Show that the given values of c are zeros of p(x) and find all other zeros for p(x)=x^3-x^2-11x+15, c=3
To show that c = 3 is a zero of p(x) = x^3 - x^2 - 11x + 15, we need to verify that p(3) = 0.
p(3) = (3)^3 - (3)^2 - 11(3) + 15
= 27 - 9 - 33 + 15
= 0
Since p(3) = 0, we have shown that c = 3 is a zero of p(x).
Now, to find the other zeros of p(x), we can use polynomial long division or synthetic division. However, since c = 3 is already a zero, we can easily factor out (x - 3) using synthetic division.
3 | 1 - 1 - 11 + 15
| 3 6 -15
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1 2 -5 0
The result is a quadratic polynomial: x^2 + 2x - 5. To find the remaining zeros, set x^2 + 2x - 5 = 0 and solve for x using factoring, completing the square, or the quadratic formula.
x^2 + 2x - 5 = 0
(x + 5)(x - 1) = 0
Setting each factor equal to zero:
x + 5 = 0 -> x = -5
x - 1 = 0 -> x = 1
Therefore, the zeros of p(x) = x^3 - x^2 - 11x + 15 are 3, -5, and 1.