A single root

We are given $p$ is a root of $x^2+2x+4=0$ which implies that $p^2+2p+4=0$ ,

We need $3p^5+6p^4+3p^3$ , dividing it by $p^2+2p+4$ we get,

$3p^5+6p^4+3p^3=(3p^3-9p)(p^2+2p+4)+18p^2+36p$

$3p^5+6p^4+3p^3=0+18p(p+2)$

$3p^5+6p^4+3p^3=-72$

The roots are $2e^{\pm\frac{i2\pi}3}=-1\pm \sqrt 3i$ by completing the squares (**) . The required expression is:-

$\begin{aligned}3\color{#20A900}{p}^3(\color{#D61F06}{p}+1)^2=& 3\left(\color{#20A900}{2e^{\pm\frac{2\pi i}3}}\right)^3(\color{#D61F06}{\cancel{-1}\pm\sqrt 3i}+\cancel{1})^2\\=&3\times 8 e^{\pm2\pi i}(-3)=\boxed{-72}\end{aligned}$

We have :- $x^{2}+2x+4=0\,\implies \, \color{#D61F06}{x^{2}=-2x-4}\,\implies \,x^{3}=-2x^{2}-4x\,\implies\,x^{3}=-2\left(\color{#D61F06}{-2x-4} \right)-4x\,\implies \, \color{#3D99F6}{x^{3}=8}$ .

Now, $x^{2}+2x+4=0\, \implies \, x^{2}+2x+1=-3\,\implies\,x^{3} \cdot (x^{2}+2x+1)= \color{#3D99F6}{x^{3}} \cdot -3\,\implies \, x^{5}+2x^{4}+x^{3}=-24\,\implies \,3x^{5}+6x^{4}+3x^{3}=3(-24)=-72$ .

If $p$ is a root to the given quadratic equation, then $\boxed{\color{#EC7300}{3p^{5}+6p^{4}+3p^{3}}}=-72$ .

Infact, we can generalize the given expression for non - negative integral exponents as :- $\huge \boxed{\color{#3D99F6}{3 \cdot p^{3n+2}+6 \cdot p^{3n+1} +3 \cdot p^{3n}\,=\,-9\cdot8^{n}}}$ .

Since $p$ is a root of $x^2+2(x+2)=0$ , then $p^2 + 2(p+2)=0$ .

$\begin{aligned} p^2 + 2(p+2) & = 0 \\ p^2 + 2p+4 & = 0 \\ \implies p^2 & = -2p-4 \\ p^3 & = -2p^2 - 4p = -2(-2p-4) - 4p = 4p+8 -4p = 8 \\ p^4 & = 8p \\ p^5 & = 8p^2 = - 16p - 32 \end{aligned}$

$\begin{aligned} \implies 3p^5 + 6p^4 + 3p^3 & = 3( - 16p - 32) + 6(8p) + 3(8) \\ & = -48p -96 + 48p + 24 \\ & = \boxed{-72} \end{aligned}$

As , $p^2+2p+4 = 0 .......(1)$

$p^2+2p+1 = -3$

$(p+1)^2 = -3$

$p = ±\sqrt{3}i-1 = 2e^{±\dfrac{2\pi}{3}}$

Multyplying eqn (1) by $3p^3$ ,

$3p^5+6p^4+12p^3 = 0$

$3p^5+6p^4+3p^3=-9p^3 = -9(8e^{±2\pi i}) = -9(8) = \boxed{-72}$

Method 1

Given that $p$ is a root of the equation, we know that

$p^2 + 2(p+2) = 0\\ p^2 + 2p + 4 = 0$

$3p^5 + 6p^4 + 3p^3\\ =3p^3(p^2+2p+1)\\ =3p^3(\color{#3D99F6}{p^2+2p+4}-3)\\ =-9p^3\\ =-9p(p^2)$

From the equation, we know that $p^2 = -2p -4$ . Substitute this in:

$-9p(p^2)\\ =-9p(-2p-4)\\ =18p^2+36p\\ =18(p^2+2p)\\ =18(\color{#3D99F6}{p^2+2p+4}-4)\\ =18(-4)\\ =\boxed{-72}$

---

Method 2

Find the value of $p$ :

$p=\dfrac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)}\\ =\dfrac{-2 \pm \sqrt{-12}}{2}\\ =-1 \pm \sqrt{-3}\\ =-1 \pm \sqrt{3}i$

Substitute any one of these roots in (you can try this with $-1 -\sqrt{3}i$ , your answer will still be the same):

$3p^5 + 6p^4 + 3p^3\\ =3p^3(p^2+2p+1)\\ =3(-1 + \sqrt{3}i)^3\left((-1+\sqrt{3}i)^2 + 2(-1 + \sqrt{3}i) + 1\right)\\ =3(-1 +3\sqrt{3}i +9 -3\sqrt{3}i)(1 - 2\sqrt{3}i - 3 - 2 + 2\sqrt{3} i + 1)\\ =3(8)(-3)\\ =\boxed{-72}$

$Since \ p\ is\ a\ root\ of\ x^2+2(x+2)=0, \ \ p^2 + 2(p+2)=0.\\ \implies \ \color{#3D99F6}{p^2=( - 2p-4),\ \ or\ \ p^2+2p= - 4}\\ 3p^5+6p^4+3p^3 \\ =3*(p^2)*p*(\color{#3D99F6}{p^2+2p+4}-3)\\ =3*(\color{#3D99F6}{-2p-4})*p*(0-3)\\ = 3*(\color{#3D99F6}{ - 2)*(p^2+2p})*(0-3)\\ = -6*(-4)*(-3)\\ = \ \ \ \ \ \ \ \ \Large \ \ \ \color{#D61F06}{-72}.$