About rational root theorem in factoring

Question

I am given $x^4-5x^2+4=0$ and told to factor it. I am not sure how to use national root theorem to do so. Since the last digit is 4, so we have $-1,-2,-4,1,2,4$ six choices? Then we try to plug in and try out individually? Or do we know quicker way to find the correct number?

Answer 1

In general, yes, try them all.

But although it is more art than science looking at the coefficients $1, -5, 4$ it's "clear" that $1+4 - 5= 0$ so $x^4=x^2 = 1$ will be a solution.

But one thing for this particular issue is that you only have $x^{even}$ terms so $(-x)^{even} = x^{even}$ so if $k$ is a solution so is $k$ so you only need to test half the values. If $1,2$ or $4$ work then $-1,-2$ or $-4$ respectively will work.

Also when you find that $x =1$ and $x=-1$ work you don't have to, although you can, divide by $x-1$ and getting $x^4 - 5x^2+1 = (x-1)(x^3+x^2-4x-4)$ and then dividing by $x-1$ to get $x^4-5x^2+1=(x-1)(x+1)(x^2-4)$.

In this case it is okay two treat $x^2$ as a single variable and note that if $x^2 = 1$ then dividing out $x^2 -1$ we get $x^4 - 5x^2 + 1=(x^2 -1)(x^2 - 4)$.

And then using the old trick (that should become automatic eventually) that $a^2 - b^2 = (a+b)(a-b)$ thatn $x^4 - 5x^2 + 1 = (x^2-1)(x^2-4)= (x-1)(x+1)(x-2)(x+2)$.

.....

My thinking... which is style, not systematic, would be.

$x^4 -5x^2 + 4$ has only even powers so let $u= x^2$ and factor $u^2 - 5u + 4$.

Habbit: Whenever I have $x^2 + \alpha x + \beta$, I look for $ab = \beta; a+b=\alpha$ solutions. Either they are fairly clear or they are not. If they are not I can then try rational root theorem. But $ab = 4$ and $ab= -5$ is is simple: $ab = 4$ means $(a,b) = (\pm 1,\pm 4)$ or $(\pm 2, \pm 2)$ and so $a+ b = \pm 5$ or $\pm 4$. So if $a+b = -5$ I have $(a,b)=(-1,-4)$.

So $u^2 - 5u + 4 = (u-1)(u-4)$.

Plug $x^2 = u$ back in.

$(x^2-1)(x^2- 4)$.

As I said, Looking for $a^2 -b^2=(a+b)(a-b)$ is automatic. But even if it wasn't it is a offshoot of $x^2 + \alpha x - \beta$ where $\alpha =0$. So I look for $-ab =\beta$ and $a+b = 0$. So $a + b = 0$ means $a=-b$. And $-ab =b^2 = \beta$ so $b = \sqrt{\beta}$ and $a=-\sqrt \beta$.

So $x^4 -5x^2 + 4 = (x-1)(x+1)(x-2)(x+2)$>

Answer 2

Hint :

Let $z=x^2$

You will have a quadratic in $z$

Answer 3

By your work easy to check that $\pm1$ and $\pm2$ are roots

and since our polynomial of fourth degree, we have no another roots, which gives the answer: $$\{1,-1,2,-2\}$$

Also, we can solve your equation by the following reasoning: $$x^4-5x^2+4=x^4-x^2-4x^2+4=$$ $$=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2),$$ which gives roots of the equation.