Question

Which polynomial function has a leading coefficient of 1, roots –2 and 7 with multiplicity 1, and root 5 with multiplicity 2?

Answer 1

Answer in factored form:
`P(x)=(x+2)(x-7)(x-5)^2`

Answer in standard form:
`P(x)=x^4-15x^3+61x^2+15x-350`

Explanation:

I don't see your choices but I can still give you a polynomial fitting your criteria. I will give the answer in both factored form and standard form.

The following results are by factor theorem:

So if x=-2 is a zero then x+2 is a factor.

If x=7 is a zero then x-7 is a factor.

If x=5 is a zero then x-5 is a factor. It says we have this factor twice. I know this because it says with multiplicity 2.

So let's put this together. The factored form of the polynomial is

A(x+2)(x-7)(x-5)(x-5)

or

`A(x+2)(x-7)(x-5)^2`

Now A can be any number satisfying a polynomial with zeros -2 and 7 with multiplicity 1, and 5 with multiplicity 5.

However, it does say we are looking for a polynomial function with leading coefficient 1 which means A=1.

`(x+2)(x-7)(x-5)^2`

Now the factored form is easy.

The standard form requires more work (multiplying to be exact).

I'm going to multiply (x+2)(x-7) using foil.

First: x(x)=x^2

Outer: x(-7)=-7x

Inner: 2(x)=2x

Last: 2(-7)=-14

--------------------Adding.

`x^2-5x-14`

I'm going to multiply
`(x-5)^2` using formula
`(u+v)^2=u^2+2uv+v^2`.

`(x-5)^2=x^2-10x+25`.

So now we have to multiply these products.

That is we need to do:

`(x^2-5x-14)(x^2-10x+25)`

I'm going to distribute every term in the first ( ) to

every term in the second ( ).

`x^2(x^2-10x+25)`

`+-5x(x^2-10x+25)`

`+-14(x^2-10x+25)`

------------------------------------------ Distributing:

`x^4-10x^3+25x^2`

`+-5x^3+50x^2-125x`

`+-14x^2+140x-350`

-------------------------------------------Adding like terms:

`x^4-15x^3+61x^2+15x-350`

Answer 2

f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

Explanation: