Polynomial factors involving inequalities

Question

How to factorise the polynomial $p(x) = x^4-2x^3 + 2x - 1$.

Hence, solve the inequality $p(x) \gt 0$ ?

Answer 1

First factor: \begin{align} p(x)=x^4-2x^3+2x-1 &=x^4-2x^3+x^2-x^2+2x-1 & \text{add and subtract $x^2$} \\ &=x^2(x^2-2x+1)-1(x^2-2x+1) & \text{group the terms}\\ &=(x^2-1)(x^2-2x+1) & \text{factor by grouping} \\ &=[(x+1)(x-1)][(x-1)^2] & \text{factor even more} \\ &=(x+1)(x-1)^3 & \text{simplify} \end{align} Next we need to find critical points, so for now set $p(x) = 0$: $$p(x)=(x+1)(x-1)^3 =0$$ which implies $(x+1)=0$ and $(x-1)^3=0$. Hence, $x=1$ and $x=-1$ are solutions to $p(x)=0$. Those will be your critical points on the number line, like this:

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And test each region separated by the critical points. For example, plug in $x=0$ into $p(x)$, if you want to test region $(-1,1)$.

• For region $(-\infty,-1)$, $p(x) > 0$.
• For region $(-1,1)$, $p(x) < 0$.
• For region $(1,\infty)$, $p(x) > 0$.

(Verify these by plugging in some $x$ values into $p(x)$!)

So, your solution set is $\boxed{(-\infty,-1)\cup(1,\infty)}$.

Answer 2

See that $x=1$ and $x=-1$ are roots of this polynomial. Then factorization would be easy. See $p(x)=(x+1)^3(x-1)$

Then for $p(x) \gt 0$, you must have $x \gt 1$ or $x \lt -1$.

Answer 3

We see an obvious root $-1$ : $P(-1)=1+2-2-1=0$

Therefore it can be factorized by $(X+1)$

$P(X)=(X+1)(X-1)^3$

Thus $P(X)>0$ when $x\in\left[-\infty,-1\right[\cup\left]1,+\infty\right[$

Answer 4

$p(x) = (x-1)^3(x+1)$. ${}{}{}{}{}$

Answer 5

The "nice" way to do it: add and subtract $x^2$ to get $$\eqalign{x^4-2x^3+2x-1 &=x^4-2x^3+x^2-x^2+2x-1\cr &=(x^4-2x^3+x^2)-(x^2-2x+1)\cr &=(x^2-1)(x^2-2x+1)\cr}$$ and the rest should be easy.

Answer 6

$P(x)=x^4-2x^3+2x-1$ $$= x^4-1^2-2x^3+2x$$ $$=(x^2)^2-1^2-2x(x^2-1)$$ $$=(x^2+1)(x^2-1)-2x(x+1)(x-1)$$ $$=(x^2+1)(x+1)(x-1)-2x(x+1)(x-1)$$ $$=(x+1)(x-1)(x^2-2x+1)$$ $$=(x+1)(x-1)(x^2-x-x+1)$$ $$=(x+1)(x-1)(x(x-1)-1(x-1))$$ $$=(x+1)(x-1)(x-1)(x-1)$$ $$=(x+1)(x-1)^3$$ Thus $x=1,-1$