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Find a polynomial of degree 3 with real coefficients and zeros of minus​3,minus​1, and​ 4, for which ​f(minus​2)equals30.

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3 votes

Answer:


f(x)=5x^(3)-65x-60

Explanation:

we know that

The roots are

x=-3,x=-4 and x=1

so

The equation of a polynomial of degree 3 with real coefficients and zeros of minus​3,minus​1, and​ 4 is equal to


f(x)=a(x+3)(x+1)(x-4)

Remember that


f(-2)=30 ----> given value

For x=-2, f(x)=30

substitute and solve for the coefficient a


30=a(-2+3)(-2+1)(-2-4)


30=a(1)(-1)(-6)


30=6a


a=5

so

The polynomial is


f(x)=5(x+3)(x+1)(x-4)

Apply distributive property


f(x)=5(x+3)(x+1)(x-4)\\f(x)=5(x+3)(x^(2)-4x+x-4)\\f(x)=5(x+3)(x^(2)-3x-4)\\f(x)=5(x^(3)-3x^(2)-4x+3x^(2)-9x-12)\\f(x)=5(x^(3)-13x-12)\\ f(x)=5x^(3)-65x-60

User BFTrick
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