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# 5.2: Factoring When the Leading Coefficient Doesn't Equal 1

Difficulty Level: Advanced Created by: CK-12
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The area of a square is 9x2+24x+16\begin{align*}9x^2 + 24x + 16\end{align*}. What are the dimensions of the square?

### Guidance

When we add a number in front of the x2\begin{align*}x^2\end{align*} term, it makes factoring a little trickier. We still follow the investigation from the previous section, but cannot use the shortcut. First, let’s try FOIL-ing when the coefficients in front of the x\begin{align*}x-\end{align*}terms are not 1.

#### Example A

Multiply (3x5)(2x+1)\begin{align*}(3x-5)(2x + 1)\end{align*}

Solution: We can still use FOIL.

FIRST 3x2x=6x2\begin{align*}3x \cdot 2x = 6x^2\end{align*}

OUTSIDE 3x1=3x\begin{align*}3x \cdot 1 = 3x\end{align*}

INSIDE 52x=10x\begin{align*}-5 \cdot 2x = -10x\end{align*}

LAST 51=5\begin{align*}-5 \cdot 1 = -5\end{align*}

Combining all the terms together, we get: 6x2+3x10x5=6x27x5\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.

#### Example B

Factor 6x2x2\begin{align*}6x^2-x-2\end{align*}.

Solution: This is a factorable trinomial. When there is a coefficient, or number in front of x2\begin{align*}x^2\end{align*}, you must follow all the steps from the investigation in the previous concept; no shortcuts. Also, m\begin{align*}m\end{align*} and n\begin{align*}n\end{align*} no longer have a product of c\begin{align*}c\end{align*} and a sum of b\begin{align*}b\end{align*}. This would not take the coefficient of x2\begin{align*}x^2\end{align*} into account. What we need to do is multiply together a\begin{align*}a\end{align*} and c\begin{align*}c\end{align*} (from ax2+bx+c\begin{align*}ax^2+bx+c\end{align*}) and then find the two numbers whose product is ac\begin{align*}ac\end{align*} and sum is b\begin{align*}b\end{align*}. Let’s use the X\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
1,12\begin{align*}-1, 12\end{align*} 11
1, -12 -11
2, -6 -4
-2, 6 4
3,4\begin{align*}{\color{red}3, -4}\end{align*} 1\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 x\begin{align*}x-\end{align*}term expanded using 3 and -4 (Step 3 from the investigation in the previous concept).

6x2x2  6x24x+3x2\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.

(6x24x)+(3x2)2x(3x2)+1(3x2)\begin{align*}& (6x^2-4x)+(3x-2)\\ & 2x(3x-2)+1(3x-2)\end{align*}

Just like in the investigation, what is in the parenthesis is the same. We now have two terms that both have (3x2)\begin{align*}(3x-2)\end{align*} as factor. Pull this factor out.

The factors of 6x2x2\begin{align*}6x^2-x-2\end{align*} are (3x2)(2x+1)\begin{align*}(3x-2)(2x + 1)\end{align*}. You can FOIL these to check your answer.

#### Example C

Factor 4x2+8x5\begin{align*}4x^2+8x-5\end{align*}.

Solution: Let’s make the steps from Example B a little more concise.

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

45=20\begin{align*}4 \cdot -5=-20\end{align*} The factors of -20 that add up to 8 are 10 and -2.

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

4x2+8x5 4x2+10x2x5\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*}

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

(4x2+10x)+(2x5)2x(2x+5)1(2x+5)(2x+5)(2x1)\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 2x\begin{align*}-2x\end{align*} before 10x\begin{align*}10x\end{align*} in Step 2?

4x22x+10x5(4x22x)(10x5)2x(2x1)+5(2x1)(2x1)(2x+5)\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 x\begin{align*}x-\end{align*}term we list first in Step 2 above.

#### Example D

Factor 12x222x20\begin{align*}12x^2-22x-20\end{align*}.

Solution: Let’s use the steps from Example C, but we are going to add an additional step at the beginning.

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

12x222x20=2(6x211x10)\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.

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

610=60 154=60, 15+4=11\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.

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

2(6x211x10)2(6x215x+4x10)\begin{align*}& 2(6x^2{\color{red}-11x}-10)\\ & 2(6x^2 {\color{red}-15x+4x}-10)\end{align*}

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

2(6x215x+4x10)2[(6x215x)+(4x10)]2[3x(2x5)+2(2x5)]2(2x5)(3x+2)\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*}

Intro Problem Revisit 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.

### Guided Practice

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

Factor the following quadratics, if possible.

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

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

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

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

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

2. 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*}

3. \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.

4. \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*}

5. \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 from the previous concept. 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*}

### Practice

Multiply the following expressions.

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

1. \begin{align*}5x^2+18x+9\end{align*}
2. \begin{align*}6x^2-21x\end{align*}
3. \begin{align*}10x^2-x-3\end{align*}
4. \begin{align*}3x^2+2x-8\end{align*}
5. \begin{align*}4x^2+8x+3\end{align*}
6. \begin{align*}12x^2-12x-18\end{align*}
7. \begin{align*}16x^2-6x-1\end{align*}
8. \begin{align*}5x^2-35x+60\end{align*}
9. \begin{align*}2x^2+7x+3\end{align*}
10. \begin{align*}3x^2+3x+27\end{align*}
11. \begin{align*}8x^2-14x-4\end{align*}
12. \begin{align*}10x^2+27x-9\end{align*}
13. \begin{align*}4x^2+12x+9\end{align*}
14. \begin{align*}15x^2+35x\end{align*}
15. \begin{align*}6x^2-19x+15\end{align*}
16. Factor \begin{align*}x^2-25\end{align*}. What is \begin{align*}b\end{align*}?
17. 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*}?

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Color Highlighted Text Notes

### Vocabulary Language: English

Greatest Common Factor

The greatest common factor of two numbers is the greatest number that both of the original numbers can be divided by evenly.

linear factors

Linear factors are expressions of the form $ax+b$ where $a$ and $b$ are real numbers.

Polynomial

A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

A quadratic polynomial is a polynomial of the 2nd degree, in other words, a polynomial with an $x^2$ term.

Trinomial

A trinomial is a mathematical expression with three terms.

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