Question 799809
B.
Given:
(1) {{{6*(x^2-4x+4)^2+(x^2-4x+4)-1}}}
Let
(2) {{{y = (x^2-4x+4)}}} in (1) to get
(3) {{{6*y^2+y-1}}} which factors into
(4) {{{(3y-1)*(2y+1)}}}
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*(x^2-4x+4)-1 = 0}}} or
(8) {{{3*x^2-12x+12-1 = 0}}} or
(9) {{{3*x^2-12x+11 = 0}}}
Using the quadratic formula, we can calcculate the two roots of (9) as
(10) {{{x = 2 +-sqrt(3)/2}}}
Now use y of (2) in (6) to get
(11) {{{2*(x^2-4x+4)+1 = 0}}} or
(12) {{{2*x^2-8x+8+1 = 0}}} or
(13) {{{2*x^2-8x+9 = 0}}}
Using the quadratic formula, we can calcculate the two roots of (9) as
(14) {{{x = 4 +-i*sqrt(2)/2}}}
Answer to B: {{{6*(x^2-4x+4)+(x^2-4x+4)-1 = 6*(x-2+sqrt(3)/2)*(x-2-sqrt(3)/2)*(x-4+i*sqrt(2)/2)*(x-4-i*sqrt(2)/2)}}}

C.
Given:
(15){{{(4j-2)^2-(2+4j)^2}}}
This is the difference of two perfect squares which factors into the product of the sum and difference of the perfect square or
(16) {{{((4j-2)+(2+4j))*((4j-2)-(2+4j))}}} or
(17) {{{(8j)*(-4)}}} or
(18){{{-32j}}}
Answer to C: -32j