a solution to this type of quadratic.

Here are some examples.

#### Example 1: Factor A Quadratic With Zero Constant Term (c = 0)

Consider the following quadratic equation:

**2x**^{2}+ 10x = 0

Note that our coefficients are a = 2, b = 10, and c = 0. Since c = 0, we have a zero constant term and we can factor out x from the quadratic (along with a 2, since a and b are both even) to get:

**2x(x + 5) = 0**

Note that the solutions of this equation are x = 0 and x = -5. The complete factorization of this quadratic is:

**2x**^{2}+ 10x = 2x(x + 5)

#### Example 2: Factor A Quadratic With Zero Constant Term (c = 0)

Consider the following quadratic equation:

**3x**^{2}– 12x = 0

Note that our coefficients are a = 3, b = -12, and c = 0. Since c = 0, we have a zero constant term and we can factor out x from the quadratic (along with a 3, since a and be are both multiples of 3) to get:

**3x(x – 4) = 0**

Note that the solutions of this equation are x = 0 and x = 4. The complete factorization of this quadratic is:

**3x**^{2}– 12x = 3x(x – 4)

### Factor A Quadratic By A Difference Of Squares

When a quadratic has a zero linear term (b = 0) and the other two terms are perfect squares (with one of them negative), we are left with a difference of squares. We can factor a difference of squares according to the formula

**A**^{2}– B^{2}= (A + B)(A – B)

Here are some examples.

#### Example 1: Factor A Quadratic By A Difference Of Squares

Consider the following quadratic equation:

**x**^{2}– 9 = 0

Note that our coefficients are a = 1, b = 0, and c = -9. Since we have a difference of squares (9 = 3^{2}), we can use the formula for a difference of squares:

**A**^{2}– B^{2}= (A + B)(A – B)

with A = x and B = 3 to get:

**A**^{2}– B^{2}= (A + B)(A – B)**x**^{2}– 3^{2}= (x + 3)(x – 3)**x**^{2}– 9 = (x + 3)(x – 3)

Note that the solutions of this equation are x = 3 and x = -3. The complete factorization of this quadratic is:

**x**^{2}– 9 = (x + 3)(x – 3)

#### Example 2: Factor A Quadratic By A Difference Of Squares

Consider the following quadratic equation:

**4x**^{2}– 100 = 0

Note that our coefficients are a = 4, b = 0, and c = -100. Since we have a difference of squares, we can use the formula for a difference of squares:

**A**^{2}– B^{2}= (A + B)(A – B)

with A = 2x and B = 10 to get:

**A**^{2}– B^{2}= (A + B)(A – B)**(2x)**^{2}– 10^{2}= (2x + 10)(2x – 10)**4x**^{2}– 100 = (2x + 10)(2x – 10)

Note that the solutions of this equation are x = 5 and x = -5. The complete factorization of this quadratic is:

**4x**^{2}– 100 = (2x + 10)(2x – 10)

Note that we also could have factored out a 4 at the beginning (to keep the numbers smaller during intermediate steps) and then multiplied by 4 again at the end.

### Factor A Quadratic By A Perfect Square Trinomial (b > 0)

When a quadratic has the form x^{2} + 2dx + d^{2} (with a = 1), we have a perfect square trinomial, which we can factor according to the formula

**x**^{2}+ 2dx + d^{2}= (x + d)(x + d)

or

**x**^{2}+ 2dx + d^{2}= (x + d)^{2}

Note that if a (the quadratic coefficient) is not equal to 1, we can divide the entire equation by a, use the formula to factor, and then multiply by a at the end.

Here are some examples.

#### Example 1: Factor A Quadratic By A Perfect Square Trinomial

Consider the following quadratic equation:

**x**^{2}+ 12x + 36 = 0

Note that our coefficients are a = 1, b = 12, and c = 36. Note that b = 12 = 2*6 and c = 36 = 6^{2}. Since we have a perfect square trinomial, we can factor by the formula:

**x**^{2}+ 2dx + d^{2}= (x + d)^{2}

with d = 6 to get:

**x**^{2}+ 2dx + d^{2}= (x + d)^{2}**x**^{2}+ 2(6)x + 6^{2}= (x + 6)^{2}**x**^{2}+ 12x + 36 = (x + 6)^{2}

Note that the solution of this equation is x = -6 (a double real root). The complete factorization of this quadratic is:

**x**^{2}+ 12x + 36 = (x + 6)^{2}

#### Example 2: Factor A Quadratic By A Perfect Square Trinomial

Consider the following quadratic equation:

**2x**^{2}+ 20x + 50 = 0

Note that all of our coefficients are even, so we can divide by 2 to get:

**x**^{2}+ 10x + 25 = 0

Note that b = 10 = 2*5 and c = 25 = 5^{2}. Since we have a perfect square trinomial, we can factor by the formula:

**x**^{2}+ 2dx + d^{2}= (x + d)^{2}

with d = 5 to get:

**x**^{2}+ 2dx + d^{2}= (x + d)^{2}**x**^{2}+ 2(5)x + 5^{2}= (x + 5)^{2}**x**^{2}+ 10x + 25 = (x + 5)^{2}

Note that the solution of this equation is x = -5 (a double real root). Since we divided by 2 at the beginning, we need to multiply by 2 at the end.

So, the complete factorization of this quadratic is:

**2(x**^{2}+ 10x + 25) = 2(x + 5)^{2}**2x**^{2}+ 20x + 50 = 2(x + 5)^{2}

### Factor A Quadratic By A Perfect Square Trinomial (b < 0)

When a quadratic has the form x^{2} – 2dx + d^{2} (with a = 1), we have a perfect square trinomial, which we can factor according to the formula

**x**^{2}– 2dx + d^{2}= (x – d)(x – d)

or

**x**^{2}– 2dx + d^{2}= (x – d)^{2}

Note that if a (the quadratic coefficient) is not equal to 1, we can divide the entire equation by a, use the formula to factor, and then multiply by a at the end.

Here are some examples.

#### Example 1: Factor A Quadratic By A Perfect Square Trinomial (b < 0)

Consider the following quadratic equation:

**x**^{2}– 8x + 16 = 0

Note that our coefficients are a = 1, b = -8, and c = 16. Note that b = -8 = 2*-4 and c = 16 = 4^{2}. Since we have a perfect square trinomial, we can factor by the formula:

**x**^{2}– 2dx + d^{2}= (x – d)^{2}

with d = 4 to get:

**x**^{2}– 2dx + d^{2}= (x – d)^{2}**x**^{2}– 2(4)x + 4^{2}= (x – 4)^{2}**x**^{2}– 8x + 16 = (x – 4)^{2}

Note that the solution of this equation is x = 4 (a double real root). The complete factorization of this quadratic is:

**x**^{2}– 8x + 16 = (x – 4)^{2}

#### Example 2: Factor A Quadratic By A Perfect Square Trinomial

Consider the following quadratic equation:

**3x**^{2}– 42x + 147 = 0

Note that all of our coefficients are multiples of 3, so we can divide by 3 to get:

**x**^{2}– 14x + 49 = 0

Note that b = -14 = 2*-7 and c = 49 = 7^{2}. Since we have a perfect square trinomial, we can factor by the formula:

**x**^{2}– 2dx + d^{2}= (x – d)^{2}

with d = 7 to get:

**x**^{2}– 2dx + d^{2}= (x – d)^{2}**x**^{2}– 2(7)x + 7^{2}= (x – 7)^{2}**x**^{2}– 14x + 49 = (x – 7)^{2}

Note that the solution of this equation is x = 7 (a double real root). Since we divided by 3 at the beginning, we need to multiply by 3 at the end.

So, the complete factorization of this quadratic is:

**3(x**^{2}– 14x + 49) = 3(x – 7)^{2}**3x**^{2}– 42x + 147 = 2(x + 5)^{2}

### Factor A Quadratic In The General Case (a = 1)

When a quadratic has the form x^{2} + bx + c (with a = 1), we can often factor by finding values d and e such that

**b = d + e****c = de**

Let’s try some examples to see how this works.

#### Example 1: Factor A Quadratic In The General Case (a = 1)

Consider the following quadratic equation:

**x**^{2}+ 7x + 6 = 0

Since a = 1, b = 7, and c = 6, we can try to factor by finding values of c and d such that:

**7 = d + e****6 = de**

The factorizations of 6 are 1*6 and (-1)*(-6). Since we want d + e to be positive, we choose d = 1 and e = 6 (positive values).

This gives us d + e = 1 + 6 = 7.

So, we can factor by the formula

**x**^{2}+ (d + e)x + de = (x + d)(x + e)

With d = 1 and e = 6, this gives us

**x**^{2}+ (1 + 6)x + (1)(6) = (x + 1)(x + 6)**x**^{2}+ 7x + 6= (x + 1)(x + 6)

Note that the solutions of this equation are x = -1 and x = -6. So, the complete factorization of this quadratic is:

**x**^{2}+ 7x + 6= (x + 1)(x + 6)

#### Example 2: Factor A Quadratic In The General Case (a = 1)

Consider the following quadratic equation:

**x**^{2}+ x – 12 = 0

Since a = 1, b = 1, and c = -12, we can try to factor by finding values of c and d such that:

**1 = d + e****-12 = de**

The factorizations of -12 are 1*(-12), (-1)*12, 3*(-4), and (-3)*4. Since we want d + e to be 1, we know that we want d = -3 and e = 4.

This gives us d + e = -3 + 4 = 1.

So, we can factor by the formula

**x**^{2}+ (d + e)x + de = (x + d)(x + e)

With d = -3 and e = 4, this gives us

**x**^{2}+ (-3 + 4)x + (-3)(4) = (x + -3)(x + 4)**x**^{2}+ x – 12 = (x – 3)(x + 4)

Note that the solutions of this equation are x = 3 and x = -4. So, the complete factorization of this quadratic is:

**x**^{2}+ x – 12 = (x – 3)(x + 4)

### Factor A Quadratic In The General Case (a Not Equal To 1)

This case is very similar to the previous one, except that we must divide through by a to get a monic quadratic (one with a leading coefficient of 1).

Let’s look at an example.

#### Example: Factor A Quadratic In The General Case (a Not Equal to 1)

Consider the following quadratic equation:

**5x**^{2}+ 20x + 15 = 0

Since all of the coefficients are multiples of 5, we begin by dividing both sides by 5 to get:

**x**^{2}+ 4x + 3 = 0

Since a = 1, b = 4, and c = 3, we can try to factor by finding values of c and d such that:

**4 = d + e****3 = de**

The factorizations of 3 are 1*3 and (-1)*(-3). Since we want d + e to be positive, we choose d = 1 and e = 3 (positive values).

This gives us d + e = 1 + 3 = 4.

So, we can factor by the formula

**x**^{2}+ (d + e)x + de = (x + d)(x + e)

With d = 1 and e = 3, this gives us

**x**^{2}+ (1 + 3)x + (1)(3) = (x + 1)(x + 3)**x**^{2}+ 4x + 3 = (x + 1)(x + 3)

Note that the solutions of this equation are x = -1 and x = -3. Since we divided by 5 at the beginning, we must multiply by 5 now.

So, the complete factorization of this quadratic is:

**5(x**^{2}+ 4x + 3) = 5(x + 1)(x + 3)**5x**^{2}+ 20x + 15 = 5(x + 1)(x + 3)

## Conclusion

Now you know about the methods you can use to factor a quadratic in several different cases. You also have some examples to help you see how these methods work.

You might also want to check out my article on how to solve a quadratic equation by completing the square.

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

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

~Jonathon