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Find the ordered pair $$(a,b)$$ of real numbers for which $$(ax + b)(x^5 + 1) - (5x + 1)$$ is divisible by $$x^2 + 1.$$

May 15, 2019

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Find the ordered pair $$(a,b)$$ of real numbers for which $$(ax + b)(x^5 + 1) - (5x + 1)$$ is divisible by $$x^2 + 1$$.

$$\begin{array}{|rcll|} \hline && \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} \\ &=& ax^6+bx^5+ax+b-5x-1 \\ &=&\mathbf{ ax^6+bx^5+(a-5)x +(b-1) } \\\\ &=& (rx^4+sx^3+tx^2+ux+v)(x^2+1) \\ &=& rx^6+sx^5+tx^4+ux^3+vx^2+rx^4+sx^3+tx^2+ux+v \\ &=&\mathbf{ \underbrace{r}_{=a}x^6+\underbrace{s}_{=b}x^5+ \underbrace{(t+r)}_{=0}x^4+\underbrace{(u+s)}_{=0}x^3+\underbrace{(v+t)}_{=0}x^2 + \underbrace{u }_{=a-5}x+\underbrace{v}_{=b-1} } \\ \hline v &=& b-1 \\ u &=& a-5 \\ r &=& a \\ s &=& b \\\\ t+r&=& 0 \quad | \quad r = a \\ t+a &=& 0 \\ \mathbf{t} &=& \mathbf{-a} \\\\ u +s &=& 0 \quad | \quad u=a-5,\ s = b \\ a-5+b &=& 0 \\ \mathbf{a +b} &=& \mathbf{5} \qquad ( 1) \\\\ v+t &=& 0 \quad | \quad v=b-1,\ t = -a \\ b-1-a &=& 0 \\ \mathbf{b-a} &=& \mathbf{1} \qquad ( 2) \\ \hline \end{array}$$

$$\mathbf{a=\ ?} \\ \mathbf{b=\ ?}$$

$$\begin{array}{|lrcll|} \hline (1)+(2): &a +b + b-a &=& 5+1 \\ & 2b &=& 6 \\ & \mathbf{b } &=& \mathbf{3} \\ \hline (1)-(2): &a +b + -(b-a) &=& 5-1 \\ & 2a &=& 4\\ & \mathbf{a } &=& \mathbf{2} \\ \hline \end{array}$$

$$(a,b) =(2,3)$$

$$\begin{array}{|rcll|} \hline && \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} \\ &=& (2x + 3)(x^5 + 1) - (5x + 1) \\ &=& \mathbf{2x^6+3x^5-3x+2} \\ \hline \end{array}$$

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May 16, 2019