Question

Solve: x^3+1=x^2+x

Answer 1

First make everything equal to zero

x^3 - x^2 - x + 1 = 0

Can we factor that?

Answer 2

im not sure

Answer 3

After you've done with @perl has mentioned, factor by grouping.\[\color{red}{x^3-x^2}+\color{blue}{(-x+1)}=0\]

Answer 4

x=1 and x=0 ?

Answer 5

not quite.

Answer 6

\[x^3+1=x^2+x\]Subtract \(-x^2\) and \(-x\) from both sides of your equation.\[x^3 -x^2-x+1=x^2-x^2+x-x\]\[x^3-x^2-x+1=0\]\[\color{red}{x^3-x^2}+\color{blue}{(-x+1)}=0\]\[\color{red}{x^2(x-1)}\color{blue}{-1(x-1)}=0\]

Answer 7

how'd u go from (-x+1) to -1(x-1)?

Answer 8

Notice how you have a factor that is the same in both groups? \[x^2\color{red}{(x-1)}-1\color{red}{(x-1)}=0\]

Answer 9

Well,... pull out a -1 from \((-x+1)\)

Answer 10

(x-2)(x^2-1)?

Answer 11

i mean (x-1)(x^2-1)

Answer 12

when you multiply \((-1)(x)\) and \((-1)(-1)\) you will end up with \((-x+1)\).

Answer 13

so then x=1 and x = -+isqrt1

Answer 14

From \(x^2\color{red}{(x-1)}-1\color{red}{(x-1)}=0\) we factor out a \((x-1)\).

That leaves us with: \((x-1)(x^2-1)\)

Answer 15

\[(x-1)(x^2-1)=0\]\[x-1=0 \implies x=1\]\[x^2-1=0 \implies x^2=1 \implies x= \pm 1\]

Answer 16

thank you for your help!! i see what i did wrong

Answer 17

The only time you would get \(x=\pm i\sqrt{1}\) would be if you had this.\[x^2+1=0\implies x^2=-1\\~~~~~~~~~~~~~~~~~~ \implies x=\pm\sqrt{-1} \ \\~~~~~~~~~~~~~~~~~~ \implies x=\pm\sqrt{-1}\sqrt{1} = \pm i\sqrt{1} \]

Answer 18

okay, i see :)

Answer 19

awesome :D