Question

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.

Answer 1

nah don' t do that

Answer 2

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

Answer 3

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?

Answer 4

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?

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

ok then lets multiply

Answer 9

So how do I find my factors for this problem?

Answer 10

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

Answer 11

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

Answer 12

Okay yes that's what I got

Answer 13

that is the answer to A

Answer 14

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

Answer 15

So the complex roots only? I see.

Answer 16

that is what is says right?

Answer 17

Yep, thanks.

Answer 18

Do I multiply this by FOIL?

Answer 19

i wouldn't

Answer 20

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

Answer 21

so don't distribute the x to the roots?

Answer 22

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

Answer 23

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 \[3^2+1^2=10\]

Answer 24

Oh okay I've never seen that before.

Answer 25

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

Answer 26

So did you distribute the x first, then subtract?

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

yw