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

Answer:Answer in factored form: [tex]P(x)=(x+2)(x-7)(x-5)^2[/tex]Answer in standard form: [tex]P(x)=x^4-15x^3+61x^2+15x-350[/tex]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 isA(x+2)(x-7)(x-5)(x-5)or[tex]A(x+2)(x-7)(x-5)^2[/tex]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.[tex](x+2)(x-7)(x-5)^2[/tex]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^2Outer: x(-7)=-7xInner: 2(x)=2xLast: 2(-7)=-14--------------------Adding.[tex]x^2-5x-14[/tex]I'm going to multiply [tex](x-5)^2[/tex] using formula [tex](u+v)^2=u^2+2uv+v^2[/tex].[tex](x-5)^2=x^2-10x+25[/tex].So now we have to multiply these products.That is we need to do:[tex](x^2-5x-14)(x^2-10x+25)[/tex]I'm going to distribute every term in the first ( ) to every term in the second ( ).[tex]x^2(x^2-10x+25)[/tex][tex]+-5x(x^2-10x+25)[/tex][tex]+-14(x^2-10x+25)[/tex]------------------------------------------ Distributing:[tex]x^4-10x^3+25x^2[/tex][tex]+-5x^3+50x^2-125x[/tex][tex]+-14x^2+140x-350[/tex]-------------------------------------------Adding like terms:[tex]x^4-15x^3+61x^2+15x-350[/tex]