document.write( "Question 474281: Factor Completely\r
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Algebra.Com's Answer #325286 by MathLover1(20850) $\"\"$ $\"About$

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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

$\"16%2Ax%5E2-16%2Ax%2B4\"$ Start with the given expression.

$\"4%284x%5E2-4x%2B1%29\"$ Factor out the GCF $\"4\"$ .

Now let's try to factor the inner expression $\"4x%5E2-4x%2B1\"$

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Looking at the expression $\"4x%5E2-4x%2B1\"$ , we can see that the first coefficient is $\"4\"$ , the second coefficient is $\"-4\"$ , and the last term is $\"1\"$ .

Now multiply the first coefficient $\"4\"$ by the last term $\"1\"$ to get $\"%284%29%281%29=4\"$ .

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

To find these two numbers, we need to list all of the factors of $\"4\"$ (the previous product).

Factors of $\"4\"$ :

1,2,4

-1,-2,-4

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

These factors pair up and multiply to $\"4\"$ .

1*4 = 4
2*2 = 4
(-1)*(-4) = 4
(-2)*(-2) = 4

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	4	1+4=5
2	2	2+2=4
-1	-4	-1+(-4)=-5
-2	-2	-2+(-2)=-4

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

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

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

$\"4x%5E2%2Bhighlight%28-2x-2x%29%2B1\"$ Replace the second term $\"-4x\"$ with $\"-2x-2x\"$ .

$\"%284x%5E2-2x%29%2B%28-2x%2B1%29\"$ Group the terms into two pairs.

$\"2x%282x-1%29%2B%28-2x%2B1%29\"$ Factor out the GCF $\"2x\"$ from the first group.

$\"2x%282x-1%29-1%282x-1%29\"$ Factor out $\"1\"$ 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.

$\"%282x-1%29%282x-1%29\"$ Combine like terms. Or factor out the common term $\"2x-1\"$

$\"%282x-1%29%5E2\"$ Condense the terms.

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So $\"4%284x%5E2-4x%2B1%29\"$ then factors further to $\"4%282x-1%29%5E2\"$

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Answer:

So $\"16%2Ax%5E2-16%2Ax%2B4\"$ completely factors to $\"4%282x-1%29%5E2\"$ .

In other words, $\"16%2Ax%5E2-16%2Ax%2B4=4%282x-1%29%5E2\"$ .

Note: you can check the answer by expanding $\"4%282x-1%29%5E2\"$ to get $\"16%2Ax%5E2-16%2Ax%2B4\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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