Asked by Fred
The polynomial P(x) is given by P(x) = ax^3 + 15x^2 + cx - 72, where a and c are constants. The three zeros of P(x) are -3, 2 and B. Find the value of B.
Answers
Answered by
Steve
P(x) = a(x+3)(x-2)(x-B) = ax^3+15x^2+cx-72
So, you have
a(x^3+(1-B)x^2-(6+B)x+6B) = ax^3+15x^2+cx-72
ax^3+a(1-B)x^2-a(6+B)x+6aB = ax^3+15x^2+cx-72
a(1-B) = 15
-a(6+B) = c
6aB = -72
Now solve for B. I think you'll find more than one solution.
So, you have
a(x^3+(1-B)x^2-(6+B)x+6B) = ax^3+15x^2+cx-72
ax^3+a(1-B)x^2-a(6+B)x+6aB = ax^3+15x^2+cx-72
a(1-B) = 15
-a(6+B) = c
6aB = -72
Now solve for B. I think you'll find more than one solution.
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