Question

What is a quartic function with only the two real zeros given? x=-4 and x=-1

Answer 1

the two zero's can be used to make the four factors for the quartic function as both are double zeros. I have expanded the function below.

(x+4)^2*(x+1)^2

(x^2+8x+16)*(x^2+2x+1)


(x^4+10x^3+33x^2+40x+16)

Answer 2

Answer:

The quartic function is:

                    $f(x)=x^4+10x^3+33x^2+40x+16$

Step-by-step explanation:

Quartic function--

A quartic function is a polynomial function with degree 4.

Now we know that for any quartic funtcion with just two roots "a" and "b" the equation of the function is calculated as:

$f(x)=(x-a)^2(x-b)^2$

Here let a= -4 and b= -1

Hence, the equation of  a quartic function is calculated as follows:

$f(x)=(x-(-4))^2(x-(-1))^2\\\\f(x)=(x+4)^2(x+1)^2$

Now as we know that:

$(a+b)^2=a^2+b^2+2ab$

Hence,

$(x+4)^2=x^2+16+8x$

and

$(x+1)^2=x^2+1+2x$

Hence,

$f(x)=(x^2+16+8x)(x^2+1+2x)\\\\\\f(x)=x^2(x^2+1+2x)+16(x^2+1+2x)+8x(x^2+1+2x)\\\\\\f(x)=x^4+x^2+2x^3+16x^2+16+32x+8x^3+8x+16x^2$

on combining like terms we have:

                   $f(x)=x^4+10x^3+33x^2+40x+16$