document.write( "Question 110211: Can someone help me solve this. I have to factor trinomials using the AC method.
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Algebra.Com's Answer #80331 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

$\"20%2Ax%5E2-20%2Ax-15\"$ Start with the given expression.

$\"5%284x%5E2-4x-3%29\"$ Factor out the GCF $\"5\"$ .

Now let's try to factor the inner expression $\"4x%5E2-4x-3\"$

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

Now multiply the first coefficient $\"4\"$ by the last term $\"-3\"$ to get $\"%284%29%28-3%29=-12\"$ .

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 $\"2x-6x\"$ . Remember, $\"2\"$ and $\"-6\"$ add to $\"-4\"$ . So this shows us that $\"2x-6x=-4x\"$ .

$\"4x%5E2%2Bhighlight%282x-6x%29-3\"$ Replace the second term $\"-4x\"$ with $\"2x-6x\"$ .

$\"%284x%5E2%2B2x%29%2B%28-6x-3%29\"$ Group the terms into two pairs.

$\"2x%282x%2B1%29%2B%28-6x-3%29\"$ Factor out the GCF $\"2x\"$ from the first group.

$\"2x%282x%2B1%29-3%282x%2B1%29\"$ Factor out $\"3\"$ 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.

$\"%282x-3%29%282x%2B1%29\"$ Combine like terms. Or factor out the common term $\"2x%2B1\"$

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So $\"5%284x%5E2-4x-3%29\"$ then factors further to $\"5%282x-3%29%282x%2B1%29\"$

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Answer:

So $\"20%2Ax%5E2-20%2Ax-15\"$ completely factors to $\"5%282x-3%29%282x%2B1%29\"$ .

In other words, $\"20%2Ax%5E2-20%2Ax-15=5%282x-3%29%282x%2B1%29\"$ .

Note: you can check the answer by expanding $\"5%282x-3%29%282x%2B1%29\"$ to get $\"20%2Ax%5E2-20%2Ax-15\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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