A polynomial product

The right hand side of the equation factors as $x^8+3x^4-4=(x^4+4)(x^4-1)$ . The left-hand factor is a special case of the Sophie Germain Identity, and as such factors as $(x^2+2x+2)(x^2-2x+2)$ . Applying a difference of squares twice, the right-hand factor becomes $(x^2+1)(x+1)(x-1)$ . Combining these two factors, we obtain $x^8+3x^4-4=\\ (x^2+2x+2)(x^2-2x+2)(x^2+1)(x+1)(x-1)$

Because we are substituting the same value into each of the $f_i$ , it doesn't matter which factor we assign to each $f_i$ . Therefore, we can say that $|f_1 (x)|+|f_2(x)|+|f_3(x)|+|f_4(x)|+|f_5(x)|=\\ |x^2+2x+2|+|x^2-2x+2|+|x^2+1|+|x+1|+|x-1|$ When $x=1$ , we find that $|f_1 (1)|+|f_2(1)|+|f_3(1)|+|f_4(1)|+|f_5(1)|=\\ |5|+|1|+|2|+|2|+|0|=\boxed{10}.$

We factorize:

$x^8+3x^4-4=(x^4+4)(x^4-1)$

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

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

Those five factors are, in some order, the five functions we want. Plugging in 1 shows that our answer is 1+5+2+2= $\boxed{10}$ .

$\begin{aligned} x^8+3x^4-4&=(x^4+4)(x^4-1) \\ &=(x^4+4)(x^2-1)(x^2+1) \\ &=(x^4+4)(x-1)(x+1)(x^2+1) \\ &=(x^4+4x^2+4-4x^2)(x-1)(x+1)(x^2+1) \\ &=[(x^2+2)^2-(2x)^2](x-1)(x+1)(x^2+1) \\ &=(x^2-2x+2)(x^2+2x+2)(x-1)(x+1)(x^2+1) \\ \end{aligned}$

It does not really matter what $f_1,f_2,f_3,f_4,f_5$ represents.

$\therefore|f_1(1)|+|f_2(1)|+|f_3(1)|+|f_4(1)|+|f_5(1)|=1+5+0+2+2=10$

The crucial step here is to be able to factor the given product.

We have $x^{8} + 3x^{4} -4$ . Obviously, this factors as $(x^{4} -1)(x^{4} +4)$ .

Then we can still factor the first factor: $x^{4} -1$ factors as $(x^{2}+1)(x^{2}-1)$

Factoring the second factor will give us $(x+1)(x-1)(x^{2}+1)(x^{4}+4)$ all in all.

For us to have five factors, we should be able to factor either $x^{2} +1$ or $x^{4}+4$ such that the factors will have integer coefficients as stated in the problem.

Trying to factor $x^{2} +1$ will give us nothing, so we have $x^{4}+4$

This step isn't that obvious, but some manipulation will lead us to the answer. Adding and subtracting $4x^{2}$ to the expression will give us

$(x^{4} + 4x^{2} +4) - 4x^{2}$ . Note that the first expression is a Perfect Square Trinomial , we have

$(x^{2}+2)^{2}-4x^{2}$ . Now this is a difference of two squares.

The expression factors as $(x^{2}+2+2x)(x^{2}+2-2x)$ . Then we have

$(x+1)(x-1)(x^{2}+1)(x^{2}+2x+2)(x^{2}-2x+2)$ all in all.

So the functions are each of the factors. Substituting 1 will give us:

$2 + 0 + 2 + 5 + 1 = 10$

$x^8+3x^4-4=(x^4+4)(x^4-1)=(x^4+4)(x^2+1)(x^2-1)=(x^4+4)(x^2+1)(x+1)(x-1)$ $x^4+4=x^4+4x^2+4-4x^2=(x^2+2)^2-(2x)^2=(x^2+2-2x)(x^2+2+2x)$ $f_n(x)={(x^2+2-2x),(x^2+2+2x),(x^2+1),(x+1),(x-1)}$ therefore the answer is $1+5+2+2+0=\boxed{10}$ .

Solution: because of f1(x)×f2(x)×f3(x)×f4(x)×f5(x)

=x^8+〖3x〗^4−4 are 5 non-constant polynomials

so f1(x)×f2(x)×f3(x)×f4(x)×f5(x)=x^8+〖3x〗^4−4= (x^4+4)(x^2+1)(x+1)(x-1)(a)

f1(0)×f2(0)×f3(0)×f4(0)×f5(0)=0^8+0^4−4

= -4 = (0^4+4)(0^2+1)(0+1)(0-1) (a)

-4= (-4) (a)

(a)=1

So, ∣f1(1)∣+∣f2(1)∣+∣f3(1)∣+∣f4(1)∣+∣f5(1)∣

= ∣1^4+4∣+∣1^2+4∣+∣ 1+1∣+∣ 1-1∣+∣1∣

= 10

I try to factorize the equation given, i get ( $x^{4} + 4$ )( $x^{2} + 1$ )( $x+1$ )( $x-1$ ),then i think of one more factor, $x+1$ and $x-1$ is totally out, $x^{2}+1$ ... erm... no idea~

Then i try $x^4+4$ . wonder why i think of this, i try completing the square and i get... $x^{4}+4$ =[( $x^{2} + 2$ )^2] - $4x^{2}$

And since $4x^{2}$ is perfect square, then use the difference between square formula, and i get... ( $x^{2} + 2 + 2x$ )( $x^{2} + 2 - 2x$ )

So the equation can be factorize to ( $x^{2} + 2x + 2$ ) ( $x^{2} - 2x + 2$ ) ( $x^{2} + 1$ ) ( $x+1$ ) ( $x-1$ )

Then substitute 1 and modulus and sum up, i get 5 + 1 + 2 + 2 + 0 = 10