Polynomial of lowest degree with zeros of 4/3(multiplicity 2) and -1/5(multiplicity 1) and with f(0)=-48

Question

Polynomial of lowest degree with zeros of 4/3(multiplicity 2) and -1/5(multiplicity 1) and with f(0)=-48

Answer 1

f(x) = a(x-4/3)^2(x+1/5)

f(x) = a(x^2 - 8x/3 + 16/9)(x+1/5)

replace x with 0

f(0)= -48 = a(16/9)(1/5)

a = -48(9/16)(5) = -3(9)(5)=-135

f(x)=[-3(9)(x^^2 ) +72x -48](5x+1)

= (-27x^2 +72x -48)(5x+1)

f(x) = -135x^3 +133x^2 -168x -48

constant term -48 is definitely correct

cubic coefficient, -135, looks good

with luck the coefficients for linear & quadratic terms are correct, but you may want to check them, redo the calculations

lowest possible degree is 3 = sum of zero multiplicities

Answer 2

P(x) = a(x - 4/3)²(x + 1/5)

-48 = a(0 - 4/3)²(x + 1/5)

Can you solve for a and answer?