Find a polynomial function of degree 5. -1 as a zero multiplicity of 2. 3 as a zero multiplicity of 2. and 0 as a zero multiplicity of 1. Leading coefficient is 4.

Question

whpalmer4 · Answer

Okay, if you have a polynomial $P(x)$ and you want to write it in factored form, with roots $r_1, r_2, ...,r_n$, it can be written 
$$P(x) = a(x-r_1)(x-r_2)...(x-r_n)$$where $a$ is a constant to arrange the polynomial to pass through a specified point, or have a given coefficient — it doesn't alter the roots, and can be chosen as 1 for convenience in many cases.

If the multiplicity of a given root is > 1, then you need to raise that root's factor to its multiplicity as a power.  For example, a root of 3 with multiplicity 2 will be $(x-3)^2$ in the factored polynomial.

hero · Answer

In other words, Compute f(x) = 4x(x + 1)^2(x - 3)^2

whpalmer4 · Answer

Yeah, enjoy the algebra :-)

anonymous · Answer

So then 4(x^2+2x+1+x^2+6x+9)=0

whpalmer4 · Answer

No, you have to multiply those together...

whpalmer4 · Answer

$$4(x^2+2x+1)(x^2-6x+9) = $$

anonymous · Answer

How do you even foil that?..

whpalmer4 · Answer

Not even that — we all dropped the $x-0$ term!

$$4x(x^2+2x+1)(x^2-6x+9)$$

whpalmer4 · Answer

Distributive property...this is why I hate FOIL as a teaching method, no one knows what to do for bigger multiplications.

hero · Answer

I had it originally. I just copied someone else's mistake

hero · Answer

I don't know why they ever taught FOIL to begin with.

whpalmer4 · Answer

:-)

$$(a+b+c)(d+e+f) = a(d+e+f) + b(d+e+f) + c(d+e+f)$$

don't forget to multiply that all by $4x$ when you're done!

whpalmer4 · Answer

Yeah, I don't understand that either, it's not like no one has ever seen the distributive property by the time they get to multiplying polynomials!

whpalmer4 · Answer

Anyhow, @pita0001, as the problem says, you'll end up with a polynomial function of degree 5, which means you'll need to have an $x^5$ term in there.

whpalmer4 · Answer

These "construct a polynomial" problems do a bit of testing of understanding, but mostly they exercise your algebraic manipulation skills :-)

anonymous · Answer

Thank you both for your help..

whpalmer4 · Answer

The only other bit I can think of that this problem didn't include is remembering that if you have a polynomial with only real coefficients, any complex roots must come in conjugate pairs, $a\pm bi$.  You'll undoubtedly see (if you haven't already) problems where they only give you one of the complex roots and you need to know how to construct the other.

whpalmer4 · Answer

(and that you need to include it, of course)