# Thread: Hey, I've got another exercise ...

1. ## Hey, I've got another exercise ...

We have the polynomial $\displaystyle p(x) = x^3 + 7x^2 + (s + 12)x + s$. This $\displaystyle p(x)$ should not have the complex roots (it needs to have just real roots from real number). Find the values $\displaystyle s$.
How I find this value $\displaystyle s$?

2. ## Re: Hey, I've got another exercise ...

I would probably be inclined to try to factorise it by grouping...

\displaystyle \begin{align*} x^3 + 7x^2 + \left( s + 12 \right) x + s &= x^3 + \left( s + 12 \right) s + 7x^2 + s \\ &= x \left( x^2 + s + 12 \right) + 7 \left( x^2 + \frac{s}{7} \right) \end{align*}

so now in order to have a common factor, that means we have to have

\displaystyle \begin{align*} s + 12 &= \frac{s}{7} \\ 7s + 84 &= s \\ 6s &= -84 \\ s &= -14 \end{align*}

Now just to check:

\displaystyle \begin{align*} x^3 +7x^2 + \left( -14 + 12 \right) x - 14 &= x^3 + 7x^2 - 2x - 14 \\ &= x^3 - 2x + 7x^2 - 14 \\ &= x \left( x^2 - 2 \right) + 7 \left( x^2 - 2 \right) \\ &= \left( x^2 - 2 \right) \left( x + 7 \right) \\ &= \left( x - \sqrt{2} \right) \left( x + \sqrt{2} \right) \left( x + 7 \right) \end{align*}

So \displaystyle \begin{align*} s = -14 \end{align*} definitely works. I don't know if there are any other possible values you could have for s though.

3. ## Re: Hey, I've got another exercise ...

Let $\displaystyle x = y-\dfrac{7}{3}$. Then:

\displaystyle \begin{align*}x^3+7x^2+(s+12)x+s & = \left(y-\dfrac{7}{3}\right)^3+7 \left(y-\dfrac{7}{3}\right)^2+(s+12)\left(y-\dfrac{7}{3}\right)+s \\ & = y^3+\left(s-\dfrac{13}{3}\right)y-\dfrac{36s+70}{27}=0\end{align*}

Next, let

$\displaystyle 3ab = s-\dfrac{13}{3}$
$\displaystyle a^3-b^3 = \dfrac{36s+70}{27}$

So, solving for $\displaystyle b^3$ will give a quadratic. From that, I think you can tell what values are going to be complex or not...

4. ## Re: Hey, I've got another exercise ...

Why do you say let $\displaystyle x = y - \dfrac{7}{3}$? Also for $\displaystyle s = 0$ works ...