Question

$find \: the \: remainder \: when \\ {x}^{3} + 3 {x}^{2} + 3x + 1 \\ \\ is \: divided \: by$

$1)\: \: \: x - \frac{1}{2}$
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Answer 1

$find \: the \: remainder \: when \\ {x}^{3} + 3 {x}^{2} + 3x + 1 \\ \\ 1)\: \: \: x - \frac{1}{2} \\ \\ solution. \\ \\ x - \frac{1}{2} = 0 \\ \\ x = \frac{1}{2} \: \: (given) \\ \\p(x) = {x}^{3} + 3 {x}^{2} + 3x + 1 \\ \\ p( \frac{1}{2} ) = ( \frac{1}{2} ) {}^{3} + 3( \frac{1}{2} ) {}^{2} + 3( \frac{1}{2} ) + 1 \\ \\ = \frac{1}{8} + 3( \frac{1}{4} ) + \frac{3}{2} + 1 \\ \\ = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + \frac{1}{1} \\ \\ = \frac{1 + 6 + 12 + 8}{8} \\ \\ = \frac{27}{8}= 3.375 \: \: \: is \: the \: answer...$

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