SOLUTION: This is factorization of polynomials, I just wanted to ask how this is factorized? Thank you. (4+a)(4-a)-12-3a

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Question 1098071: This is factorization of polynomials, I just wanted to ask how this is factorized? Thank you.
(4+a)(4-a)-12-3a

Found 3 solutions by Theo, ikleyn, MathTherapy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
start with (4+a) * (4-a) - 12 - 3a

(4+a) * (4-a) = 16 - a^2

expression becomes 16 - a^2 - 12 - 3a

rearrange in descending order of degree to get -a^2 - 3a - 12 + 16

combine like terms to get -a^2 - 3a + 4

set this equal to 0 to get -a^2 - 3a + 4 = 0

it's easier to factor if the coefficient of the x^2 term is positive so multiply both sides of the equation by -1 to get:

a^2 + 3a - 4 = 0

factor this quadratic equation to get (a+4) * (a-1) = 0

solve for a to get a = -4 or a = 1.

you are basically simplifying it and combining like terms and you are left with a quadratic equation which you factor in the usual ways.

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
(4+a)(4-a)-12-3a = (4+a)*(4-a) - 3*(4+a) = (4+a)*(4-a-3) = (4+a)*(1-a).

That is all.

This method is called "factoring by grouping".

Answer by MathTherapy(10551)   (Show Source):
You can put this solution on YOUR website!

This is factorization of polynomials, I just wanted to ask how this is factorized? Thank you.
(4+a)(4-a)-12-3a

(4 + a)(4 - a) - 12 - 3a
(4 + a)(4 - a) - 3(4 + a) ----- Factoring - 12 - 3a
(4 - a)(4 + a) - 3(4 + a) ----- Rearranging 1st product
[(4 - a) - 3](4 + a)
(4 - a - 3)(4 + a) =====>