SOLUTION: Find the real zeros of the polynomial function P(x) = x(x + 1)(x2 - 81). A. 0, -1, 9, -9 B. 0, -1, 9 C. -1, 9, -9 D. 0, 1, 9

Algebra ->  Graphs -> SOLUTION: Find the real zeros of the polynomial function P(x) = x(x + 1)(x2 - 81). A. 0, -1, 9, -9 B. 0, -1, 9 C. -1, 9, -9 D. 0, 1, 9       Log On


   

Question 89996: Find the real zeros of the polynomial function P(x) = x(x + 1)(x2 - 81).
A. 0, -1, 9, -9
B. 0, -1, 9
C. -1, 9, -9
D. 0, 1, 9


Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the real zeros of:
p%28x%29+=+x%28x%2B1%29%28x%5E2-81%29
To find the zeros, just set p(x) = 0 and, since the function is already factored, apply the zero products principle:
x%28x%2B1%29%28x%5E2-81%29+=+0 therefore:
x+=+0 or x%2B1+=+0 or x%5E2-81+=+0
The roots are:
x+=+0
x+=+-1
x%5E2+=+81 so x+=+9 or x+=+-9