Compute the largest root of $x^4-x^3-5x^2+2x+6$

Question

I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval $]2,\tfrac52[$. The answer is $\frac{1+\sqrt{13}}{2}$. But how can I derive that exact result?

Furthermore, to know that it is the largest root, I must find the other two roots, must not I? By sign shift I know one root between $1$ and $2$, but this method cannot give me the third root (the graph touches the $x$-axis). Maybe calculus? Using calculus we get a degree polynomial of degree which also is hard to find roots to. We get $p'(-1)=5$ and $p'(-2)=-22$ so we have a stationary point in $]-2,-1[$ and similarly we get one in $]1,2[$. Since it is two stationary points it must have one maximum and one minimum point. But how do I know which is which, I mean I cannot plug in a value in the second derivative.

Can somebody help me? Thanks in advance

score 3 · Accepted Answer · answered May 01 '15 at 17:38

3

One way is to set $$x^4-x^3-5x^2+2x+6=(x^2+ax+b)(x^2+cx+d)$$ where $a,b,c,d$ are integers such that $|b|\gt |d|$. Having $$-1=c+a$$ $$-5=d+ac+b$$ $$2=ad+bc$$ $$6=bd$$ will give you $a=-1,b=-3,c=0,d=-2$, i.e. $$x^4-x^3-5x^2+2x+6=(x^2-x-3)(x^2-2).$$

answered May 01 '15 at 17:38

mathlove

139,939

Looks sweet! Factorization worked too though. but this is what may benefit in future. Nice work! – Kushashwa Ravi Shrimali May 01 '15 at 17:55
@KushashwaRaviShrimali: I don't understand your comment. This is factorization! – TonyK May 01 '15 at 18:05
I'm sorry, from that I meant, factorization by evaluating roots. (check out my answer). Tbh, I got my wording wrong there. Pardon me for that. I just meant, another way to do it. – Kushashwa Ravi Shrimali May 01 '15 at 18:06

score 2 · Answer 2 · answered May 01 '15 at 17:34

If you already know that $\frac{1+\sqrt{13}}2$ is a root, you could try to divide $p(x)$ by $x^2-x-3$ ("the" polynomial with $\frac{1+\sqrt{13}}2$ as root). This will give you another quadratic polynomial, so that you can inded find all roots of $p$ explicitly.

score 1 · Answer 3 · answered May 01 '15 at 17:35

1

You can see that: $$ x^4−x^3−5x^2+2x+6=(x^2-2)(x^2-x-3)$$

answered May 01 '15 at 17:35

Elaqqad

13,725

score 0 · Answer 4 · answered May 01 '15 at 17:49

You've : $p(x) = x^4 - x^3 - 5x^2 + 2x + 6 $ .

Rearrange the given polynomial as : $$ p(x) =(x^4 - 5x^2 + 6) - x(x^2 - 2) $$

For the first term, let us factor it separately. Say, $x^2 = t$ . So, you've :

$$\begin{align} f(x) =&\ x^4 - 5x^2 + 6\\ f(x) =&\ t^2 - 5t + 6 \end{align}$$

Roots of $f(x)$ are : $t = 3 \ \text{OR} \ 2 $ . Now, plugging-in $t = x^2$ back, we get: $x = \pm \sqrt{3},\pm \sqrt{2}$ .

So, we can write $f(x)$ as : $f(x) = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$

And thus, $p(x)$ can be written as:

$$\begin{align} p(x) =& (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2}) - x(x^2 -2) \\ =& (x^2 - 3)(x^2 -2 ) -x(x^2-2) \\ =& (x^2-2)(x^2 - x - 3) \end{align}$$

Therefore, you've : $p(x) = (x^2 - 2)(x^2 - x - 3)$

You can proceed from here, I guess. Hope it helps.

Compute the largest root of $x^4-x^3-5x^2+2x+6$

4 Answers4