SOLUTION: Question: Find all the real and complex zeros of the polynomial x^4+8x^3+8x^2=72x-153 Possible Answers: (a) none of these (b) 4, -4, 3-2i, 3+2i (c) 3, -3, -4-i, -4+i (d) 3,

Algebra ->  College  -> Linear Algebra -> SOLUTION: Question: Find all the real and complex zeros of the polynomial x^4+8x^3+8x^2=72x-153 Possible Answers: (a) none of these (b) 4, -4, 3-2i, 3+2i (c) 3, -3, -4-i, -4+i (d) 3,      Log On


   



Question 30358: Question:
Find all the real and complex zeros of the polynomial x^4+8x^3+8x^2=72x-153
Possible Answers:
(a) none of these
(b) 4, -4, 3-2i, 3+2i
(c) 3, -3, -4-i, -4+i
(d) 3, -3, -4-2i, -4 + 2i
Thanks!

Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
The problem should be
Find all the real and complex zeros of the polynomial x^4+8x^3+8x^2-72x-153
Putting f(x)=x^4+8x^3+8x^2-72x-153 = 0
we observe by inspection that 3 and -3 are roots of f(x) =0
Therefore (x-3) and (x+3) are factors
And (x+3)(x-3) = x^2-9
Dividing x^4+8x^3+8x^2-72x-153 by x^2-9
we get x^2+8x+17 = 0
x^2+8x = -17
The left hand side is of the form a^2+2ab
and we require b^2 to make it a perfect square
Here a = x and b = 4
Adding b^2 = 4^2 = 16 to both the sides
x^2+8x+16 = -17+16
(x+4)^2 = -1
That is (x+4)^2 = i^2
Taking square root
(x+4) = +i or -i
That is x = -4+i or x = -4-i
Therefore the four roots are 3,-3,-4-i and -4+i
which is your choice (c) 3, -3, -4-i, -4+i