# Determining the number of real roots of a certain function

I apologize in advance for asking a homework question, but I have genuinely no idea on how to approach part b). The question is as follows:

(a) Show that the polynomial expression $$x^4 -x^2 + x +\frac{5}{4}$$ can be written in the form $$(x^2-a)^2 + (x-b)^2$$

(b) Hence, determine the number of real roots of $$y=6x^5 - 10x^3 +15x^2 + 300x +30$$

I know how to do part a) and got $$(x^2-1)^2+(x+\frac{1}{2})^2=x^4 -x^2 + x +\frac{5}{4}$$, but I am unsure of how to do part b) at all. It would be great if in your answer you could explain some of the logic and reasoning behind each step.

Thank you

• Hint. Look at the derivative of the quintic. When you solve the problem you can post an answer to your own question, and accept it. – Ethan Bolker Apr 20 at 15:28
• Interesting I get $x^4-x^2+x+10=0$ I'll see what I can do from there, thanks! – sab hoque Apr 20 at 15:30
• Observe that $$y'(x)= 30 \left (x^4-x^2+x + \frac 5 4 \right ) + 300 - \frac {75} {2} = 30 (x^2-1)^2 + 30 \left (x+\frac 1 2\right )^2 + \frac {525} {2} >0.$$ So $y(x)$ is strictly increasing. – Dbchatto67 Apr 20 at 15:32
• Exactly one root @sab hoque. – Dbchatto67 Apr 20 at 15:34
• @sab hoque: Standard welcome for new member. Welcome aboard. – dantopa Apr 20 at 15:40

## 1 Answer

An interesting observation to make is that when the polynomial is differentiated, a very nice term comes out \begin{align} y^\prime &= 30x^4 -30x^2 +30x +300 \\ &=30\left(x^4 -x^2 +x +10\right) \\ &=30\left((x^2-1)^2 +\left(x+\frac{1}{2}\right)^2 +\frac{35}{4}\right)>0\tag{from part (a)} \end{align} Therefore the function $$y=6x^5 - 10x^3 +15x^2 + 300x +30$$ is monotonically increasing (strictly increasing) for all $$x\in R$$. This suggests that the curve is of the sort: A curve of this form can only have one root

• You should also mention that a polynomial of odd degree has at least one root by intermediate value theorem. But in this case since the polynomial is strictly increasing so it can have at most one real root. Combining these two arguments it follows that the given polynomial $y(x)$ has exactly one real root. – Dbchatto67 Apr 20 at 16:10
• @Dbchatto67 I'm not sure what intermediate value theorem is, but I assume it is that $f(-1)<0$ and $f(0)>0$ so there must be a root in between? – sab hoque Apr 20 at 16:17
• Precisely it is. See here en.m.wikipedia.org/wiki/Intermediate_value_theorem – Dbchatto67 Apr 20 at 16:21