Question

Find a polynomial of degree 3 with real coefficients and zeros of -3, -1, and 4, for which f(-2)=24

Answer 1

`f(x)=4x^(3)-52x-48`

Explanation:

we know that

The roots of the polynomial are values of x when the value of the polynomial is equal to zero

x=-3 ----> (x+3)=0

x=-1 ----> (x+1)=0

x=4 ----> (x-4)=0

so

The equation of the polynomial is

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

Remember that

f(-2)=24

That means

For x=-2

f(x)=24

substitute the value of x and the value of y and solve for the coefficient a

`24=a(-2+3)(-2+1)(-2-4)`

`24=a(1)(-1)(-6)`

`24=6a`

`a=4`

substitute

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

Applying distributive property

Convert to expanded form

`f(x)=4(x+3)(x+1)(x-4)\\\\f(x)=4(x+3)(x^(2) -4x+x-4)\\\\f(x)=4(x+3)(x^(2)-3x-4)\\\\f(x)=4(x^(3)-3x^(2)-4x+3x^(2) -9x-12)\\\\f(x)=4(x^(3)-13x-12)\\\\f(x)=4x^(3)-52x-48`