Find all other zeros for p(x)=x^3-x^2-11x+15 c=3
To find the other zeros of the polynomial, we can use synthetic division and the given zero, c=3.
Performing synthetic division with 3 as the divisor:
3 │ 1 -1 -11 15
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3 6 15
The result is 3x^2 + 6x + 15.
Now, we need to find the zeros of the quadratic equation, 3x^2 + 6x + 15 = 0.
Using the quadratic formula,
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 6, and c = 15:
x = (-6 ± √(6^2 - 4(3)(15))) / (2(3))
x = (-6 ± √(36 - 180)) / 6
x = (-6 ± √(-144)) / 6
Since the discriminant (-144) is negative, there are no real solutions for this quadratic equation. Therefore, there are no additional zeros for the polynomial p(x) = x^3 - x^2 - 11x + 15.