Show that the given values of c are zeros of p(x) and find all other zeros of p(x). P(x)=x^3+2x^2-9x-18, c=-2
To show that c = -2 is a zero of p(x), we substitute -2 into the polynomial p(x) and show that it equals zero.
p(-2) = (-2)^3 + 2(-2)^2 - 9(-2) - 18
= -8 + 2(4) + 18 - 18
= -8 + 8 + 0
= 0
Since p(-2) equals 0, we have shown that c = -2 is a zero of p(x).
To find the other zeros of p(x), we can use long division or synthetic division to divide p(x) by the binomial x - c. In this case, c = -2, so we divide p(x) by x + 2.
Using synthetic division:
-2 | 1 2 -9 -18
-2 0 18
____________________
1 0 -9 0
The quotient is 1x^2 + 0x - 9. So, the other zeros of p(x) are the zeros of the quadratic polynomial x^2 - 9. We can solve for these zeros by setting x^2 - 9 equal to zero:
x^2 - 9 = 0
(x - 3)(x + 3) = 0
This gives us two additional zeros: x = 3 and x = -3.
Therefore, the zeros of p(x) are -2, 3, and -3.