Find a polynomial equation that satisfies the given conditions.

Question

Degree 3; the coefficient of x^3 is 1; three roots are 2,-2, and 4

Answer 1

the three zeros are

x=-2

x+2 = 0. x+2 is the corresponding factor

x=2

x-2 = 0 x-2 is the corresponding factor

x=4

x-4 =0 x-4 is the corresponding factor

Multiply the three factors together

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

=(x^2- 4)(x-4)

y = x^3 -4x^2 -4x +16

check the answer:

f(4) = 4^3 -4(4^2) -4(4)+16=0

f(2) = 2^3 -4(2^2)-4(2)+16 = 0

f(-2) =(-2)^3 -4((-2)^2) -4(-2) +16 = 0

Answer 2

You can build a polynomial from its roots by stating the polynomial as

a(x-2)(x+2)(x-4) = y

This guarantees all the roots yield y=0. The constant a can then be set by another constraint (like a point that the polynomial goes through or specifying a coefficient). Here a=1 so the x³ term has a one in front of it when you multiply the polynomial out.

Multiply out (x²-4)(x-4) = y

Please consider a tutor. Take care.