Question

find the remainder when a²x² +ax + 1 is divided by ax +1
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Answer 1

$Let \: p(x) = a^{2}x^{2} + ax + 1$

$If \: p(x) \:is \: divided \: by \: (ax + 1 ) \:then$

$the \: remainder \: is \: p\Big( \frac{-1}{a}\Big)$

$p\Big( \frac{-1}{a}\Big) = a^{2}\times \Big( \frac{-1}{a}\Big)^{2} + a \times \Big( \frac{-1}{a}\Big) +1$

$= a^{2} \times \frac{1}{a^{2}} -1 + 1$

$= 1$

Therefore.,

$\red{ Remainder } \green { = 1 }$

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Answer 2

ANSWER✔

$\dashrightarrow P(x)= a^2x^2+ax+1$

$\dashrightarrow g(x)= ax+1\\ \implies ax+1=0\\ \implies x= \dfrac{-1}{a}$

$\implies p\bigg( \dfrac{-1}{a} \bigg) = a^2 \bigg( \dfrac{-1}{a} \bigg)^2+ a\bigg( \dfrac{-1}{a} \bigg) +1$

$\implies p\bigg( \dfrac{-1}{a} \bigg)= a^2 \bigg( \dfrac{1}{a^2} \bigg) + \cancel{a}\:\:\bigg( \dfrac{-1}{\cancel{a}} \bigg) +1$

$\implies p\bigg( \dfrac{-1}{a} \bigg)= \cancel{a^2} \bigg( \dfrac{1}{\cancel{a^2}} \bigg) + \cancel{a}\:\:\bigg( \dfrac{-1}{\cancel{a}} \bigg)+1$

$\implies 1+(-1)+1$

$\implies 1-1+1$

$\implies 1$

$\large{\boxed{\bf{ \star\:\: remainder=1 \:\: \star}}}$

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