Given P(x) = a(b-x)(x+c)(dx + 4), where a, b, c and d are constants. Given the first term of the expansion of P(x) is –12x³ and has a zero of -3. Find P(x) if the
polynomial leaves a remainder of 48 when divided by x and a remainder of
- 312x +312 when divided by
4x -x²-3.

really need help.. please

true

explanation:

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

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation:

the correct answer is the last option