# How do you solve the polynomial inequality and state the answer in interval notation given 3x^2+2x<x^4?

Sep 11, 2017

$x < 0$ or $x > 2$

#### Explanation:

We seek $x$ such that:

$3 {x}^{2} + 2 x < {x}^{4}$

The steps to solving any nonlinear inequality are:

• Rewrite the inequality so that there is a zero on the right side, in the form $f \left(x\right) < 0$ or $f \left(x\right) > 0$ (or $f \left(x\right) \le 0$ or $f \left(x\right) \ge 0$).
• Factorise the function, $f \left(x\right)$ if possible.
• Find the critical values where a sign change can occur by finding the roots of $f \left(x\right) = 0$.
• Determine the sign of $f \left(x\right)$ in the intervals formed by the critical values.

The solution will be those intervals in which the function has the correct signs satisfying the inequality.

We can rearrange the equation as follows:

$3 {x}^{2} + 2 x < {x}^{4} \implies {x}^{4} - 3 {x}^{2} - 2 x > 0$

Which we can write as:

$f \left(x\right) > 0$ where $f \left(x\right) = {x}^{4} - 3 {x}^{2} - 2 x$

Next we solve the equation $f \left(x\right) = 0$:

${x}^{4} - 3 {x}^{2} - 2 x = 0$
$\therefore x \left({x}^{3} - 3 x - 2\right) = 0$

We can factorise the cubic by inspection noting that $x = 2$ is a solution thus $x - 2$ is a factor, and so:

$x \left(x - 2\right) {\left(x + 1\right)}^{2} = 0$

And so we have three roots:

$x = - 1 , 0 , 2$

We must now examine the sign of $f \left(x\right)$ in each interval, partitioned by these roots, i.e:

 x lt -1; -1 lt x lt 0; 0 lt x lt 2; x gt 2

The easiest way to do this is to look at the sign of each factor (or component) of $f \left(x\right)$ via a sign chart

 {: ( ul("factor"), ul(x lt -1), ul(-1 lt x lt 0), ul(0 lt x lt 2), ul(x gt 2) ), ( x, -, -, +, + ), ( x-2, -, -, -, + ), ( (x+1)^2, +, +, +, + ), ( x(x-2)(x+1)^2, +,+ ,- ,+ ) :}

So, overall, we have established that the function, $f \left(x\right)$ is positive in all intervals with the exception of of $0 < x < 2$, so the solution of the inequality is:

$x < 0$ or $x > 2$

We can visualize the solutions graphically: