The area of a rectangle is \begin{align*}x^2-3x-28\end{align*}. What are the length and width of the rectangle?

### Watch This

Watch the first few examples in this video, until about 12:40.

Khan Academy: Factoring Quadratic Expressions

### Guidance

A **quadratic equation** has the form \begin{align*}ax^2+bx+c\end{align*}, where \begin{align*}a \ne 0\end{align*} (If \begin{align*}a = 0\end{align*}, then the equation would be linear). For all quadratic equations, the 2 is the largest and only exponent. A quadratic equation can also be called a **trinomial** when all three terms are present.

There are four ways to solve a quadratic equation. The easiest is **factoring**. In this concept, we are going to focus on factoring when \begin{align*}a = 1\end{align*} or when there is no number in front of \begin{align*}x^2\end{align*}. First, let’s start with a review of multiplying two factors together.

#### Example A

Multiply \begin{align*}(x+4)(x-5)\end{align*}.

**Solution:** Even though this is not a quadratic, the product of the two **factors** will be. Remember from previous math classes that a factor is a number that goes evenly into a larger number. For example, 4 and 5 are factors of 20. So, to determine the larger number that \begin{align*}(x + 4)\end{align*} and \begin{align*}(x-5)\end{align*} go into, we need to multiply them together. One method for multiplying two polynomial factors together is called FOIL. To do this, you need to multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and the LAST terms together and then combine like terms.

Therefore \begin{align*}(x+4)(x-5)=x^2-x-20\end{align*}. We can also say that \begin{align*}(x+4)\end{align*} and \begin{align*}(x-5)\end{align*} are factors of \begin{align*}x^2-x-20\end{align*}.

### More Guidance

Now, we will “undo” the multiplication of two factors by factoring. In this concept, we will only address quadratic equations in the form \begin{align*}x^2+bx+c\end{align*}, or when \begin{align*}a = 1\end{align*}.

#### Investigation: Factoring

1. From the previous example, we know that \begin{align*}(x+m)(x+n)=x^2+bx+c\end{align*}.

FOIL \begin{align*}(x+m)(x+n)\end{align*}.

\begin{align*}(x+m)(x+n) \Rightarrow x^2+\underbrace{nx+mx}_{bx}+\underbrace{mn}_{c}\end{align*}

2. This shows us that the **constant** term, or \begin{align*}c\end{align*}, is equal to the product of the constant numbers inside each factor. It also shows us that the **coefficient** in front of \begin{align*}x\end{align*}, or \begin{align*}b\end{align*}, is equal to the sum of these numbers.

3. Group together the first two terms and the last two terms. Find the Greatest Common Factor, or GCF, for each pair.

\begin{align*}& (x^2+nx)+(mx+mn)\\ & x(x+n)+m(x+n)\end{align*}

4. Notice that what is inside both sets of parenthesis in Step 3 is the same. This number, \begin{align*}(x + n)\end{align*}, is the GCF of \begin{align*}x(x + n)\end{align*} and \begin{align*}m(x + n)\end{align*}. You can pull it out in front of the two terms and leave the \begin{align*}x + m\end{align*}.

\begin{align*}& x(x+n)+m(x+n)\\ & (x+n)(x+m)\end{align*}

We have now shown how to go from FOIL-ing to factoring and back. Let’s apply this idea to an example.

#### Example B

Factor \begin{align*}x^2+6x+8\end{align*}.

**Solution:** Let’s use the investigation to help us.

\begin{align*}x^2+6x+8=(x+m)(x+n)\end{align*}

So, from Step 2, \begin{align*}b\end{align*} will be equal to the sum of \begin{align*}m\end{align*} and \begin{align*}n\end{align*} and \begin{align*}c\end{align*} will be equal to their product. Applying this to our problem, \begin{align*}6 = m + n\end{align*} and \begin{align*}8 = mn\end{align*}. To organize this, use an “\begin{align*}X\end{align*}”. Place the sum in the top and the product in the bottom.

The green pair above is the only one that also adds up to 6. Now, move on to Step 3 from our investigation. We need to rewrite the \begin{align*}x-\end{align*}term, or \begin{align*}b\end{align*}, as a sum of \begin{align*}m\end{align*} and \begin{align*}n\end{align*}.

\begin{align*}& \quad \ x^2+{\color{red}6x}+8\\ & \qquad \quad {\color{red} \swarrow}{\color{red} \searrow}\\ & x^2+{\color{red}4x+2x}+8\\ & (x^2+4x)+(2x+8)\\ & x(x+4)+2(x+4)\end{align*}

Moving on to Step 4, we notice that the \begin{align*}(x + 4)\end{align*} term is the same. Pull this out and we are done.

Therefore, the factors of \begin{align*}x^2+6x+8\end{align*} are \begin{align*}(x+4)(x+2)\end{align*}. You can FOIL this to check your answer.

#### Example C

Factor \begin{align*}x^2+12x-28\end{align*}.

**Solution:** We can approach this problem in exactly the same way we did Example B. This time, we will not use the “\begin{align*}X\end{align*}.” What are the factors of -28 that also add up to 12? Let’s list them out to see:

\begin{align*}-4 \cdot 7, 4 \cdot -7, 2 \cdot -14, {\color{red}-2 \cdot 14}, 1 \cdot -28, -1 \cdot 28\end{align*}

The \begin{align*}{\color{red}\mathbf{red}}\end{align*} pair above is the one that works. Notice that we only listed the factors of *negative* 28.

\begin{align*}& \ \ x^2+{\color{red}12x}-28\\ & \qquad \ \ {\color{red} \swarrow}{\color{blue} \searrow}\\ & x^2{\color{red}-2x+14x}-28\\ & (x^2-2x)+(14x-28)\\ & x(x-2)+14(x-2)\\ & (x-2)(x+14)\end{align*}

By now, you might have a couple questions:

- Does it matter which \begin{align*}x-\end{align*}term you put first? NO, order does not matter. In the previous example, we could have put \begin{align*}14x\end{align*} followed by \begin{align*}-2x\end{align*}. We would still end up with the same answer.
- Can I skip the “expanded” part (Steps 3 and 4 in the investigation)? YES and NO. Yes, if \begin{align*}a = 1\end{align*} No, if \begin{align*}a \ne 1\end{align*} (the next concept). If \begin{align*}a = 1\end{align*}, then \begin{align*}x^2+bx+c=(x+m)(x+n)\end{align*} such that \begin{align*}m + n = b\end{align*} and \begin{align*}mn = c\end{align*}. Consider this a shortcut.

#### Example D

Factor \begin{align*}x^2-4x\end{align*}.

**Solution:** This is an example of a quadratic that is not a trinomial because it only has two terms, also called a **binomial**. There is no \begin{align*}c\end{align*}, or constant term. To factor this, we need to look for the GCF. In this case, the largest number that can be taken out of both terms is an \begin{align*}x\end{align*}.

\begin{align*}x^2-4x=x(x-4)\end{align*}

Therefore, the factors are \begin{align*}x\end{align*} and \begin{align*}x-4\end{align*}.

**Intro Problem Revisit** Recall that the area of a rectangle is \begin{align*}A = lw\end{align*}, where *l* is the length and *w* is the width. To find the length and width, we can therefore factor the area \begin{align*}x^2-3x-28\end{align*}.

What are the factors of –28 that add up to –3? Testing the various possibilities, we find that \begin{align*}-7 \cdot 4 = -28\end{align*} and \begin{align*}-7 + 4 = -3\end{align*}.

Therefore, \begin{align*}x^2-3x-28\end{align*} factors to \begin{align*}(x - 7)(x + 4)\end{align*}, and one of these factors is the rectangle's length while the other is its width.

### Guided Practice

1. Multiply \begin{align*}(x-3)(x + 8)\end{align*}.

Factor the following quadratics, if possible.

2. \begin{align*}x^2-9x+20\end{align*}

3. \begin{align*}x^2+7x-30\end{align*}

4. \begin{align*}x^2+x+6\end{align*}

5. \begin{align*}x^2+10x\end{align*}

#### Answers

1. FOIL-ing our factors together, we get:

\begin{align*}(x-3)(x+8)= x^2+8x-3x-24 = x^2+5x-24\end{align*}

2. Using the “\begin{align*}X\end{align*},” we have:

From the shortcut above, \begin{align*}-4 + -5 = -9\end{align*} and \begin{align*}-4 \cdot -5 = 20\end{align*}.

\begin{align*}x^2-9x+20=(x-4)(x-5)\end{align*}

3. Let’s list out all the factors of -30 and their sums. The sums are in red.

\begin{align*}-10 \cdot 3 \ {\color{red}(-7)}, -3 \cdot 10 \ {\color{red}(7)}, -2 \cdot 15 \ {\color{red}(13)}, -15 \cdot 2 \ {\color{red}(-13)}, -1 \cdot 30 \ {\color{red}(29)}, -30 \cdot 1 \ {\color{red}(-29)}\end{align*}

From this, the factors of -30 that add up to 7 are -3 and 10. \begin{align*}x^2+7x-30 = (x-3)(x+10)\end{align*}

4. There are no factors of 6 that add up to 1. If we had -6, then the trinomial would be factorable. But, as is, this is not a factorable trinomial.

5. The only thing we can do here is to take out the GCF. \begin{align*}x^2+10x=x(x+10)\end{align*}

### Explore More

Multiply the following factors together.

- \begin{align*}(x+2)(x-8)\end{align*}
- \begin{align*}(x-9)(x-1)\end{align*}
- \begin{align*}(x+7)(x+3)\end{align*}

Factor the following quadratic equations. If it cannot be factored, write *not factorable*. You can use either method presented in the examples.

- \begin{align*}x^2-x-2\end{align*}
- \begin{align*}x^2+2x-24\end{align*}
- \begin{align*}x^2-6x\end{align*}
- \begin{align*}x^2+6x+9\end{align*}
- \begin{align*}x^2+8x-10\end{align*}
- \begin{align*}x^2-11x+30\end{align*}
- \begin{align*}x^2+13x-30\end{align*}
- \begin{align*}x^2+11x+28\end{align*}
- \begin{align*}x^2-8x+12\end{align*}
- \begin{align*}x^2-7x-44\end{align*}
- \begin{align*}x^2-8x-20\end{align*}
- \begin{align*}x^2+4x+3\end{align*}
- \begin{align*}x^2-5x+36\end{align*}
- \begin{align*}x^2-5x-36\end{align*}
- \begin{align*}x^2+x\end{align*}

**Challenge** Fill in the \begin{align*}X\end{align*}’s below with the correct numbers.