Find a polynomial of degree 3 with a real coefficient that satisfies the given condition. Zeros: -2, 1, 0; f(2)=32
1 answer:
Its factors would be
(x+2)*(x-1)*(x+0)
x^2 +x -2
x^3 + 0 + x^2 + 0 -2x +0
Equation: x^3 + x^2 -2x
f(2) = 8 + 4 -4
2x^3 + 2x^2 -4x +0
f(2) = 16 + 8 -8
3x^3 + 3x^2 -6x +0
f(2) = 24 +12 -12
4x^3 + 4x^2 -8x +0
f(2) = 32 +16 -16
So, the equation is:
4x^3 + 4x^2 -8x = 0
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Answer: Parallel
Step-by-step explanation:
Step-by-step explanation:
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hope it helps.
Using the Quadratic formula
your answer would be A and C
The answer to your question is a,c,d