The area of a square is \begin{align*}9x^2 + 24x + 16\end{align*}

### Factoring Quadratic Functions

When we add a number in front of the \begin{align*}x^2\end{align*}**cannot** use the shortcut. First, let’s try FOIL-ing when the coefficients in front of the \begin{align*}x-\end{align*}

Let's multiply \begin{align*}(3x-5)(2x + 1)\end{align*}

We can still use FOIL.

FIRST \begin{align*}3x \cdot 2x = 6x^2\end{align*}

OUTSIDE \begin{align*}3x \cdot 1 = 3x\end{align*}

INSIDE \begin{align*}-5 \cdot 2x = -10x\end{align*}

LAST \begin{align*}-5 \cdot 1 = -5\end{align*}

Combining all the terms together, we get: \begin{align*}6x^2+3x-10x-5 = 6x^2-7x-5\end{align*}

Now, let’s work backwards and factor a trinomial to get two factors. Remember, you can always check your work by multiplying the final factors together.

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

This is a factorable trinomial. When there is a coefficient, or number in front of \begin{align*}x^2\end{align*}, you must follow all the steps you are familiar with to factor; no shortcuts. Also, \begin{align*}m\end{align*} and \begin{align*}n\end{align*} no longer have a product of \begin{align*}c\end{align*} and a sum of \begin{align*}b\end{align*}. This would not take the coefficient of \begin{align*}x^2\end{align*} into account. What we need to do is multiply together \begin{align*}a\end{align*} and \begin{align*}c\end{align*} (from \begin{align*}ax^2+bx+c\end{align*}) and then find the two numbers whose product is \begin{align*}ac\end{align*} and sum is \begin{align*}b\end{align*}. Let’s use the \begin{align*}X\end{align*} to help us organize this.

Now, we can see, we need the two factors of -12 that also add up to -1.

Factors |
Sum |
---|---|

\begin{align*}-1, 12\end{align*} | 11 |

1, -12 | -11 |

2, -6 | -4 |

-2, 6 | 4 |

\begin{align*}{\color{red}3, -4}\end{align*} | \begin{align*}{\color{red}-1}\end{align*} |

-3, 4 | 1 |

The factors that work are 3 and -4. Now, take these factors and rewrite the \begin{align*}x-\end{align*}term expanded using 3 and -4.

\begin{align*}& \quad \ 6x^2 {\color{red}-x}-2\\ & \qquad \ \ {\color{red} \swarrow}{\color{red} \searrow}\\ & 6x^2{\color{red}-4x+3x}-2\end{align*}

Next, group the first two terms together and the last two terms together and pull out any common factors.

\begin{align*}& (6x^2-4x)+(3x-2)\\ & 2x(3x-2)+1(3x-2)\end{align*}

What is in the parenthesis is *the same*. We now have two terms that both have \begin{align*}(3x-2)\end{align*} as factor. Pull this factor out.

The factors of \begin{align*}6x^2-x-2\end{align*} are \begin{align*}(3x-2)(2x + 1)\end{align*}. You can FOIL these to check your answer.

Now, let's factor \begin{align*}4x^2+8x-5\end{align*}.

Let’s make the steps we followed in the previous problem a little more concise.

Step 1: Find \begin{align*}ac\end{align*} and the factors of this number that add up to \begin{align*}b\end{align*}.

\begin{align*}4 \cdot -5=-20\end{align*} The factors of -20 that add up to 8 are 10 and -2.

Step 2: Rewrite the trinomial with the \begin{align*}x-\end{align*}term expanded, using the two factors from Step 1.

\begin{align*}& \quad \ 4x^2+{\color{red}8x}-5\\ & \qquad \quad \ {\color{red} \swarrow}{\color{red} \searrow}\\ & 4x^2+{\color{red}10x-2x}-5\end{align*}

Step 3: Group the first two and second two terms together, find the GCF and factor again.

\begin{align*}& (4x^2+10x)+(-2x-5)\\ & 2x(2x+5)-1(2x+5)\\ & (2x+5)(2x-1)\end{align*}

Alternate Method: What happens if we list \begin{align*}-2x\end{align*} before \begin{align*}10x\end{align*} in Step 2?

\begin{align*}& 4x^2-2x+10x-5\\ & (4x^2-2x)(10x-5)\\ & 2x(2x-1)+5(2x-1)\\ & (2x-1)(2x+5)\end{align*}

This tells us it does not matter which \begin{align*}x-\end{align*}term we list first in Step 2 above.

Let's factor \begin{align*}12x^2-22x-20\end{align*}.

Let’s use the steps from the previous problem, but we are going to add an additional step at the beginning.

Step 1: Look for any common factors. Pull out the GCF of all three terms, if there is one.

\begin{align*}12x^2-22x-20 = 2(6x^2-11x-10)\end{align*}

This will make it much easier for you to factor what is inside the parenthesis.

Step 2: Using what is inside the parenthesis, find \begin{align*}ac\end{align*} and determine the factors that add up to \begin{align*}b\end{align*}.

\begin{align*}6 \cdot -10 = -60 \ {\color{red}\rightarrow -15 \cdot 4 = -60}, \ -15+4=-11\end{align*}

The factors of -60 that add up to -11 are -15 and 4.

Step 3: Rewrite the trinomial with the \begin{align*}x-\end{align*}term expanded, using the two factors from Step 2.

\begin{align*}& 2(6x^2{\color{red}-11x}-10)\\ & 2(6x^2 {\color{red}-15x+4x}-10)\end{align*}

Step 4: Group the first two and second two terms together, find the GCF and factor again.

\begin{align*}& 2(6x^2-15x+4x-10)\\ & 2 \left[(6x^2-15x)+(4x-10)\right]\\ & 2 \left[ 3x(2x-5)+2(2x-5)\right]\\ & 2(2x-5)(3x+2)\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the dimensions of the square.

The dimensions of a square are its length and its width, so we need to factor the area \begin{align*}9x^2 + 24x + 16\end{align*}.

We need to multiply together \begin{align*}a\end{align*} and \begin{align*}c\end{align*} (from \begin{align*}ax^2+bx+c\end{align*}) and then find the two numbers whose product is \begin{align*}ac\end{align*} and whose sum is \begin{align*}b\end{align*}.

Now we can see that we need the two factors of 144 that also add up to 24. Testing the possibilities, we find that \begin{align*}12 \cdot 12 = 144\end{align*} and \begin{align*}12 + 12 = 24\end{align*}.

Now, take these factors and rewrite the \begin{align*}x-\end{align*}term expanded using 12 and 12.

\begin{align*}& \quad \ 9x^2 {\color{red}+24x}+16\\ & \qquad \ \ {\color{red} \swarrow}{\color{red} \searrow}\\ & 9x^2{\color{red}+12x+12x}+16\end{align*}

Next, group the first two terms together and the last two terms together and pull out any common factors.

\begin{align*}& (9x^2+12x)+(12x+16)\\ & 3x(3x+4)+4(3x+4)\end{align*}

We now have two terms that both have \begin{align*}(3x+4)\end{align*} as factor. Pull this factor out.

The factors of \begin{align*}9x^2+24x+16\end{align*} are \begin{align*}(3x + 4)(3x + 4)\end{align*}, which are also the dimensions of the square.

#### Example 2

Multiply \begin{align*}(4x-3)(3x + 5)\end{align*}.

FOIL: \begin{align*}(4x-3)(3x + 5) = 12x^2 +20x-9x-15 = 12x^2+11x-15\end{align*}

**Factor the following quadratics, if possible.**

#### Example 3

#### \begin{align*}15x^2-4x-3\end{align*}

Use the steps from the examples above. There is no GCF, so we can find the factors of \begin{align*}ac\end{align*} that add up to \begin{align*}b\end{align*}.

\begin{align*}15 \cdot -3 = -45 \end{align*} The factors of -45 that add up to -4 are -9 and 5.

\begin{align*}& 15x^2 {\color{red}-4x}-3\\ & (15x^2 {\color{red}-9x)}+({\color{red}5x}-3)\\ & 3x(5x-3)+1(5x-3)\\ & (5x-3)(3x+1)\end{align*}

#### Example 4

\begin{align*}3x^2+6x-12\end{align*}

\begin{align*}3x^2+6x-12\end{align*} has a GCF of 3. Pulling this out, we have \begin{align*}3(x^2+2x-6)\end{align*}. There is no number in front of \begin{align*}x^2\end{align*}, so we see if there are any factors of -6 that add up to 2. There are not, so this trinomial is not factorable.

#### Example 5

\begin{align*}24x^2-30x-9\end{align*}

\begin{align*}24x^2-30x-9\end{align*} also has a GCF of 3. Pulling this out, we have \begin{align*}3(8x^2-10x-3)\end{align*}. \begin{align*}ac = -24\end{align*}. The factors of -24 than add up to -10 are -12 and 2.

\begin{align*}& 3(8x^2{\color{red}-10x}-3)\\ & 3 \left[(8x^2{\color{red}-12x})+({\color{red}2x}-3)\right]\\ & 3 \left[ 4x(2x-3)+1(2x-3)\right]\\ & 3(2x-3)(4x+1)\end{align*}

#### Example 6

\begin{align*}4x^2+4x-48\end{align*}

\begin{align*}4x^2+4x-48\end{align*}. has a GCF of 4. Pulling this out, we have \begin{align*}4(x^2+x-12)\end{align*}. This trinomial does not have a number in front of \begin{align*}x^2\end{align*}, so we can use the shortcut we are familiar with. What are the factors of -12 that add up to 1?

\begin{align*}& 4(x^2+x-12)\\ & 4(x+4)(x-3)\end{align*}

### Review

Multiply the following expressions.

- \begin{align*}(2x-1)(x + 5)\end{align*}
- \begin{align*}(3x + 2)(2x-3)\end{align*}
- \begin{align*}(4x + 1)(4x-1)\end{align*}

Factor the following quadratic equations, if possible. If they cannot be factored, write *not factorable*. Don’t forget to look for any GCFs first.

- \begin{align*}5x^2+18x+9\end{align*}
- \begin{align*}6x^2-21x\end{align*}
- \begin{align*}10x^2-x-3\end{align*}
- \begin{align*}3x^2+2x-8\end{align*}
- \begin{align*}4x^2+8x+3\end{align*}
- \begin{align*}12x^2-12x-18\end{align*}
- \begin{align*}16x^2-6x-1\end{align*}
- \begin{align*}5x^2-35x+60\end{align*}
- \begin{align*}2x^2+7x+3\end{align*}
- \begin{align*}3x^2+3x+27\end{align*}
- \begin{align*}8x^2-14x-4\end{align*}
- \begin{align*}10x^2+27x-9\end{align*}
- \begin{align*}4x^2+12x+9\end{align*}
- \begin{align*}15x^2+35x\end{align*}
- \begin{align*}6x^2-19x+15\end{align*}
- Factor \begin{align*}x^2-25\end{align*}. What is \begin{align*}b\end{align*}?
- Factor \begin{align*}9x^2-16\end{align*}. What is \begin{align*}b\end{align*}? What types of numbers are \begin{align*}a\end{align*} and \begin{align*}c\end{align*}?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.2.