How to factor this cubic polynomial?

Question

I am asking for detailed steps how to factor the cubic polynomial

$${x^3-3x+2}$$

The solution is

$${(x-1)^2(x+2)}$$

Answer 1

So we are factorising:

$$ x^3 - 3x + 2 $$

The value $ 1 $ satisfies the equation, so we should having $ (x-1) $ as one of our factors. Let's try to take that out step by step.

Step 1: Try to take $ (x-1) $ common directly. The first term should always be $ x^3 $ as it is there in your function. So you have $ x^2 $ outside.

$$ x^3 - 3x + 2 = x^2 ( x - 1 ) + something $$

So when you take $ (x-1) $ common first, on expanding, you would have an extra $ -x^2 $ which we need to get rid of. So you will add another $ x^2 $ to that. And there, try to take out $ (x-1) $ as a factor again taking $ x $ outside. So you have:

$$ x^2 (x-1) + x(x-1) + something $$

Now you have $ x^3 - x .. $ . But our expression contains $ - 3x $ , so we subtract $ 2x $ to make it match, and again take $ (x-1) $ common. You get:

$$ x^2 (x-1) + x(x-1) - 2(x-1) = (x^2 + x - 2)(x-1) $$

Pretty much good. That matches our original function. Now time to factorise that $ (x^2 + x - 2) $ , this time do it yourself. You can factor out $ (x-1) $ again! So our final result would be:

$$ x^3 - 3x + 2 = (x-1)(x-1)(x+2) = (x-1)^2 (x+2) $$

Answer 2

$x=1$ is one solution of your equation $$x^3-3x+2=0$$ then you can divide by $x-1$ and you will get a quadradic equation $$x=1$$ is a solution since $$1^3-3+2=0$$

Answer 3

Usually, we will use factor theorem and division.

For this one, we can also factorize it as

\begin{align*} x^3-3x+2&=x(x^2-1)-2(x-1)\\ &=x(x-1)(x+1)-x(x-1)\\ &=(x-1)[x(x+1)-2]\\ &=(x-1)(x-1)(x+2) \end{align*}

Answer 4

$-2$ is a root of $x^3-3x+2$, which says that $x^3-3x+2$ is divided by $x+2$.

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

Answer 5

Note that candidates for rational solutions for $$x^3-3x+2=0$$ are $ x=1, -1, 2, -2.$

Upon checking those candidates we find that $ x=1$ and $x=-2$ are solutions. That implies $$(x-1)(x+2)=x^2+x-2$$ is a factor of $ x^3-3x+2$. Divide $ x^3-3x+2$ by $x^2+x-2$ and find the quotient which is $x-1.$ Therefore $$ x^3-3x+2=(x-1)^2(x+2)$$ with roots at $x=1$ and $x=-2.$

Answer 6

$$x^3-3x+2=x^3-3x+3-1=-3(x-1)+x^3-1=-3(x-1)+(x-1)(x^2+x+1)$$

$$x^3-3x+2=(x-1)(x^2+x-2)$$

Then find the roots of $x^2+x-2$