Write the equation of the polynomial function with the following real zeros and multiplicity if (2,16) is a point on the graph

Question

zero at x=-2 with a with a multiplicity of 2
zero at x=1 with a multiplicity of 3
zero at x=0 with a multiplicity of 1

I did x(x+2)^2 (x-1)^3
16 =a^2(2+2)^2 (2-1)^3
16=a^22(4)^2 (1)^3
16=a^2(16) 1
16=a32 a=1/2
final answer is y=1/2(x+2)^2(x-1)^3 x

Is this correct? Thank you for checking my work

Answer 1

I wrote part of it incorrect up above.
Use this work.
I did x(x+2)^2 (x-1)^3
16 =a(2)(2+2)^2 (2-1)^3
16=a(2) (4)^2 (1)^3
16=a(2)(16) 1
16=a32 a=1/2
final answer is y=1/2(x+2)^2(x-1)^3 x

Answer 2

Yes : ) This is correct : )

Answer 3

Thank you

Answer 4

Your opening function should have been
y = ax(x+2)^2 (x-1)^3

you went from
x(x+2)^2 (x-1)^3 to
16 =a(2)(2+2)^2 (2-1)^3

all of sudden resurrecting an equal sign and the "a".

Just a case of "bad mathematical form". Remember, Math is a precise language.