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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.

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2 Answers 2

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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

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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 $0$

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    $\begingroup$ @Luan, so the solutions are $-2,1,0,2$ $\endgroup$ May 10, 2013 at 14:42
  • $\begingroup$ There is no need to repeat the 2, is? Yes, thank you! $\endgroup$
    – Thums
    May 10, 2013 at 14:44
  • $\begingroup$ @LuanCristianThums, it's $2$ and $-2,$ they are not same! $\endgroup$ May 10, 2013 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, 2013 at 14:49

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