How to factor $x^4 +3x -2$?

Question

I have figured out there is two roots between $0$ and $1 ,-1$ and $-2$ for $x^4 +3x -2 = 0$.

Therefore there should be two factors $(x + a)$ and $(y - b)$ where $a,b \in R^+$. But how to find these $a$ and $b$?

When they found I can find the next factor in $ax^2+bx+c$ form and can check for further factors easily.

Answer 1

Hints:

If it factors, you know the form will $(x^2 + bx \pm 1)(x^2 + cx \mp 2)$. You need a sum of $3$ and need for the cubic term to cancel out..

Now, can you use that and figure out the factors and find $b$ and $c$?

Result: $(x^2-x+2) (x^2+x-1)$

Answer 2

For it to have some "nice" linear factors, the roots must be one of $\pm 1,\pm 2$ (this is due to the rational root theorem). You can quickly check that these are not the roots. The next bet is quadratic factors, i.e., $$(x^4+3x-2) = (x^2+ax+b)(x^2+cx+d)$$ Expanding the right hand side gives us \begin{align} a+c & = 0\\ b + d + ac & = 0\\ ad+bc & = 3\\ bd & = - 2 \end{align} This gives us the factors to be $$(x^2-x+2) \text{ and } (x^2+x-1)$$

Answer 3

Over $\rm\,\Bbb Q\!:\:$ Rational Root Test excludes linear factors. Testing for quadratic factors is easy since the constant term is prime, which greatly constrains possible factors. We prove a more general case. Yours is $\rm\: p = 2,\:$ so $\rm\:3 = b = -a(p\!+\!1) = -3a\:\Rightarrow\:a = -1,\:$ so $\rm\:2 = p = 1\!+\!sa^2 = 1\!+\!s,\:$ so $\rm\:s = 1.$

Lemma $\ $ If $\rm\ f(x) = x^4 + b x-p\ $ has a quadratic factor $\rm\,\in \Bbb Q[x],\:$ and $\rm\:p\:$ is prime then

$$\rm f(x)\, =\, x^4\! -\! a(p\!+\!1)\, x\! -\! p\, =\, (s\,x^2\! +\! a\,x\!+\!p)(\color{#c00}s\,x^2\!-\!a\,x\!\color{#0a0}{-\!1}),\quad p = 1\!+\!sa^2,\ \ s =\pm1 $$

Proof $\ $ Invoking Gauss's Lemma, we may assume that it splits into monic quadratics $\rm\,\in\Bbb Z[x].\:$ Since $\rm\:p\:$ is prime, one of the factors has constant term $\rm\:\pm p,\:$ so scaling it by $\rm\,\pm1\,$ it will have the form $\rm\:s\,x^2\! +\! a\,x\!+\!p,\,\ s=\pm1 \:$. Comparing coef's, its cofactor must have leading coef $\rm\,= \color{#c00}s,\:$ constant coef $= \color{#0a0}{-1},\:$ and linear coef $\rm\ a' = - a,\:$ since the coef of $\rm\:x^3\:$ in the product $\rm\, = (a+a')s = 0.$ Finally, multiplying the two factors and comparing coef's yields the result. $\ \ $ QED