document.write( "Question 888536: Solve equation by using factoring without using the quadratic formula.
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Algebra.Com's Answer #537444 by richwmiller(17219) $\"\"$ $\"About$

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12x^2+28x-5=0\r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"12x%5E2%2B28x-5\"$ , we can see that the first coefficient is $\"12\"$ , the second coefficient is $\"28\"$ , and the last term is $\"-5\"$ .

Now multiply the first coefficient $\"12\"$ by the last term $\"-5\"$ to get $\"%2812%29%28-5%29=-60\"$ .

Now the question is: what two whole numbers multiply to $\"-60\"$ (the previous product) and add to the second coefficient $\"28\"$ ?

To find these two numbers, we need to list all of the factors of $\"-60\"$ (the previous product).

Factors of $\"-60\"$ :

1,2,3,4,5,6,10,12,15,20,30,60

-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60

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

These factors pair up and multiply to $\"-60\"$ .

1*(-60) = -60
2*(-30) = -60
3*(-20) = -60
4*(-15) = -60
5*(-12) = -60
6*(-10) = -60
(-1)*(60) = -60
(-2)*(30) = -60
(-3)*(20) = -60
(-4)*(15) = -60
(-5)*(12) = -60
(-6)*(10) = -60

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"28\"$ :

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First Number	Second Number	Sum
1	-60	1+(-60)=-59
2	-30	2+(-30)=-28
3	-20	3+(-20)=-17
4	-15	4+(-15)=-11
5	-12	5+(-12)=-7
6	-10	6+(-10)=-4
-1	60	-1+60=59
-2	30	-2+30=28
-3	20	-3+20=17
-4	15	-4+15=11
-5	12	-5+12=7
-6	10	-6+10=4

From the table, we can see that the two numbers $\"-2\"$ and $\"30\"$ add to $\"28\"$ (the middle coefficient).

So the two numbers $\"-2\"$ and $\"30\"$ both multiply to $\"-60\"$ and add to $\"28\"$

Now replace the middle term $\"28x\"$ with $\"-2x%2B30x\"$ . Remember, $\"-2\"$ and $\"30\"$ add to $\"28\"$ . So this shows us that $\"-2x%2B30x=28x\"$ .

$\"12x%5E2%2Bhighlight%28-2x%2B30x%29-5\"$ Replace the second term $\"28x\"$ with $\"-2x%2B30x\"$ .

$\"%2812x%5E2-2x%29%2B%2830x-5%29\"$ Group the terms into two pairs.

$\"2x%286x-1%29%2B%2830x-5%29\"$ Factor out the GCF $\"2x\"$ from the first group.

$\"2x%286x-1%29%2B5%286x-1%29\"$ Factor out $\"5\"$ 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.

$\"%282x%2B5%29%286x-1%29\"$ Combine like terms. Or factor out the common term $\"6x-1\"$

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Answer:

So $\"12%2Ax%5E2%2B28%2Ax-5\"$ factors to $\"%282x%2B5%29%286x-1%29\"$ .

In other words, $\"12%2Ax%5E2%2B28%2Ax-5=%282x%2B5%29%286x-1%29\"$ .

Note: you can check the answer by expanding $\"%282x%2B5%29%286x-1%29\"$ to get $\"12%2Ax%5E2%2B28%2Ax-5\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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