Question

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

Answer 1

Answer:

$f(x)=5x^{3}-65x-60$

Step-by-step 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

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