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:

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

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

Step-by-step 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

Answer:

f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

Step-by-step explanation: