document.write( "Question 719352: How do you factor 4m^2 +14m +1 \n" ); document.write( "

Algebra.Com's Answer #441324 by MathLover1(20849) $\"\"$ $\"About$

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

Looking at the expression $\"4x%5E2%2B14x%2B1\"$ , we can see that the first coefficient is $\"4\"$ , the second coefficient is $\"14\"$ , 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 $\"14\"$ ?

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 $\"14\"$ :

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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 there are no pairs of numbers which add to $\"14\"$ . So $\"4x%5E2%2B14x%2B1\"$ cannot be factored.

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Answer:

So $\"4%2Ax%5E2%2B14%2Ax%2B1\"$ doesn't factor at all (over the rational numbers).

So $\"4%2Ax%5E2%2B14%2Ax%2B1\"$ is prime.

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