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,
Accepted Solution
A:
Answer:f(x) = [tex]\frac{1}{4}[/tex] x³ - 3x - 4Step-by-step 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, thusf(x) = a(x + 2)²(x - 4) ← a is a multiplierTo find a substitute (0, 16) into the equation16 = a(4)(16), thus64a = 16a = [tex]\frac{16}{64}[/tex] = [tex]\frac{1}{4}[/tex]f(x) = [tex]\frac{1}{4}[/tex](x + 2)²(x - 4) ← expand factors = [tex]\frac{1}{4}[/tex](x² + 4x + 4)(x - 4) = [tex]\frac{1}{4}[/tex](x³ + 4x² + 4x - 4x² - 16x - 16) = [tex]\frac{1}{4}[/tex](x³- 12x - 16) = [tex]\frac{1}{4}[/tex] x³ - 3x - 4