document.write( "Question 124267This question is from textbook Precalculus
\n" ); document.write( ": I'm stumped, can you help me answer this question?
\n" ); document.write( "The zeros of the function f(x)=3x^2+4x-4 can be found by factoring as follows: (x+2)(3x-2)
\n" ); document.write( "Is this True or False?
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Algebra.Com's Answer #91054 by MathLover1(20849) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( " $\"True\"$ \r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"3x%5E2%2B4x-4\"$ , we can see that the first coefficient is $\"3\"$ , the second coefficient is $\"4\"$ , and the last term is $\"-4\"$ .

Now multiply the first coefficient $\"3\"$ by the last term $\"-4\"$ to get $\"%283%29%28-4%29=-12\"$ .

Now the question is: what two whole numbers multiply to $\"-12\"$ (the previous product) and add to the second coefficient $\"4\"$ ?

To find these two numbers, we need to list all of the factors of $\"-12\"$ (the previous product).

Factors of $\"-12\"$ :

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to $\"-12\"$ .

1*(-12) = -12
2*(-6) = -12
3*(-4) = -12
(-1)*(12) = -12
(-2)*(6) = -12
(-3)*(4) = -12

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"4\"$ :

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First Number	Second Number	Sum
1	-12	1+(-12)=-11
2	-6	2+(-6)=-4
3	-4	3+(-4)=-1
-1	12	-1+12=11
-2	6	-2+6=4
-3	4	-3+4=1

From the table, we can see that the two numbers $\"-2\"$ and $\"6\"$ add to $\"4\"$ (the middle coefficient).

So the two numbers $\"-2\"$ and $\"6\"$ both multiply to $\"-12\"$ and add to $\"4\"$

Now replace the middle term $\"4x\"$ with $\"-2x%2B6x\"$ . Remember, $\"-2\"$ and $\"6\"$ add to $\"4\"$ . So this shows us that $\"-2x%2B6x=4x\"$ .

$\"3x%5E2%2Bhighlight%28-2x%2B6x%29-4\"$ Replace the second term $\"4x\"$ with $\"-2x%2B6x\"$ .

$\"%283x%5E2-2x%29%2B%286x-4%29\"$ Group the terms into two pairs.

$\"x%283x-2%29%2B%286x-4%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%283x-2%29%2B2%283x-2%29\"$ Factor out $\"2\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28x%2B2%29%283x-2%29\"$ Combine like terms. Or factor out the common term $\"3x-2\"$

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Answer:

So $\"3%2Ax%5E2%2B4%2Ax-4\"$ factors to $\"%28x%2B2%29%283x-2%29\"$ .

In other words, $\"3%2Ax%5E2%2B4%2Ax-4=%28x%2B2%29%283x-2%29\"$ .

Note: you can check the answer by expanding $\"%28x%2B2%29%283x-2%29\"$ to get $\"3%2Ax%5E2%2B4%2Ax-4\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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