How to Factor Second Degree Polynomials (Quadratic Equations)

Reviewed by Grace Imson, MA

Last Updated: July 6, 2024 Fact Checked

A polynomial contains a variable (x) raised to a power, known as a degree,^{[1]
X
Research source} and several terms and/or constants. To factor a polynomial means to break the expression down into smaller expressions that are multiplied together. These skills are Algebra I and above, and may be difficult to understand if your math skills are not at this level.

Method 1

Method 1 of 7:

Starting Out

Download Article

The standard format for the quadratic equation is:^{[2]
X
Research source}

ax² + bx + c = 0
Start by ordering the terms in your equation from highest to lowest power, just like this standard format. For example, take:

6 + 6x² + 13x = 0
We will re-order this expression so it's easier to work with by simply moving the terms around:

6x² + 13x + 6 = 0
Factoring the polynomial will result in two smaller expressions which can be multiplied to produce the original polynomial:^{[3]
X
Research source}

6x² + 13x + 6 = (2x + 3)(3x + 2)
In this example, (2x +3) and (3x + 2) are factors of the original expression, 6x² + 13x + 6.

Advertisement
Multiply the factors you identified. Then combine like terms and you're done. Start with:^{[4]
X
Research source}

(2x + 3)(3x + 2)
Let's test it, multiplying the terms using FOIL (first - outer - inner - last), obtaining:^{[5]
X
Research source}

6x² + 4x + 9x + 6
From here, we can add 4x and 9x together since they're like terms. We know our factors are correct because we get the equation we started with:

6x² + 13x + 6

Method 2

Method 2 of 7:

Trial and Error

Download Article

If you have a fairly simple polynomial, you might be able to figure out the factors yourself just from sight. For instance, after practice, many mathematicians are able to know that the expression 4x² + 4x + 1 has the factors (2x + 1) and (2x + 1) just from having seen it so much. (This will obviously not be as easy with more complicated polynomials.) For this example, let's use a less common expression:

3x² + 2x - 8

Using the expression format ax² + bx + c = 0, identify the a and c terms and list out what factors they have. For 3x² + 2x - 8, that means:

a = 3 and has one set of factors: 1 * 3
c = -8 and has four sets of factors: -2 * 4, -4 * 2, -8 * 1, and -1 * 8.
You'll be filling in the constants for each expression into the space you've made:

( x )( x )
For the a term in our example, 3x², there is only one possibility for our example:

(3x )(1x )
Let's say we chose 8 and 1. Write it in:

(3x 8)(x 1)
Depending on the signs in the original expression, it's possible to figure out what the signs for the constants should be. Let's call the two constants for our two factors h and k:

If ax² + bx + c then (x + h)(x + k)
If ax² - bx - c or ax² + bx - c then (x - h)(x + k)
If ax² - bx + c then (x - h)(x - k)
For our example, 3x² + 2x - 8, the signs must be:(x - h)(x + k), giving us the two factors:

(3x + 8) and (x - 1)
A quick first test to run is to see if the middle term is at least the correct value. If it is not, you may have chosen the wrong c factors. Let's test our answer:

(3x + 8)(x - 1)
By multiplication, we arrive at:

3x² - 3x + 8x - 8
Simplifying this expression by adding the like terms (-3x) and (8x), we get:

3x² - 3x + 8x - 8 = 3x² + 5x - 8
We know now that we must have identified the wrong factors:

3x² + 5x - 8 ≠ 3x² + 2x - 8
In our example, let's try 2 and 4 instead of 1 and 8:

(3x + 2)(x - 4)
Now our c term is a -8, but our outside/inside product (3x * -4) and (2 * x) is -12x and 2x, which will not combine to make the correct b term of +2x.

-12x + 2x = 10x
10x ≠ 2x
Let's try moving the 2 and 4 around:

(3x + 4)(x - 2)
Now, our c term (4 * 2 = 8) is still okay, but the outside/inside products are -6x and 4x. If we combine them:

-6x + 4x = 2x
2x ≠ -2x We're pretty close to the 2x we were aiming for, but it's the wrong sign.
We're going to stick with the same order, but swap which one has the minus:

(3x - 4)(x + 2)
Now the c term is still okay, and the outside/inside products are now (6x) and (-4x). Since:

6x - 4x = 2x
2x = 2x We can now recognize the positive 2x from the original problem. These must be the correct factors.

Method 3

Method 3 of 7:

Decomposition

Download Article

This method will identify all possible factors of the a and c terms and use them to figure out what the factors should be. If the numbers are very large or other guesswork-type methods seem like they'd take too long, use this method.^{[6]
X
Research source} Let's use the example:

6x² + 13x + 6

In this example, a is 6 and c is also 6.

6 * 6 = 36
We're looking for two numbers that are factors of the a * c product we identified and also add up to the b term (13).

4 * 9 = 36
4 + 9 = 13
Let's use k and h to represent the two numbers we got, 4 and 9:

ax² + kx + hx + c
6x² + 4x + 9x + 6
Organize the equation so that you can factor out the greatest common factor of the first two terms and the last two terms. Both factored groups should be the same. Add the Greatest Common Factors together and enclose them in parentheses next to the factored group; the result will be your two factors:^{[7]
X
Research source}

6x² + 4x + 9x + 6
2x(3x + 2) + 3(3x + 2)
(2x + 3)(3x + 2)

Method 4

Method 4 of 7:

Triple Play

Download Article

Similar to the decomposition method, the 'triple play' method^{[8]
X
Research source} examines possible factors of the product of the a and c terms and uses them to figure out what b must be. Consider for this example the equation:

8x² + 10x + 2

As with the decomposition method, this is going to help us identify candidates for the b term. In this example, a is 8 and c is 2.

8 * 2 = 16
This step is identical to the decomposition method - we're testing and rejecting candidates for the constants. The product of the a and c terms is 16, and the c term is 10:

2 * 8 = 16
8 + 2 = 10
Take our two numbers from the previous step - let's call them h and k - and put them into this expression:

((ax + h)(ax + k))/ a

Here, we'd get:

((8x + 8)(8x + 2)) / 8
In this example, we're seeing if (8x + 8) or (8x + 2) can be divided by 8. (8x + 8) is divisible by 8, so we'll divide this term by a and leave the other one as is.

(8x + 8) = 8(x + 1)
The term we're saving out of here is what's left after dividing by the a term:(x + 1)
In this example, the second term, has a GCF of 2, since 8x + 2 = 2(4x + 1). Combine this answer with the term you identified in the previous step. These are your equation's factors.

2(x + 1)(4x + 1)

Method 5

Method 5 of 7:

Difference of Two Squares

Download Article

Some coefficients in polynomials may be identified as a 'squares', or the product of two numbers. Identifying these squares allows you to factor some polynomials much faster.^{[9]
X
Research source} Consider the equation:

27x² - 12 = 0

In this case, we can see that 27 and 12 are both divisible by 3, so we'll separate that out:

27x² - 12 = 3(9x² - 4)
To use this method you should be able to take the square root of the terms evenly. (Notice that we'll have left out the negative signs - since these numbers are squares they may be products of positive or two negative numbers)

9x² = 3x * 3x and 4 = 2 * 2
We'll take the a and c values from our step above - a = 9 and c = 4, then find their square roots - √a = 3 and √c = 2. These are the coefficients for the factor expressions:

27x² - 12 = 3(9x² - 4) = 3(3x + 2)(3x - 2)

Method 6

Method 6 of 7:

Quadratic Formula

Download Article

If all else fails and the equation will not factor evenly use the quadratic formula.^{[10]
X
Research source} Consider the example:

x² + 4x + 1 = 0

x = -b ± √(b² - 4ac)
---------------------
2a
We obtain the expression:

x = -4 ± √(4² - 4•1•1) / 2
You should get two x values. As shown above, we obtain two answers:

x = -2 + √(3) or x = -2 - √(3)
Plug the x values you obtained into two polynomial expressions as the constants.^{[11]
X
Research source} These will be your factors. If we call our two answers h and k, we're writing two factors like so:

(x - h)(x - k)
In this case, our final answer is:

(x - (-2 + √(3))(x - (-2 - √(3)) = (x + 2 - √(3))(x + 2 + √(3))

Method 7

Method 7 of 7:

Using a Calculator

Download Article

If you are allowed to use one, a graphic calculator makes the factoring process much easier, especially on standardized tests. These instructions are for a TI graphing calculator. We will use the example equation:

y = x² − x − 2

You will use the equation solver, also known as the [Y = ] screen.
Once you've entered your equation, press [GRAPH] - you should see a smooth arc representing your equation (and it will be an arc since we are dealing with polynomials).
3
Locate where the arc intersects the x axis. Since polynomial equations are traditionally written as ax² + bx + c = 0, these are the two x values that cause the expression to equal zero:

(-1, 0), (2, 0)
x = -1, x = 2
- If you cannot identify where your graph crosses the x axis by sight, press [2nd] and then [TRACE]. Press [2] or select "zero". Slide the cursor to the left of an intersect and press [ENTER]. Slide the cursor to the right of an intersect and press [ENTER]. Slide the cursor as close as possible to the intersect and press [ENTER]. The calculator will find the x value. Do this for the other intersect also.
If we term our two x-values h and k, the expression we will be using is:

(x - h)(x - k) = 0
Thus, our two factors must be:

(x - (-1))(x - 2) = (x + 1)(x - 2)

Community Q&A

Search

Add New Question

Question

Is there a difference between ax^2 + bx + c = 0 and ax^2 + bx + c ?

Orangejews

Community Answer

The first is an equation. It is true for 2 choices of x, and false for all others. The second is an expression. You can evaluate it for any number x and see what number it equals, but you can't say "solve it," since without an = sign, there is nothing to solve.

Thanks! We're glad this was helpful.
Thank you for your feedback.
If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow
Yes No

Not Helpful 2 Helpful 8
Question

What are the steps to find the solution set of x+4+9-x=5?

Community Answer

First, do all the calculations concerning like terms. You'll get (x-x)+(4+9)=5. Calculate, and you'll get 13=5, which is impossible. Assuming through equation is x^2+4+9+x=5, we get x^2+1x+8=0. Since we cannot find its factors by the a&c or triple play method, we use the quadratic equation and get the roots as {-1+(1-32)^0.5} and {-1-(1-32)^0.5}. Simplified, [{(-31)^0.5}-1] and [{(-31)^0.5}-1]. Not real numbers.

Thanks! We're glad this was helpful.
Thank you for your feedback.
If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow
Yes No

Not Helpful 1 Helpful 4
Question

Which is the fastest method to find the value of x, when x^2 - 11x - 87 = 0?

Donagan

Top Answerer

Use the quadratic formula (Method 5 above).

Thanks! We're glad this was helpful.
Thank you for your feedback.
If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow
Yes No

Not Helpful 1 Helpful 6

See more answers

Ask a Question

200 characters left

Include your email address to get a message when this question is answered.

Submit

Tips

If you have a TI-84 calculator (graphing) there is a program named SOLVER that will solve a quadratic equation. It will also solve any other degree polynomial.

Thanks
Helpful 0 Not Helpful 0
If a term does not exist the coefficient is 0. It will be helpful to rewrite the equation if this occurs, e.g. x² + 6 = x² + 0x + 6.

Thanks
Helpful 0 Not Helpful 0
If you factored your polynomial using the quadratic formula and got answers with radicals, you may want to convert the x values to fractions in order to check them.

Thanks
Helpful 0 Not Helpful 0

Show More Tips

Submit a Tip

All tip submissions are carefully reviewed before being published

Name

Please provide your name and last initial

Submit

Thanks for submitting a tip for review!

Warnings

If you are learning this concept in a math class, pay attention to what your teacher advises and do not just use your favorite method. Your teacher may ask you to use a specific method on the test or not allow graphing calculators.

Thanks
Helpful 12 Not Helpful 4