Write a polynomial with rational coefficients having roots 3, 3+i, and 3-i.
Part 1: Write the factors (in the form x-a) that are associated with the roots (a) given in the problem.
Part 2: Multiply the 2 factors with complex terms to produce a quadratic expression.
Part 3: Multiply the quadratic expression you just found by the 1 remaining factor to find the resulting cubic polynomial.

- anonymous

- schrodinger

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- anonymous

nah don' t do that

- anonymous

\[(x-(3+i))(x-(3-i))\] is a pain to multiply there is a much easier wy

- anonymous

you want a quadratic with zeros \(3+i\) and \(3-i\) there is an easy way to find it
also a real real easy way
which would you like to use?

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## More answers

- anonymous

Well I am pretty sure I can find the easier way, but the packet I'm doing requires me to do it the hard way, unfortunately. Can you help me with the harder way?

- anonymous

what difference does it make? you have to end up with a quadratic, might as well just write it

- anonymous

Well each of the steps is graded separately as a different question. So if I skip the first two steps (or do them differently) then I get points taken off.

- anonymous

if the zeros are \(a\pm bi\) the quardratic is
\[x^2-2ax+(a^2+b^2)\]

- anonymous

ok then lets multiply

- anonymous

So how do I find my factors for this problem?

- anonymous

if a root is \(r\) then a factor is \((x-r)\)

- anonymous

your roots are 3, 3 + i, 3 - i so your factors are
\[(x-3)(x-(3+i))(x-(3-i))\]

- anonymous

Okay yes that's what I got

- anonymous

that is the answer to A

- anonymous

for B they want you to multiply
\[(x-(3+i))(x-(3-i))\]

- anonymous

So the complex roots only? I see.

- anonymous

that is what is says right?

- anonymous

Yep, thanks.

- anonymous

Do I multiply this by FOIL?

- anonymous

i wouldn't

- anonymous

well ok i guess, but keep \(3+i\) and \(3-i\) together
lets do the first outer inner last business, but to it the easy way

- anonymous

so don't distribute the x to the roots?

- anonymous

first is a no brainer, it is \(x^2\)

- anonymous

lets skip to the last
when you multiply a complex number \(a+bi\) by its complex conjugate \(a-bi\) you get the real number \(a^2+b^2\) to the "last" is \[3^2+1^2=10\]

- anonymous

Oh okay I've never seen that before.

- anonymous

then "outer and inner"\[-(3+i)x-(3-i)x\]combine like terms, the \(i\) part goes bye bye and you are left with \(-6x\)

- anonymous

So did you distribute the x first, then subtract?

- anonymous

if you haven't seen that then you should learn it now
\[(a+bi)(a-bi)=a^2+abi-abi-b^2i^2=a^2+b^2\]

- anonymous

yes i would call it "distribute then combine like terms" but it is the same thing

- anonymous

Okay thanks. Okay and so the last part is to multiply the quadratic we found by x-3?

- anonymous

final answer
\[x^2-6x+10\] as promised

- anonymous

yes i will let you do that yourself, it is easy enough right?

- anonymous

Yeah it'll take me a while because I'm rusty but I can do it. Thank you so much!

- anonymous

yw

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