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

Question

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

Answer 1

I would start by multiplying factors containing the 3 given roots, and then multiply by an unknown real number a:

 

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

 

f(-2) = 24 = a(-2+3)(-2+1)(-2-4) = a(1)(-1)(-6) = 6a

 

6a = 24, so a = 24/6 = 4

 

Final answer:

 

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

 

Note this has 3 roots: -3,-1,4; and f(-2) = 24, and the coefficients are real.

Answer 2

Find the equation of the polynomial with roots (–√3, 0) and (7i, 0), real coefficients, and passing through the point (1, 200)