Find a polynomial function of degree 3 with real coefficients that has the given zeros. - 2, 3, -4

Polynomial functions' standard form is f(x)=ax^n+bx^n-1...+k

However, polynomials can also be written in intercept form

f(x)=(x-x1)(x-x2)(x-x3)...(x-xn) for roots/zeros x1,x2,x3,...,xn

With the three numbers that you have -2, 3, -4 you can write function

f(x)=(x-(-2))(x-3)(x-(-4))

Then you can simplify and distribute

f(x)=(x+2)(x-3)(x+4)

f(x)=(x^2-x-6)(x+4)

f(x)=x^3+3x^2-10x-24

Which would be the answer

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

step-by-step explanation:

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