Question

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

Answer 1

Answer:

$f(x)=4x^{3}-52x-48$

Step-by-step 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

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