SOLUTION: Factor the polynomial and use the factored form to find the real zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) P(x) = 1/6(2x4 + 3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the polynomial and use the factored form to find the real zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) P(x) = 1/6(2x4 + 3      Log On


   

Question 1185932: Factor the polynomial and use the factored form to find the real zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)
P(x) = 1/6(2x4 + 3x3 − 16x − 24)^2
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

P%28x%29+=+%281%2F6%29%282x%5E4+%2B+3x%5E3-16x-+24%29%5E2
%281%2F6%29%282x%5E4+%2B+3x%5E3-16x-+24%29%5E2=0+will be if
2x%5E4+%2B+3x%5E3-16x-+24=0
%282x%5E4+%2B+3x%5E3%29-%2816x%2B+24%29=0
x%5E3%282x+%2B+3%29-8%282x%2B+3%29=0
%282x+%2B+3%29+%28x%5E3+-+8%29+=+0........use 8=2%5E3
%282x+%2B+3%29+%28x%5E3+-+2%5E3%29+=+0 .......use the rule for factoring the difference of the cubes
%282x%2B3%29%28x-2%29%28x%5E2%2B2x%2B4%29=0......... zeros are

if %282x%2B3%29=0 => x=-3%2F2
if %28x-2%29+=0 =>+x=2
for+x%5E2%2B2x%2B4=0 use quadratic formula
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29 .....in your case a=1, b=2, c=4
x=%28-2%2B-sqrt%282%5E2-4%2A1%2A4%29%29%2F%282%2A1%29
x=%28-2%2B-sqrt%284-16%29%29%2F2
x=%28-2%2B-sqrt%28-12%29%29%2F2
x=%28-2%2B-sqrt%28-4%2A3%29%29%2F2
x=%28-2%2B-2sqrt%283%29%2Ai%29%2F2........simplify
x=%28-1%2B-sqrt%283%29%2Ai%29
roots: -1%2Bsqrt%283%29%2Ai, -1-sqrt%283%29%2Ai
all roots will be
real roots:
x=-3%2F2
x=2
complex roots:
-1%2Bsqrt%283%29%2Ai
+-1-sqrt%283%29%2Ai