+296 Find the ordered pair \((a, b)\) of real numbers for which
\((ax + b)(x^5 + 1) - (5x + 1)\)
is divisible by \(x^{2}+1.\)
+296 Thanks so much! I will try to figure out any simpler ways, great job on your answer! I suspect that guest is the one who is falsely posting answers, as melody as stated on one of my other posts.
+26388 Find the ordered pair \((a, b)\) of real numbers for which
\((ax + b)(x^5 + 1) - (5x + 1)\)
is divisible by \(x^{2}+1\).
\(\begin{array}{|lrcll|} \hline & P(x) &=& Q(x) (x^{2}+1) \\ & \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} &=& \mathbf{Q(x) (x^{2}+1)} \\ & && \boxed{x^2+1 = 0\\ x^2 = -1\\ x=\pm i } \\ \hline x=i: & (ai + b)(i^5 + 1) - (5i + 1) &=& Q(i)\times 0 \quad | \quad i^5 = i \\ & (ai + b)(i + 1) - (5i + 1) &=& 0 \\ & (ai + b)(i + 1) - 5x - 1 &=& 0 \\ & ai^2+i(a+b-5) +b-1 &=& 0 \quad | \quad i^2 = -1 \\ & -a+i(a+b-5) +b-1 &=& 0 \\ (1) & \mathbf{-a+b-1+i(a+b-5)} &=& \mathbf{0} \\ \hline x=-i: & \left(a(-i) + b\right)\left((-i)^5 + 1\right) - \left(5(-i) + 1\right) &=& Q(-i) \times 0 \quad | \quad (-i)^5= -i \\ & (-ai + b)(-i + 1) - (-5i + 1) &=& 0 \\ & (-ai + b)(-i + 1) 5i-1 &=& 0 \\ & ai^2-i(a+b-5) +b-1 &=& 0 \quad | \quad i^2 = -1 \\ & -a-i(a+b-5) +b-1 &=& 0 \\ (2) & \mathbf{-a+b-1-i(a+b-5)} &=& \mathbf{0} \\ \hline (1)+(2):& (-a+b-1)+i(a+b-5) \\ & + (-a+b-1)-i(a+b-5)&=& 0 \\ & 2(-a+b-1) &=& 0 \\ & -a+b-1 &=& 0 \\ &\mathbf{b} &=& \mathbf{1+a} \\ \hline (1)-(2):& (-a+b-1)+i(a+b-5) \\ &- \Big( (-a+b-1)-i(a+b-5) \Big) &=& 0 \\ & (-a+b-1)+i(a+b-5) \\ & -(-a+b-1)+i(a+b-5) &=& 0 \\ & 2i(a+b-5) &=& 0 \\ & a+b-5 &=& 0 \\ &\mathbf{b} &=& \mathbf{5-a} \\ \hline &b=1+a &=& 5-a \\ & 1+a &=& 5-a \\ & 2a &=& 4 \\ & \mathbf{a} &=& \mathbf{2} \\\\ & b&=&1+a \\ & b&=&1+2 \\ & \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}\)
