The area of a square is \begin{align*}9x^2 + 24x + 16\end{align*} . What are the dimensions of the square?
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James Sousa: Ex: Factor Trinomials When A is NOT Equal to 1 - Grouping Method
Guidance
When we add a number in front of the @$\begin{align*}x^2\end{align*}@$ 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 @$\begin{align*}x-\end{align*}@$ terms are not 1.
Example A
Multiply @$\begin{align*}(3x-5)(2x + 1)\end{align*}@$
Solution: 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.
Example B
Factor @$\begin{align*}6x^2-x-2\end{align*}@$ .
Solution: 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 from the investigation in the previous concept; 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 (Step 3 from the investigation in the previous concept).
@$$\begin{align*}& \quad \ 6x^2 {\color{red}-x}-2\\ & \qquad \ \ {\color{red} \swarrow}{\color{red} \searrow}\\ & 6x^2{\color{red}-4x+3x}-2\end{align*}@$$
Next, group the first two terms together and the last two terms together and pull out any common factors.
@$$\begin{align*}& (6x^2-4x)+(3x-2)\\ & 2x(3x-2)+1(3x-2)\end{align*}@$$
Just like in the investigation, 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.
Example C
Factor @$\begin{align*}4x^2+8x-5\end{align*}@$ .
Solution: Let’s make the steps from Example B a little more concise.
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.
2. Rewrite the trinomial with the @$\begin{align*}x-\end{align*}@$ term expanded, using the two factors from Step 1.
@$$\begin{align*}& \quad \ 4x^2+{\color{red}8x}-5\\ & \qquad \quad \ {\color{red} \swarrow}{\color{red} \searrow}\\ & 4x^2+{\color{red}10x-2x}-5\end{align*}@$$
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.
Example D
Factor @$\begin{align*}12x^2-22x-20\end{align*}@$ .
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.
@$$\begin{align*}12x^2-22x-20 = 2(6x^2-11x-10)\end{align*}@$$
This will make it much easier for you to factor what is inside the parenthesis.
2. Using what is inside the parenthesis, find @$\begin{align*}ac\end{align*}@$ 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.
3. Rewrite the trinomial with the @$\begin{align*}x-\end{align*}@$ term expanded, using the two factors from Step 2.
@$$\begin{align*}& 2(6x^2{\color{red}-11x}-10)\\ & 2(6x^2 {\color{red}-15x+4x}-10)\end{align*}@$$
4. Group the first two and second two terms together, find the GCF and factor again.
@$$\begin{align*}& 2(6x^2-15x+4x-10)\\ & 2 \left[(6x^2-15x)+(4x-10)\right]\\ & 2 \left[ 3x(2x-5)+2(2x-5)\right]\\ & 2(2x-5)(3x+2)\end{align*}@$$
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. @$\begin{align*}4x^2+4x-48\end{align*}@$
Answers
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?
@$$\begin{align*}& 4(x^2+x-12)\\ & 4(x+4)(x-3)\end{align*}@$$
Explore More
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 @$\begin{align*}c\end{align*}@$ ?