Question

name the integral polynomial function of degree 7 whose zeros are: 0 of multiplicity 2, 1 of multiplicity 3, -5 of multiplicity 1 and -1/3 of multiplicity 1 express your answer in factored form. ( i don't understand what it means.)

Answer 1

They want a seventh degree polynomial and they have given you seven roots.
0 is a root of the polynomial and it has multiplicity 2 meaning the root is repeated twice. That means x^2 is one factor.

Answer 2

root of 1 with multiplicity 3
root 1 implies (x-1) is a factor
multiplicity of 3 means (x-1)^3 is a factor.

Answer 3

i got this: x^2 (x-3)^3 (x+5) (3x+1)

Answer 4

Wait, it has to be (x-1)^3

Answer 5

ok thanks! but what are integral polynomial function?

Answer 6

When multiplied out, the coefficients have to be integers.

Had you made the last factor (x +1/3) = (3x+1)/3 then the coefficients when multiplied out will not be integers.

Answer 7

So you have to drop the 3 in the denominator and make the last factor (3x+1)

Answer 8

i confused...

Answer 9

so the answer will be still (3x+1) right?

Answer 10

-1/3 of multiplicity 1
implies (x + 1/3) is a factor
which can be simplified to (3x+1)/3.
So the polynomial is: x^2 (x-3)^3 (x+5) (3x+1)/3

If you had done like that, then when you multiplied out all the factors, the coefficients will not be integers. So you have to drop the 3 in the denominator and say the polynomial is x^2 (x-3)^3 (x+5) (3x+1). This when multiplied out will have integer coefficients.

Answer 11

oh i see! thank you for the explanation!

Answer 12

You are welcome.

Answer 13

can i ask you another question?