SOLUTION: Factor completely 4x^3 + 4x^2

SOLUTION: Factor completely 4x^3 + 4x^2 - 80x

Question 173192This question is from textbook Introductory Algebra
: Factor completely
4x^3 + 4x^2 - 80x This question is from textbook Introductory Algebra

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Start with the given expression

Factor out the GCF

Now let's focus on the inner expression

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Looking at we can see that the first term is and the last term is where the coefficients are 1 and -20 respectively.

Now multiply the first coefficient 1 and the last coefficient -20 to get -20. Now what two numbers multiply to -20 and add to the middle coefficient 1? Let's list all of the factors of -20:

Factors of -20:
1,2,4,5,10,20

-1,-2,-4,-5,-10,-20 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -20
(1)*(-20)
(2)*(-10)
(4)*(-5)
(-1)*(20)
(-2)*(10)
(-4)*(5)

note: remember, the product of a negative and a positive number is a negative number

Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1

First Number	Second Number	Sum
1	-20	1+(-20)=-19
2	-10	2+(-10)=-8
4	-5	4+(-5)=-1
-1	20	-1+20=19
-2	10	-2+10=8
-4	5	-4+5=1

From this list we can see that -4 and 5 add up to 1 and multiply to -20

Now looking at the expression , replace with (notice adds up to . So it is equivalent to )

Now let's factor by grouping:

Group like terms

Factor out the GCF of out of the first group. Factor out the GCF of out of the second group

Since we have a common term of , we can combine like terms

So factors to

So this also means that factors to (since is equivalent to )

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So our expression goes from and factors further to

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Answer:

So factors to

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