SOLUTION: factoring a quadratic with leading coefficient greater than one z^2+12z+20

Question 91429: factoring a quadratic with leading coefficient greater than one
z^2+12z+20
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

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

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

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

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

Factors of :

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20

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

These factors pair up and multiply to .

1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20

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

First Number	Second Number	Sum
1	20	1+20=21
2	10	2+10=12
4	5	4+5=9
-1	-20	-1+(-20)=-21
-2	-10	-2+(-10)=-12
-4	-5	-4+(-5)=-9

From the table, we can see that the two numbers and add to (the middle coefficient).

So the two numbers and both multiply to and add to

Now replace the middle term with . Remember, and add to . So this shows us that .

Replace the second term with .

Group the terms into two pairs.

Factor out the GCF from the first group.

Factor out 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.

Combine like terms. Or factor out the common term

===============================================================

Answer:

So factors to .

In other words, .

Note: you can check the answer by expanding to get or by graphing the original expression and the answer (the two graphs should be identical).

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