Is Vieta's practical?

neclect the x=0 solution. $4*0^2+1=1$ which doesnt affect the product. $f(x)=5x^4+4x^3+3x^2+2x+1=5(x-a_1)(x-a_2)(x-a_3)(x-a_4)$ . remember this:the basic definition of a polynomial.we have the given expression equal to $25(4^4(\frac{i}{2}-a_1)(-\frac{i}{2}-a_1)(\frac{i}{2}-a_2)(-\frac{i}{2}-a_2)...(\frac{i}{2}-a_{4})(-\frac{i}{2}-a_{4}))=4^4(5(\frac{i}{2}-a_1)(\frac{i}{2}-a_2)....(\frac{i}{2}-a_{4}))(5(-\frac{i}{2}-a_1)(-\frac{i}{2}-a_2)....(-\frac{i}{2}-a_{4}))$ this is equal to by our above def: $4^4f(\frac{i}{2})f(-\frac{i}{2})$ it is not hard to compute, you can use AGP or differentiate GP. either ways, the answer will just be $4^4(\frac{-8+9i}{16})(\frac{-8-9i}{16})=8^2+9^2=\boxed{145}$

(Adapted from Aareyan Manzoor's method. Rewritten for clarity.)

Notice that the given equation has a root $x=0$ , which can be ignored since in the given expression, $4(0)^2+1 = 1$ , which is the multiplicative identity. Hence let $a_5 = 0$ , and $\displaystyle f(x) = \sum_{m=1}^5 mx^{m-1} = 5 \prod_{n=1}^4 (x-a_n)$ .

Applying the identity $(a+ib)(a-ib) = a^2+b^2$ , we can reform the given expression:

$\begin{aligned} 25 \prod_{n=1}^4 (4a_n^2+1) &= 25 \prod_{n=1}^4 (2a_n+i)(2a_n-i) \\ &= 25 \prod_{n=1}^4 \left[ 2 \left( -\frac{i}{2}-a_n \right) \right] \prod_{n=1}^4 \left[ 2 \left( \frac{i}{2}-a_n \right) \right] \\ &= \left[ 2^4 f \left( -\frac{i}{2} \right) \right] \left[ 2^4 f \left( \frac{i}{2} \right) \right] \\ &= 2^8 f \left( -\frac{i}{2} \right) f \left( \frac{i}{2} \right). \end{aligned}$

As has been noted, the above may be computed easily. For variety's sake, I will present a method besides AGP/differentiating GP, which uses the aforementioned identity:

$\begin{aligned} 2^8 f \left( -\frac{i}{2} \right) f \left( \frac{i}{2} \right) &= 2^8 \left[ \sum_{m=1}^5 m\left(-\frac{i}{2}\right)^{m-1} \right] \left[ \sum_{m=1}^5 m\left(\frac{i}{2}\right)^{m-1} \right] \\ &= 2^8 \left[ 5\left(\frac{1}{2}\right)^4 - 3\left(\frac{1}{2}\right)^2 + 1 + 4i\left(\frac{1}{2}\right)^3 - 2i\left(\frac{1}{2}\right)\right] \left[ 5\left(\frac{1}{2}\right)^4 - 3\left(\frac{1}{2}\right)^2 + 1 - 4i\left(\frac{1}{2}\right)^3 + 2i\left(\frac{1}{2}\right)\right] \\ &= 16^2 \left[ \left( \frac{5}{16} - \frac{3}{4} + 1 \right)^2 + \left( \frac{4}{8} - \frac{2}{2} \right)^2 \right] \\ &= \left( 5 - 12 + 16 \right)^2 + \left( 8 - 16 \right)^2 \\ &= 9^2+8^2 \\ &= \boxed{145}. \end{aligned}$

Here's another solution without using complex numbers:
One of the roots of the polynomial is $0$ which leaves us with $5x^4+4x^3+3x^2+2x+1 = 5(x-a_1)(x-a_2)(x-a_3)(x-a_4)$

$\implies \frac{d}{dx}(1+x+...+x^5)=\frac{5x^6-6x^5+1}{(x-1)^2}=5(x-a_1)(x-a_2)(x-a_3)(x-a_4)$

$5x^6-6x^5+1 = 5(x-1)^2(x-a_1)(x-a_2)(x-a_3)(x-a_4)$
Replacing $x$ in $5x^6-6x^5+1$ with $\sqrt{x}$ and rationalising, we get $25x^6-36x^5+10x^3+1 = 25(x-1)^2(x-a_1^2)(x-a_2^2)(x-a_3^2)(x-a_4^2)$
Substitute $x= {-1\over 4}$ to get the required value.

$\begin{aligned} \sum_{m=1}^5 mx^m & = 0 \\ \Rightarrow x + 2x^2 + 3x^3 + 4x^4 + 5x^5 & = 0 \\ x(5x^4 + 4x^3 + 3x^2 + 2x + 1) & = 0 \end{aligned}$

This implies that one of the roots is $0$ and let it be $a_5 = 0$ . Then we have:

\begin{aligned} 25 \prod_{m=1}^5 (4a_m^2 +1) & = 25 \prod_{m=1}^\color{#3D99F6}{4} (4a_m^2 +1) \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Since }4a_5^2 + 1 = 1} \\ & = 25 \prod_{m=1}^4 (2a_m^2 +i) (2a_m^2 -i) \\ & = 25 \prod_{m=1}^4 (2a_m^2 +i) \prod_{m=1}^4 (2a_m^2 -i) \\ & = 25 \left(16\prod_{m=1}^4 a_m +8i\sum_{cyc}a_1a_2a_3 -4\sum_{cyc}a_1a_2 -2i\sum_{m=1} ^4 a_m + 1 \right) \\ & \quad \quad \quad \times \left(16\prod_{m=1}^4 a_m - 8i\sum_{cyc}a_1a_2a_3 -4\sum_{cyc}a_1a_2 + 2i\sum_{m=1} ^4 a_m + 1 \right) \\ & = 25 \left(16 \color{#3D99F6}{\left(\frac{1}{5}\right)} +8i\color{#3D99F6}{\left(-\frac{2}{5}\right)} -4 \color{#3D99F6}{\left(\frac{3}{5}\right)} -2i\color{#3D99F6}{\left(-\frac{4}{5}\right)} + 1 \right) \\ & \quad \quad \quad \times \left(16 \color{#3D99F6}{\left(\frac{1}{5}\right)} - 8i\color{#3D99F6}{\left(-\frac{2}{5}\right)} -4 \color{#3D99F6}{\left(\frac{3}{5}\right)} + 2i\color{#3D99F6}{\left(-\frac{4}{5}\right)} + 1 \right) \quad \quad \small \color{#3D99F6}{\text{By Vieta's formulas}} \\ & = \frac{25}{25}(16 - 16i - 12 + 8i + 5) (16 + 16i - 12 - 8i + 5) \\ & = (9 - 8i)(9+8i) = 81 + 64 = \boxed{145} \end{aligned}