Question

Create the equation of a cubic in standard form that has a double zero at -2, another zero at 4, and a y-intercept of 16,​

Answer 1

f(x) =
`(1)/(4)` x³ - 3x - 4

Explanation:

Given that x = - 2 has multiplicity 2 and x = 4 are zeros then

(x + 2)² and (x - 4) are the factors of the polynomial and the polynomial is the product of the factors, thus

f(x) = a(x + 2)²(x - 4) ← a is a multiplier

To find a substitute (0, 16) into the equation

16 = a(4)(16), thus

64a = 16

a =
`(16)/(64)` =
`(1)/(4)`

f(x) =
`(1)/(4)`(x + 2)²(x - 4) ← expand factors

=
`(1)/(4)`(x² + 4x + 4)(x - 4)

=
`(1)/(4)`(x³ + 4x² + 4x - 4x² - 16x - 16)

=
`(1)/(4)`(x³- 12x - 16)

=
`(1)/(4)` x³ - 3x - 4