One Step Harder

I kinda see that $p(x) = 7-x^{2}$ for $x = 1,-2,3,-4$ . Every time $|x|$ increases by $1$ , $p(x)$ decreases by $3,5,7$ . That is some sight of quadratic sequence with no terms of $x$ .

Which means $p(x) - 7 + x^{2} = 0$ has roots $x = 1,-2,3,-4$ .

Therefore, $p(x)-7+x^{2} = c(x-1)(x+2)(x-3)(x+4)$ for some constant $c$ .

And $p(x)$ is also monic, that means $c = 1$ .

$p(x) = (x-1)(x+2)(x-3)(x+4) - x^2 + 7$

Plug in $x = 4$ we get $p(4) = 3\times 6\times 1 \times 8 - 4^{2} + 7 = \boxed{135}$ ~~~

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Just plus 4 to 135 if you say that the answer is not 135. =_="

Let $p(x) = x^4+ax^3+bx^2+cx+d$ , then:

$\begin{aligned} &p(1) &=&1+a+b+c+d &=&6 \\ &p(-2) &=&16-8a+4b-2c+d&=&3 \\ &p(3) &= &81+ 27a +9b +3c +d&=&-2 \\ &p(-4) &= &256-64a +16b -4c +d&= &-9 \end{aligned}$

This implies that: $\begin{bmatrix} 1&1&1&1 \\ -8&4&-2&1 \\ 81&27&3&1 \\ 256&-64&-4&1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}=\begin{bmatrix} 5 \\ -13 \\ -83 \\ -265 \end{bmatrix}$

Solve the matrix equation above to get the coefficient $a$ , $b$ , $c$ and $d$ . You may find the inverse matrix to solve or use Gaussian elimination method as follows.

$\left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 27 & 9 & 3 & 1 & | & -83 \\ -8 & 4 & -2 & 1 & | & -13 \\ 1 & 1 & 1 & 1 & | & 5 \end{matrix} \right]$

$\Rightarrow \begin{matrix} R_1 &\rightarrow R_1 \\ R_2-27R_4 &\rightarrow R_2 \\ R_1-8R_3 &\rightarrow R_3 \\ (8R_3+R_3)/3 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 0 & -18 & -24 & -26 & | & -218 \\ 0 & -16 & 12 & -7 & | & -161 \\ 0 & 4 & 2 & 3 & | & 9 \end{matrix} \right]$

$\Rightarrow \begin{matrix} R_1 &\rightarrow R_1 \\ R_2/2 &\rightarrow R_2 \\ R_3+48R_4 &\rightarrow R_3 \\ 2R_2+9R_4 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 0 & -9 & -12 & -13 & | & -109 \\ 0 & 0 & 20 & 5 & | & -125 \\ 0 & 0 & -30 & -25 & | & -355 \end{matrix} \right]$

$\Rightarrow \begin{matrix} R_1 &\rightarrow R_1 \\ R_2 &\rightarrow R_2 \\ R_3 &\rightarrow R_3 \\ -(2R_4+3R_3)/35 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 0 & -9 & -12 & -13 & | & -109 \\ 0 & 0 & 20 & 5 & | & -125 \\ 0 & 0 & 0 & 1 & | & 31 \end{matrix} \right]$

$\Rightarrow \begin{matrix} R_1 &\rightarrow R_1 \\ R_2 &\rightarrow R_2 \\ (R_3-5R_4)/20 &\rightarrow R_3 \\ R_4 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 0 & -9 & -12 & -13 & | & -109 \\ 0 & 0 & 1 & 0 & | & -14 \\ 0 & 0 & 0 & 1 & | & 31 \end{matrix} \right]$

$\Rightarrow \begin{matrix} R_1 &\rightarrow R_1 \\ -(R_2+12R_3+13R_4)/9 &\rightarrow R_2 \\ R_3 &\rightarrow R_3 \\ R_4 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} -64 & 16 & -4 & 1 & | & -265 \\ 0 & 1 & 0 & 0 & | & -14 \\ 0 & 0 & 1 & 0 & | & -14 \\ 0 & 0 & 0 & 1 & | & 31 \end{matrix} \right]$

$\Rightarrow \begin{matrix} -(R_1-16R_2+4R_3-R_4)/64 &\rightarrow R_1 \\ R_2 &\rightarrow R_2 \\ R_3 &\rightarrow R_3 \\ R_4 &\rightarrow R_4 \end{matrix} \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & 0 & | & -14 \\ 0 & 0 & 1 & 0 & | & -14 \\ 0 & 0 & 0 & 1 & | & 31 \end{matrix} \right]$

$\Rightarrow a = 2 \quad b = -14 \quad c = - 14 \quad d = 31$

$\Rightarrow p(x) = x^4+2x^3-14x^2-14x+31$

$\Rightarrow p(4) =256 + 2(64) - 14(16)-14(4)+31 = 256+128-224-56+31 = 135$

Therefore, $|p(4)+4| = 135+4 = \boxed{139}$