Question

factor x^5-2x^4+8x^2-13x+6 completely over the set of complex numbers

Answer 1

You asked that question already. I told you how to do it. Is there something that didn't make sense to you?

Answer 2

i really need to see it step by step to fully understand

Answer 3

How many of the steps that I gave you did you complete?

What are the possible rational factors of that polynomial?

Answer 4

plus or minus 1,2,3,6

Answer 5

Right. So now you need to go and plug each of those values into \[f(x) = x^5-2x^4+8x^2-13x+6\]and see which ones make \(f(x) = 0\).

Answer 6

1 makes 0

Answer 7

Let's try the first candidate, \(x = 1\):

\[f(1) = (1)^5-2(1)^4+8(1)^2 -13(1) + 6 = 1-2+8-13 +6= 0\]
So that means \((x-1)\) is a factor of \(f(x)\)

Answer 8

Our next step is to divide \(f(x)\) by \((x-1)\) to get a simplified polynomial which has the same remaining factors as \(f(x)\). What do you get when you do that?

Answer 9

x^4-x^3-x^2+7x+1?

Answer 10

close. last term is incorrect (+1), should be (-6)

Answer 11

ok now here is where idk what to do next

Answer 12

You know by the RRT that the only way we could have f(x) end with +6 is by having a factor or factors that multiplied to 6, and so if we divide off a factor ending in 1, our poly must still end with 6, right?

Answer 13

It's worth figuring out where you made the division mistake, but let's go on.
Our new polynomial is \[x^4-x^3-x^2+7x-6\]It has the same set of possible factors: \(\pm 1, \pm 2, \pm 3, \pm 6\)

Let's try \(x=1\) and see if \((x-1)\) is a repeated factor.

\[(1)4-(1)^3-(1)^2+7(1)-6 = 1-1-1+7-6 = 0 \]That means that \((x-1)\) is still a factor of our simplified polynomial, so we divide it by \((x-1\).

Answer 14

Uh, \((x-1)\)

What does that give us for our even simpler polynomial?

Answer 15

it gives us 0 and x^3-x+6

Answer 16

Correct! Now, once again the potential factors are \(\pm 1, \pm 2, \pm 3, \pm 6\). This time, \(x=1\) is not a zero:
\[(1)^3-1+6=6\]so we have to try the others.

Answer 17

ok 2?

Answer 18

yes, \(x = 2\) is a zero of the new, slimmer, more glamorous polynomial :-)

So we divide it by \((x-2)\) and what do we get?

Answer 19

x^2+2x-3

Answer 20

No, that's incorrect.

Answer 21

sorry x^2+2x+3

Answer 22

nevermind i dont know

Answer 23

i get a remainder of 12 it should be 0 if x-2 is a factor

Answer 24

x-2 is not a factor it is x+2 that is a factor

Answer 25

Ah, let's check something:
our polynomial is \[x^3-x+6\]
\[(2)^3-(x)+6=8-2+6 = 12\]
So 2 isn't a root after all
\[(-2)^3-(-2)+6=-8+2+6 = 0\]
so (x+2) is the factor, not (x-2)

Answer 26

That's why we couldn't get the division right!

Answer 27

lol so now we factor right?

Answer 28

well use the quadratic formula

Answer 29

So now you have x^2-2x+3 and you can use the quadratic or complete the square or whatever to get the two remaining complex roots

Answer 30

but we will only have a total of 4 roots when there should be 5 we have one neg root one pos root and 2 com roots. we should either have 4,2,or 0 pos real roots and 2 or 0 neg real roots.

Answer 31

No, 5 roots were found: we had 1 as a root twice, remember.

Answer 32

You can see this if you graph the curve. At a root where the multiplicity is an even number (such as x=1 here), the curve touches the x-axis and curves away without crossing. At an odd multiplicity, the curve crosses the x-axis.