Question

the polynomial x^5-3x^4+2x^3-2x^2+3x+1 is denoted by f(x). when f(x) is divided by (x^2+1), the remainder is 2x. find all the real roots of the equation f(x)=2x

Answer 1

\[f(x)=(x^2+1)(something)+2x\]

Answer 2

\(x^2+1\) has no real roots, so it will have to be the real roots of the "something"

Answer 3

x^3-3x^2+x+1=something?

Answer 4

i have no idea, but i can check

Answer 5

thanks

Answer 6

great
however you got it, it is right

Answer 7

they are saying find the real roots of the equation f(x) = 2x. What does that mean?

Answer 8

so your job is now to find the roots of \(x^3-3x^2+x+1\)

Answer 9

i also dont know

Answer 10

which is not that hard, because one of the roots is 1

Answer 11

so you can factor as \((x-1)(x^2-2x-1)\) and solve for the second part via the quadratic formula

Answer 12

what is the meaning of f(x)=2x?

Answer 13

\[f(x)=2x\] means exactly that
we know one answer is 1, this means \(f(1)=2\)

Answer 14

find a number for which the function evaluated at the number is the same as two times the number

Answer 15

for example if \(f(3)=6\) what would be a solution
it is not, that is just an example

Answer 16

f(x) is not ( x³ - 3x² + x +1 ) it is

x^5 - 3x^4 + 2x³ -2x² +3x +1

Answer 17

right, and \(f(1)=1-3+2-2+3+1=2\)
and also \(2\times 1=2\)

Answer 18

which means if \(x=1\) then \(f(x)=2x\)

Answer 19

I understand what you are doing here but just a bit confused with the wording of the problem that is all

Answer 20

you still have to find the roots of \(x^2-2x-1\)

Answer 21

sort of get it now

Answer 22

yeah it is kind of confusing
you can think of it as \(f(x)=x^5 - 3x^4 + 2x³ -2x² +3x +1
\) and \(g(x)=2x\) solve
\[f(x)=g(x)\]

Answer 23

or for that matter solve
\[x^5 - 3x^4 + 2x³ -2x² +3x +1
=2x\]

Answer 24

that is exactly what I did and we get x^5 - 3x^4 + 2x^3 - 2x^2 + x +1 = 0