Squares R.I.P

We see that $f(x)=5x^2$ for $x=1,2,3$ , now this implies that $f(x)=a(x-1)(x-2)(x-3)+5x^2$ we see that $f(x)$ has constant term as $6a$ therefore $3x \times f(x)$ has no constant term and nor does $2x^2$ but $[f(x)]^2$ is a $6$ degree polynomial with constant term as $(6a)^2$

Now the product of roots of $[f(x)]^{2} + 3x \ f(x) + 2x^{2} = 0$ is $\dfrac{\text{Constant term}}{\text{coefficient of leading term}}$ i.e. $\dfrac{(6a)^2}{a^2}=\color{#3D99F6}{\boxed{\color{#20A900}{\boxed{(\color{#69047E}{(36)}}}}}$

Now in the given expression, it is pretty obvious that $3x \ f(x)$ has no constant number output and neither does $2x^{2}$

Now $[f(x)]^{2}\ = \ f(x)\times f(x)$

$f(x) \ = \ ax^{3} \ + \ bx^{2} \ + \ cx \ + \ d$

Thus the constant term in $[f(x)]^{2}$ must be $d^{2}$ as it must not contain any non zero power of $x$ as it's coefficient.

Thus the product of roots in this $6$ degree polynomial must be $\frac { { d }^{ 2 } }{ { a }^{ 2 } }$ as the leading coefficient will be $a^{2}$

Now we are given $f(1)$ , $f(2)$ and $f(3)$ .

Thus

$a\quad +\quad b\quad +\quad c\quad +\quad d\quad =\quad 5\\ 8a\quad +\quad 4b\quad +\quad 2c\quad +\quad d\quad =\quad 20\\ 27a\quad +\quad 9b\quad +\quad 3c\quad +\quad d\quad =\quad 45\\$

Eliminating $b$ and $c$ using the three equations, we get the ratio of $d$ to $a$ as $-6$

Thus $\frac { { d }^{ 2 } }{ { a }^{ 2 } } \quad =\quad 36\quad$

Consider polynomial (x-1)(x-2)(x-3)+5x^2 F(1)=5 F(2)=20 F(3)=45 Simplifying f(x)=x^3-6x^2+11x-6 Using this we get +36 as the constant term of equation., which is equal to product of roots

$f(x) = ax^3 + bx^2 + cx + d$ is a cubic polynomial implies that $g(x) = [f(x)]^2 + 3xf(x) + 2x^2$ is a $6^{th}$ degree polynomial with leading coefficient $a^2$ and constant coefficient $d^2$ . By the Vieta relations, the product of the roots of $g$ is $\dfrac{d^2}{a^2}$ .

$f(1) = 1, f(2) = 20$ , and $f(3) = 45$ give rise to the following system of equations:

$\begin{cases} a + b + c + d = 5 \\ 8a + 4b + 2c + d = 20\\ 27a + 9b + 3c + d = 45 \end{cases}$

Multiply the first equation by $3$ , the second equation by $-3$ , and add all the equations of the new system to get $d = -6a$ , which implies $\dfrac{d^2}{a^2} = 36$ .

Let us assume f(x)= x^3 +ax^2+ bx+ c 1+b+c+d=5....... 8+4b+2c+d=20.......... 27+9b+3c+d=45........... We get b=-1 c=11 d=-6. Equation-x^3 - x^2 +11x-6. so product of roots -: Constant term/coefficient of x^6 =36/1=36 *we will get the constant term only from [f(x)]^2. Also x^6 is generated from [f(x)]^2 ONLY.