2
$\begingroup$

First of all, I tried to search online but I didn't find any explanation. The following equations, I just don't know how I could solve'em, the question asks for the solutions.

$(x²-4)*(2x-2)*x=0$

$x^4-x^3-x^2+x=0$

I don't think I ever saw equations like this on high school, should I apply baskara? It looks like a quartic equation but all the quartic equations I searched for contains a constant after the "$x$". Can you help me, please?

Please pardon me for the noob (I think) question, I'm sure it will look simple after you explain it but I really couldn't find a way to solve it yet.

$\endgroup$
4
$\begingroup$

For the first one, if the product of three factors is 0 then at least one of these is 0; so you have to solve three (easy) equations.

For the second one, write it as $x(x^3−x^2−x+1)=0$ and note that the sum in parentheses is 0 when $x=1$, so you may factor $x^3−x^2−x+1 = (x-1)(x^2+Ax+1)$. Find the value of $A$ and solve the 2nd degree equation

$\endgroup$
2
$\begingroup$

HINT:

Don't need to multiply

The equation reduces to $(x-2)(x+2)(x-1)x=0$

As the product is $0,$ at least one of the multiplier must be $0$

$\endgroup$
  • 1
    $\begingroup$ @Luan, so the solutions are $-2,1,0,2$ $\endgroup$ – lab bhattacharjee May 10 '13 at 14:42
  • $\begingroup$ There is no need to repeat the 2, is? Yes, thank you! $\endgroup$ – Thums May 10 '13 at 14:44
  • $\begingroup$ @LuanCristianThums, it's $2$ and $-2,$ they are not same! $\endgroup$ – lab bhattacharjee May 10 '13 at 14:45
  • $\begingroup$ YES, of course. I need to sleep a little, didn't notice the negative signal... thank you. $\endgroup$ – Thums May 10 '13 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.