Let $f$ be a polynomial function of degree 3 satisfying $f(1) = 3$, $f(2) = 5$, and $f(3) = 7$. If the product of the roots of the equation $(f(x))^2 + 4x f(x) + 3x^2 = 0$ is 4 and the sum of all possible values of $f(4)$ is $k$, then find $[k]$. [Note: $[y]$ denotes the greatest integer function less than or equal to $y$.]

Road-map of the solution

1. Represent the family of cubics that pass through the three given points. (One free parameter remains.)
2. Rewrite the big expression as a product to see two simpler cubic equations.
3. Use Vieta's formula on each cubic to link the parameter to the product of their roots.
4. Match the given product (4) to find all allowed parameter values.
5. Evaluate $f(4)$ for every allowed parameter and add the results.

1. Cubic family through three points

A cubic $f(x)$ can be written as
$f(x) = 2x + 1 + t\,(x-1)(x-2)(x-3)$
Why? $2x+1$ is the simple straight line that already goes through the three points. Multiplying by the factor $(x-1)(x-2)(x-3)$ keeps those points fixed (because it vanishes at each of them) and the constant $t$ lets us tilt the cubic up or down—this is the one free knob.

2. Factor the complicated expression

Set
$(f(x))^2 + 4x f(x) + 3x^2 = 0$
Treat $f(x)$ as $y$ for a moment:
$y^2 + 4xy + 3x^2 = (y + x)(y + 3x).$
Hence a root x must satisfy either
$f(x) + x = 0 \quad\text{or}\quad f(x) + 3x = 0.$
Each is a cubic equation in $x` because$f$ is cubic.

3. Products of roots via Vieta

Let $g(x) = f(x)+x, \qquad h(x)=f(x)+3x.$
Both share the same leading part $t(x-1)(x-2)(x-3)$, so their leading coefficient is $t$.
For any cubic $a x^3 + \dots + d$, the product of roots is $-\dfrac{d}{a}$. Evaluate constants: $g(0)=1-6t, \quad h(0)=1-6t.$
Hence $\text{Product of roots of }g = 6 - \frac{1}{t}, \qquad \text{Product of roots of }h = 6 - \frac{1}{t}.$
The big equation’s six roots combine the two sets, so $\left(6 - \frac{1}{t}\right)^2 = 4.$

4. Solve for $t$

Take square roots:
$6 - \frac{1}{t} = \pm 2.$ Two possibilities:

• $6 - \dfrac{1}{t} = 2 \;\Rightarrow\; t=\dfrac{1}{4}$
• $6 - \dfrac{1}{t} = -2 \;\Rightarrow\; t=\dfrac{1}{8}$

5. Compute $f(4)$

Using $f(x)=2x+1+ t\, (x-1)(x-2)(x-3)$ and $(4-1)(4-2)(4-3)=3\times2\times1=6$: $f(4)=9+6t.$

• For $t=\dfrac{1}{4}$: $f(4)=9+\dfrac{6}{4}=10.5$
• For $t=\dfrac{1}{8}$: $f(4)=9+\dfrac{6}{8}=9.75$ Sum: $k = 10.5 + 9.75 = 20.25$

Greatest integer $\, [k] = 20$.

What is the question?

We are given a mystery curve that is drawn by a cubic‐shape formula (degree 3). We know three points on that curve: (1,3), (2,5) and (3,7).
From that cubic we build another expression:

(f(x))² + 4x·f(x) + 3x²

We are told that the x-values that make this expression zero multiply together and give 4. We must work out all possible values of the mystery cubic at x = 4, add them, and then take the greatest integer smaller than (or equal to) that sum.

How to think of it (kid analogy)

Imagine you have a rubber band stretched through three nails in a board (those three points). You can still wiggle the band a bit and keep it touching the same three nails (because a cubic has one extra degree of freedom). After each wiggle you look at a special gadget which says beep whenever

(y + x)(y + 3x) = 0

(because the big expression factors that way). The total multiplication of all x where it beeps must be 4. Only certain wiggles satisfy that rule, and we must find what height the band reaches at x = 4 for those wiggles.

Step-by-step calculation

1. General cubic through the three given points

   $$f(x)=2x+1 + t\,(x-1)(x-2)(x-3) \qquad (t\in\mathbb R)$$

2. Rewrite the given equation

   $$(f(x))^2 + 4x f(x) + 3x^2 = (f(x)+x)(f(x)+3x)=0.$$

   Therefore solve $g(x)=f(x)+x=0 \quad \text{or} \quad h(x)=f(x)+3x=0.$

3. Form $g(x)$ and $h(x)$

   $$g(x)=t(x-1)(x-2)(x-3)+3x+1,\qquad h(x)=t(x-1)(x-2)(x-3)+5x+1.$$

4. Product of roots of each cubic The leading coefficient of both $g$ and $h$ is $t$.
   Evaluate at $x=0$:

   $$g(0)=h(0)=1-6t.$$

   Hence (Vieta)

   $$\text{Product}(g)=-(1-6t)/t=6-1/t,\qquad \text{Product}(h)=6-1/t.$$

5. Condition from the problem

   $$(6-1/t)^2 = 4 \;\Longrightarrow\; 6-1/t=\pm2.$$

   [Case\ 1] $6-1/t=2 \;\Rightarrow\; t=1/4$
   [Case\ 2] $6-1/t=-2\;\Rightarrow\; t=1/8$

6. Compute $f(4)$ for each valid $t$

   $$f(4)=9+6t.$$

   • $t=1/4 \Rightarrow f(4)=9+1.5=10.5$
   • $t=1/8 \Rightarrow f(4)=9+0.75=9.75$

   Sum: $k=10.5+9.75=20.25$.

7. Greatest integer value

   $$[k] = [20.25] = 20. \quad\boxed{20}$$