Polynomials with integer coefficients

Notice that we can factor $x^5-x^2-x-1$ as follows: $\begin{aligned} x^5-x^2-x-1 &= (x^5-x) - (x^2+1) \\ &= x(x^4-1) - (x^2+1) \\ &= x(x^2-1)(x^2+1) - (x^2+1) \\ &= [x(x^2-1) - 1](x^2+1) \\ &= (x^3-x-1)(x^2+1). \end{aligned}$ Neither of these polynomials are further factorable over the integers, so $f(x)$ and $g(x)$ must be $x^3-x-1$ and $x^2+1$ in some order. We let $x=3$ in both polynomials to get the values $23$ and $10,$ respectively. Thus, $\left|f(3) + g(3)\right| = \left|23+10\right| = \boxed{33}.$

The given equation can be factorized as follows :

$x^5-x^2-x-1=x(x^4-1)-(x^2+1)$

$=x(x^2+1)(x^2-1)-(x^2+1)$

$=(x^2+1)\{x(x^2-1)-1\}=(x^2+1)(x^3-x-1)$ . Let $f(x)=x^2+1$ and $g(x)=x^3-x-1$ . Hence , $f(3)=10$ and $g(3)=23$ . Therefore , $|f(3)+g(3)|=33$ .

Note that $x=i$ and $x=-i$ are roots of $f(x)*g(x)$ , so $x^2+1$ is a factor.

Calculating $x^5-x^2-x-1=(x^2+1)(x^3-x-1)$ gives us the formulas for $f(x)$ and $g(x)$ .

This results in $f(3)+g(3)=10+23=33$

For this explanation, $f(3) = f$ and $g(3) = g$

We know that $|f + g| = \sqrt{(f + g)^{2}}$ . Therefore, we can expand this equation to $\sqrt{f^{2} + 2fg +g^{2}}$ . It is given that $fg = 230$ , so we can simplify the equation to $\sqrt{460 + f^{2} + g^{2}}$ .

Because $fg = 230$ we know that the possible pairs of $f$ and $g$ are $(1,230), (2,115), (5,46)$ , and $(10,23)$ . By plugging each one in, we get the possible answers $231, 117, 51$ , and $33$ respectively.

$4$ possible answers, $3$ guesses...we've got a 75% chance of picking the correct one =P. I'd love to know if this can be tweaked to get a definite answer!

Let's factor out $f(x) \times g(x)$ ...

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$\large{x^5-x^2-x-1}$

$\large{=x^5+x^3-x^3-x-x^2-1}$

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

$\large{=(x^2+1)(x^3-x-1) ~~~ \cdot \cdot \cdot (*)}$

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The question says that $f(x)$ and $g(x)$ are two non-constant polynomials with integer coefficients... We can see that $(*)$ can't be further factored into polynomials with integer coefficients...

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Thus, we can conclude that, one of the factors in $(*)$ is $f(x)$ and the other is $g(x)$ ... Since the question seeks the value of $|f(3)+g(3)|$ , that is, for the same value of $x$ , hence we don't have to care about which one is $f(x)$ and which one is $g(x)$ ...

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We just simple plug in $3$ instead of $x$ in both of $(x^2+1)$ and $(x^3-x-1)$ , and get...

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$\large{x^2+1=3^2+1=9+1=10}$

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$\large{x^3-x-1=3^3-3-1=27-4=23}$

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Therefore, the value of $|f(x)+g(x)|$ is $|10+23|=|33|=\large{\fbox{33}}$

If this monic polynomial had a rational root, then this root is an integer, and divides the constant coefficient. One can check that +1 and -1 aren't roots, so f and g must be a quadratic and a cubic polynomial in some order. Because of the small coefficients, it is worth trying some complex unit roots, for instance, i (and its conjugate -i) are roots, so using polynomial division, we get the factorization (x^2+1)(x^3-x-1). This is unique because x^3-x-1 is irreducible over Q. Hence f(x) and g(x) are x^2+1 and x^3-x-1, f(3)+g(3)=10+23=33.

Note that $x^5 - x^2 - x - 1$ can be factorised into $(x^2 + 1) * (x^3 - x - 1)$ or $-(x^2 + 1) * -(x^3 - x - 1)$ . It doesn't affect our final answer because the question is asking for the absolute value.

Therefore, let $f(x) = (x^2 + 1)$ and $g(x) =(x^3 - x - 1)$ or vise versa .

Substitute $x=3$ and we have $|f(x)+g(x)| = 10+23 = 33$

x^5 - x^2 - x - 1 = x^5 - x - (x^2 + 1) = x(x^4 - 1) - (x^2 + 1) = x(x^2 + 1)(x^2 - 1) - (x^2 + 1) = (x^2 + 1){x(x^2 - 1) - 1} = (x^2 + 1)(x^3 -x -1)

Hence, f(x) = (x^2 + 1) and g(x) = (x^3 -x -1) Now substitute 3 in place of x and add the two results to get 33.

F(x) x G(x) = x^5 - x^2 - x - 1 can be factored to (x^2 + 1)(x^3 - x - 1). Assuming F(x) = (x^2 + 1), F(3) is equal to 10. This would mean that G(x) = (x^3 - x - 1), so G(3) = 23. We know that our factorization is therefore accurate, as we are given that F(3) x G(3) = 230, which holds true for the above factorization. Therefore, all that we must do to find our solution is F(3) + G(3), which is (3^2 + 1 + 3^3 - 3 - 1), which is equal to 33, our solution.

If $f(x)$ has degree of 1, then it must be either $x+1$ or $x-1$, but since $f(1)g(1)$ and $f(-1)g(-1)$ are not zero, then this can't be true. Thus $f(x)$ has degree of 2 and $g(x)$ has degree of 3. WLOG, we may assume that they're both monic, since if not, we can always flip the sign of all coefficients.

Case 1: $(x^2 + ax + 1)(x^3 + bx^2 + cx - 1)$ Then from equating the coefficients, we have: $a + b = 0$ $1 + ab + c = 0$ $-1 + ac + b = -1$ $c - a = -1$ Because $b=-a=-ac$ then either $a=0$ or $c=1$

If $a=0$ then $b=0,c=-1$ If $c=1$ then $a=b=2$, contradicting the third equation.

Case 2: $(x^2+ax-1)(x^3 + bx^2 + cx + 1)$ Then we have: $a+b = 0$ $-1 + ab + c = 0$ $-b + ac + 1 = -1$ $-c + a = -1$ Because $c =1-ab = 1+a^2$ and $c=1+a$ then $a=a^2$, so $a=0$ or $a=1$. If $a=0$ then $b=0, c=1$, contradicting $-b + ac + 1 = -1$ If $a=1$ then $b=-1, c=2$, contradicting $-b + ac + 1 = -1$

So the only possibility is $(x^2 + 1)(x^3 -x -1)$ which means that $f(x) + g(x) = x^3 + x^2 - x$

We factorize the given equation:

$f(x) \times g(x) = x^{5}-x^{2}-x-1$

$= x(x^{4}-1)-(x^{2}+1)$

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

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

Let us assume $f(x) = x^{2}+1$ and $g(x) = x^{3}-x-1$ . (Note that the assignment of the polynomials could be the other way but this does not affect the final answer as we are concerning about the sum of the polynomials). Replacing $x$ with 3, we would have $f(3) = 10$ and $g(3) = 23$ . Hence the final answer is: $10+23=33$ .

P.S 1 : To be more precise we have to show that the factorization above is unique.

P.S 2 : We know that $f(3) \times g(3) = 230$ . Since both $f(x)$ and $g(x)$ are polynomials with integer coefficients we can be sure that both $f(3)$ and $g(3)$ are integers as well. So the problem is to find two integers that their multiplication results in 230. The prime divisors of 230 is 2, 5 and 23 and the whole set of the divisors of 230 is $\{1, 2, 5, 10, 23, 46, 115, 230\}$ . Clearly, there would be four possible pairs for $f(3)$ and $g(3)$ :

(1, 230), (2, 115), (5, 46) and (10, 23)

Up to here, we know that the final answer would be either 231, 117, 51 or 33. As Brilliant gives you 3 tries per question, you have a decent chance of 75% to guess the right answer :)

$f(x) \times g(x) = x^{5} - x -(x^{2} + 1)$

$f(x) \times g(x) = x(x^{4} - 1) - (x^{2} + 1)$

$f(x) \times g(x) = x(x^{2} + 1)(x^{2} - 1) - (x^{2} + 1)$

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

Hence,

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

$g(x) = x^{3} - x - 1$ and vice versa

Hence, $|f(3) + g(3)| = |3^{2} + 1 + 3^{3} - 3 - 1|$

Hence, $|f(3) + g(3)| = 33$

x^5 - x^2 - x - 1 = (x^2 + 1)(x^3 - x - 1) . We may take f(x) = x^2 + 1 and g(x) = x^3 - x - 1. So |f(3) + g(3))| = 10 + 23 = 33

after factorizing x^5−x^2−x−1 we will get that,

x^5−x^2−x−1 =>(x^2+1)(x^3-x-1) So we get that, f(x)=(x^2+1) and ,g(x)=(x^3-x-1)

so, ∣f(3)+g(3)∣=10+23=33

f(x)Xg(x)=(x^5-x^2-x-1)=(x^2+1)(x^3-x-1)

Let,f(x)=(x^2+1) and g(x)=(x^3-x-1) hence,f(3)=10;g(3)=23 then |f(3)+g(3)|=|10+23|=33

write f(x)×g(x)=x5−x2−x−1. as f(x)=x2+1,g(x)=x3 x 1 so ∣f(3)+g(3)∣=33