Factoring Quadratics: Tricks of the Trade

You’re probably getting more familiar with quadratic expressions and equations. They’re those algebraic expressions with an x^{2} that seem to pop up everywhere in your math studies. Right? I’m Drew Laiche, from Portland Math Tutor, LLC, and I’m here to help you unravel the mystery of factoring quadratics. Factoring is a crucial step in solving these and many other polynomials. With some practice, you’ll be cracking them open like a pro.

We will discuss several techniques for factoring quadratic expressions. However, before we get into the details of each technique, we need to get our expressions set up in a specific way. That is, before proceeding further we should write our expression in what’s call Standard Form.

Standard Form

Writing our quadratic expression in Standard Form makes it possible to factor or, alternatively, to use the quadratic formula (click here to learn how to use the quadratic formula, How To Solve Quadratics Using The Quadratic Formula).

Standard form of a quadratic expression is given as;

\large{ ax^2 + bx + c }

a, b, and c are constants, and a \not= 0.

1. 1. 1. 1. ax^2 is the quadratic term.
         2. bx is the linear term.
         3. c is the constant term.

Greatest Common Factor (GCF)

Before proceeding to factor a quadratic expression using any of the techniques mentioned below, it is crucial to first factor out the Greatest Common Factor (GCF) of all terms in the quadratic expression, if one exists. The GCF is the largest numeric or algebraic monomial factor of each term of the expression.

Example; 9x^2+18x+27

\large{ 9x^2+18x-27 = 9(x^2+2x-3) }

Example; 4x^3-20x^2+24x

\large{ 4x^3-20x^2+24x = 4x(x^2-5x+6) }

By factoring out the GCF first, we simplify the quadratic expression and can potentially make subsequent factoring steps easier or even possible. For instance, by factoring out the GCF, we may end up with a coefficient of one for the quadratic term, (x^2), making the quadratic easier to factor.

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Some Factoring Methods

Now we will explore a few factoring strategies that will be useful depending on the particular quadratic expression you are working with. *Note that I use phrases to refer to these factoring techniques that may or may not coincide with the phrases that your teacher uses. 

1. Your Favorite Shortcut!; a = 1

2. The Box Method; a \not = 1

3. Factor by Grouping;

4. Difference of Squares; a^2-b^2

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1. Factoring When a=1

Factoring quadratic expressions when the coefficient of the squared term (the ‘a’ value; known as the leading coefficient) is equal to one is a common occurrence. This comes up so often that it is good to have a shortcut method to factor this special case. Here are the steps you would follow:

\large{ x^2+bx+c }

1. Identify the b and c values: In our simplified quadratic expression, b is the coefficient of x and c is the constant.
   
   For example, let’s consider the expression x^2 + 7x + 10. Here, b = 7 and c = 10.
   
   
2. Find two numbers that multiply to c and add to b: You’re looking for two numbers that multiply together to give you c (the constant term) and add together to give you b (the coefficient of x).
   
   In our example, the numbers 5 and 2 both add up to b=7 and multiply to c=10.
   
   
3. Write the factored expression: Once you have found these two numbers, you can write the factored expression as (x+m)(x+n), where m and n are the numbers you found.
   
   In our example, the factored form of the expression is (x + 5)(x + 2).

At this point, I think that it would be a good practice to expand the two binomial factors to verify that the product results in the original quadratic expression that we started with. This is good to make sure we have the correct solution but also to make sure we understand how this technique works. This isn’t magic, and you should understand why you are doing what you are doing.

 \begin{aligned} \text{Check; }(x+5)(x+2)&=x^{2}+2x+5x+10 \\
&=x^{2}+7x+10 \end{aligned} 

Well, that’s the process of factoring quadratic expressions when a equals one. It’s a straightforward technique that can be applied quickly once you get the hang of it!

Example:  x^2 + 5x + 6

We’re looking for two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3. So our factored equation is (x + 2)(x + 3).

Example: x^2 - 7x + 10

We’re looking for two numbers that multiply to 10 and add to -7. Those numbers are -2 and -5. So, our factored equation is (x - 2)(x - 5).

2. Factoring When a \not =  1 (The Box Method)

For many students the Box Method is a godsend when a \ne 1. It’s also kinda fun. It’s like a Sudoku puzzle for quadratic equations! Full disclosure, you can use this method with any quadratic expression, but it is more cumbersome so it usually isn’t our first choice if we have other options.

1. 1. ax^2+bx+c
      
   2. Draw a box divided in half vertically and horizontally such that we have four quadrants.
      
   3. Put the quadratic term ax^2 into the top left quadrant and put the constant term, c into the bottom right quadrant.
      
   4. Now, find the product of the leading term and the trailing term; ax^2 \cdot c.
      
   5. Find the two factors of of this product, ax^2 \cdot c, that add to bx
      
   6. Put these two factors, one each, in the top right and bottom left quadrants.
      
   7. Now find the greatest common factor of the top row and write that outside the box to the left of the top row.
      
   8. After you factor out the GCF for the top row write the terms that would remain above the respective columns.
      
   9. This will leave one space, outside the box, to the left of the bottom row. Here you will write a term such that multiplying it times the term above the first column will result in the term in the bottom left quadrant. Likewise, the last term that you wrote outside and to the left of the bottom row if multiplied by the term above the second column will result in the term in the bottom right quadrant.
      
   10. Congratulations! You have factored the quadratic expression. Your factors will be the binomial consisting of the the two terms above the box and the other binomial will be the two terms to the left of the box.
       

Example 2x^2 + 7x + 6

We create a box, divide it into four sections. In the top left, we put our 2x^2 term, and the 3 term goes in the bottom right. We then find factors of 2x^2 \cdot 6 that add to 7x. That would be 4x \text{ and }3x. These go in the remaining squares of the box. Factor the GCF out of the top row and proceed to right the remaining factors above the two columns and to the left of the bottom row.

You should end up with;

 \large(2x+3)(x+2)

No try it on your own without looking back at these notes. Did you get the correct factorization? If not, don’t sweat it. Keep practicing.  You’ll get there.

1)

2)

3)

4)

5)

6)

Example 3x^2 - 10x - 8

We create a box, divide it into four sections. In the top left, we put our 3x^2 term, and the -8 term goes in the bottom right. We then find factors of 3x^2 \cdot -8 = -24x^2 that add to -10x. That would be -12x \text{ and } 2x. These go in the remaining squares of the box. Factor the GCF out of the top row and proceed to right the remaining factors above the two columns and to the left of the bottom row.

You should end up with:

\large(3x+2)(x-4)

3. Factoring by Grouping

Factoring by grouping is an effective technique when we have four terms in an expression or when quadratic expressions don’t lend themselves to easy factoring. Let’s look at the step-by-step process of factoring a quadratic expression using this method.

In a nutshell:  Start in Standard Form.  A quadratic polynomial expression, in one variable, can be simplified into the form of ax^2+bx+c. The first step in factoring by grouping is to rearrange this equation into ax^2+mx+nx+c such that m and n are two numbers such that mn = ac and m+n=b. This essentially breaks down the middle term bx into two terms such that bx = mx + nx.

Example:  6x^2+5x-4

1. Here, a = 6, b = 5, and c = -4.

2. We need two numbers that multiply to a \cdot c = -24 and add to b = 5. These numbers are \textcolor{red}{8} and \textcolor{blue}{-3}, so we rewrite the bx term as 5x=\textcolor{red}{8}x\textcolor{blue}{-3}x or 5x=\textcolor{blue}{-3}x\textcolor{red}{+8}x

\Large{ 6x^{2}+(\textcolor{red}{8}x\textcolor{blue}{-3}x)-4 }

\Large{ 6x^{2}+(\textcolor{blue}{-3}x\textcolor{red}{+8}x)-4 }

3. Grouping: Now group the terms in pairs. This is done by grouping the first two terms together and the last two terms together.

\Large{ (6x^{2}\textcolor{red}{+8}x)+(\textcolor{blue}{-3}x-4) }

\Large{ (6x^{2}\textcolor{blue}{-3}x)+(\textcolor{red}{8}x-4) }

4. Factor out the GCF from each group: Identify the Greatest Common Factor (GCF) within each group and factor it out. This simplifies the expression.

\Large{ 2x(3x+4)-1(3x+4) }

\Large{ 3x(2x-1)+4(2x-1) }

5. Factor by Grouping: At this point, you should notice that the binomials in parentheses are the same. You can now treat the whole binomial as a common factor and factor it out of the expression.

\Large{ (3x+4)(2x-1) }

\Large{ (2x-1)(3x+4) }

And there you have it! This is the factored form of the quadratic expression using the grouping method. This method can be useful for quadratic expressions that can’t be factored easily by simple inspection. Please note that it doesn’t matter what order you insert the terms to replace bx; you will end up with equivalent factored expressions regardless.

I hope that you can see how factoring by grouping and the box method outlined above are essentially the same process simply wrapped in different packaging. If the connection between the two methods is not clear at first, I encourage you to challenge yourself by spending a little bit of time thinking on the similarities and commonalities between the two methods.

4. Difference of Squares

The difference of squares pattern uses the concept of conjugate pairs. Conjugates are a pair of binomials that are the same, except for the operation that separates the two terms. For example, (a - b) and (a + b) are conjugates. When a pair of conjugates are multiplied, the result is always the difference of the squares of each term. Therefore, we can write a^2 - b^2 = (a - b)(a + b).

Let’s prove that this forms a mathematical identity (an equation that is true regardless of the values substituted for the variables). To prove this to ourselves, we can start with the product of the two binomials and use the distributive property to expand.

\large{ \begin{aligned} (a - b)(a + b) &= a \cdot a + a \cdot b - b \cdot a - b \cdot b \\
(a - b)(a + b) &=a^2+ab-ab-b^2 \\
(a - b)(a + b) &=a^2-b^2 
\end{aligned} }

Well, there you have it. We see that the center terms cancel out
because they are additive inverses of one another.

So if we can start with product of binomial conjugates and expand to write the expression as a difference of squares, then we should be able to start with an expression in the form of a difference of squares and then factor it as the product of binomial conjugates.

Example:  x^2 - 25

The expression x^2 - 25 is a difference of squares because x^2 and 25 are both perfect squares and they’re being subtracted.

We know that 25 = 5^2, so;

\large{x^2 - 25 = x^2 - 5^2 }

Recognizing that this fits the pattern a^2 - b^2 where a = x and b = 5. we can use the conjugate pairs (x - 5) and (x + 5) to rewrite x^2 - 25 as (x - 5)(x + 5). And there you have it, we factored the expression:

\large{ x^2 - 5^2 = (a-b)(a+b) }

This is the power of the difference of squares factoring technique. By recognizing the pattern and identifying the conjugate pair, we can quickly and easily factor the expression.

\large(x - 5)(x + 5)

Take x^2 - 25, for instance. This can be factored to (x - 5)(x + 5). It’s like magic, isn’t it?

Example; 9x^2 - 16

\large{ \begin{aligned} 9x^2 &= (3x)^2 \\
16 &= (4)^2 \\
9x^2 - 16 &= (3x)^2 - (4)^2 \\
&= (3x-4)(3x+4) \end{aligned} }

Remember, this only works when the quadratic expression is the difference of two squares.

As you continue to master these techniques, keep in mind that practice makes perfect. So take your time, practice often, and soon enough you’ll be factoring quadratic expressions like a pro!

Remember, every great mathematician once struggled with these concepts, but with perseverance and the right guidance, you’ll soon overcome your mathematical hurdles. As always, feel free to drop by Portland Math Tutor, LLC for any further assistance. Happy factoring!