# Math Help - Inequality with higher degree

1. ## Inequality with higher degree

x^4-x<0

How do I solve it?

2. ## Re: Inequality with higher degree

Have you tried to factor the LHS yet?

3. ## Re: Inequality with higher degree

I tried doing X(X^3 -1) <0 then x=0 x^3-1=0 didn't work when i checked using solution

4. ## Re: Inequality with higher degree

Maybe you can factor $\displaystyle x^3-1$ further?

Hint: $\displaystyle a^3-b^3 = (a-b)(a^2-ab+b^2)$

Sketching the graph of $f(x) = x^4-x$ will help after you have found the zeros.

5. ## Re: Inequality with higher degree

Originally Posted by RK29
x^4-x<0 How do I solve it?
Can you see that $x\not= 0~?$
So can you solve $x^3-1<0~?$
If so solve it.

6. ## Re: Inequality with higher degree

Originally Posted by RK29
I tried doing X(X^3 -1) <0 then x=0 x^3-1=0 didn't work when i checked using solution
Hint: f(1)=0 and so (x-1) is a factor.

7. ## Re: Inequality with higher degree

Originally Posted by RK29
x^4-x<0

How do I solve it?
$x^4$ is non-negative number for all $x\in\mathbb{R}$

So when, $x^4?

x must be positive, but not greater than $1$. Can you see that?

8. ## Re: Inequality with higher degree

Originally Posted by RK29
x^4-x<0

How do I solve it?
$x^4 - x < 0$

$x(x^3 - 1) < 0$

$x(x-1)(x^2 + x + 1) < 0$

note that the factor $(x^2 + x + 1) > 0$ for all x (how can you check that the factor is always positive?)

for the other two factors that can equal zero ... critical values are x = 0 and x = 1

these two critical values divide the x-value number line into three regions ...

x < 0 , 0 < x < 1 , and x > 0

test a single x-value from each of the above intervals into the original inequality to see if this value makes the inequality true or false ...

if true, then all values of x in that interval make the inequality true ... if false, then otherwise.

9. ## Re: Inequality with higher degree

$x^4-x<0$

$x^4-x=0$

This works as well just remember to put the strict inequity back in the answer.