Question
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem :
$(x^2 – 1x + 9); (x + 1)$
$(x^2 – 1x + 9); (x + 1)$
✓
Answer
p(x) = $x^2 – 7x + 9$
Divisor =$x + 1$
\therefore take $x = – 1$
\therefore By remainder theorem,
\therefore Remainder =$p(-1)$
p(x) = $x^2 – 7x + 9$
$\therefore p(-1) = (- 1)^2 – 7(- 1) + 9$
$= 1 + 7 + 9$
$\therefore Remainder =17$
Divisor =$x + 1$
\therefore take $x = – 1$
\therefore By remainder theorem,
\therefore Remainder =$p(-1)$
p(x) = $x^2 – 7x + 9$
$\therefore p(-1) = (- 1)^2 – 7(- 1) + 9$
$= 1 + 7 + 9$
$\therefore Remainder =17$
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