Are you smarter than me? 33

$f(x)$ seems a biquadratic polynomial with sum of roots as $\frac{10}{3}$ . But since two or all roots of $f(x)$ may be imaginary, we need to find out the real roots.

By hit and trial method, $3$ is a root of $f(x)$ . By factor theorem, $(x-3)$ is a factor of $f(x)$ .

$f(x)=(3x^3-x^2+x+2)(x-3)$ Again by hit and trial, $\frac{-2}{3}$ is a root.

So $f(x)=(x-3)(3x+2)(x^2-x+1)$

Hence only real roots are $3$ and $\frac{-2}{3}$ . So the sum would be $\frac{7}{3}$ .

$a=7,b=3$

$a+b=\boxed{10}$

$\Large{Hint}$ : Assume that it can be expressed as $\large{(x^2+kx+l)(x^2-kx+m)}$ then you may compare coefficients

$\large{f(x)} = 3x^4-10x^3+4x^2-x-6.$

By trial and error method , $(x-3)$ is a factor of $\large{f(x)}.$

${f(x)} = (x-3)(3x^3-x^2+x+2)$

Again by trial and error method , $\dfrac{-2}{3}$ is another root.

${f(x)} = (x-3)(3x+2)(x^2-x+1)$

We can observe that the only real roots are $3$ , $\dfrac{-2}{3}$ .

Therefore sum of the roots is $3+(\frac{-2}{3}$ ) which is $\dfrac{7}{3}.$

Here , $a = 7$ and $b = 3.$

Therefore $a + b = 7 + 3 = 10.$

Therefore the answer is $10.$

Real roots are 3 and - 2/ 3. Therefore, sum up to 7/ 3.

Answer: 10

$f(x)$ is a biquadratic polynomial, so I tried finding a roots by Rational root theorem .

I found $f(\frac{-2}{3}) = 0, f(3) = 0$

Then use Synthetic Division, you will realize that the other polynomial which is a factor of $f(x)$ is $x^2 - x + 1$ , which has no real roots.

So, $\frac{a}{b} = -\frac{2}{3} + 3 = \frac{7}{3}$

$\boxed{a+b=\color{#D61F06}{10}}$

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