SOLUTION: Factor each expression: B. 6(×^2-4×+4)^2+(×^2-4×+4)-1 C.(4j-2)^2-(2+4j)^2

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Question 799809: Factor each expression:
B. 6(×^2-4×+4)^2+(×^2-4×+4)-1
C.(4j-2)^2-(2+4j)^2
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
B.
Given:
(1) 6%2A%28x%5E2-4x%2B4%29%5E2%2B%28x%5E2-4x%2B4%29-1
Let
(2) y+=+%28x%5E2-4x%2B4%29 in (1) to get
(3) 6%2Ay%5E2%2By-1 which factors into
(4) %283y-1%29%2A%282y%2B1%29
Now set each factor of (3) equal to zero to get
(5) 3y - 1 = 0 and
(6) 2y + 1 = 0
Now use y of (2) in (5) to get
(7) 3%2A%28x%5E2-4x%2B4%29-1+=+0 or
(8) 3%2Ax%5E2-12x%2B12-1+=+0 or
(9) 3%2Ax%5E2-12x%2B11+=+0
Using the quadratic formula, we can calcculate the two roots of (9) as
(10) x+=+2+%2B-sqrt%283%29%2F2
Now use y of (2) in (6) to get
(11) 2%2A%28x%5E2-4x%2B4%29%2B1+=+0 or
(12) 2%2Ax%5E2-8x%2B8%2B1+=+0 or
(13) 2%2Ax%5E2-8x%2B9+=+0
Using the quadratic formula, we can calcculate the two roots of (9) as
(14) x+=+4+%2B-i%2Asqrt%282%29%2F2
Answer to B:
C.
Given:
(15)%284j-2%29%5E2-%282%2B4j%29%5E2
This is the difference of two perfect squares which factors into the product of the sum and difference of the perfect square or
(16) %28%284j-2%29%2B%282%2B4j%29%29%2A%28%284j-2%29-%282%2B4j%29%29 or
(17) %288j%29%2A%28-4%29 or
(18)-32j
Answer to C: -32j