Question

Solve This; Mods And Stars...

Find the remainder when x³ + 3x² + 3x + 1 is divided by
$(i) \: x - \frac{1}{2}$
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Answer 1

$\: \huge{{\mathfrak\red{⛄answer⛄}}}$

$x - \frac{1}{2}$

$x² + 3x² + 3x +1\: \: \: \div \: \: \: x - \frac{1}{2}$

put divisors = 0

$x - \frac{1}{2} = 0$

$x ➭ \frac{1}{2}$

Let p(x)= x³ + 3x² + 3x + 1

$putting \: \: \: x = \frac{1}{2}$

p=$</strong><strong>(\frac{1}{2})=( \frac{1}{2})^{3} +3( \frac{1}{2})^{2}+3( \frac{1}{2})+1$

$= \frac{1}{8} + 3( \frac{1}{4} ) + \frac{3}{2} + 1$

$= \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$

$= \frac{1 + 3(2) + 3(4) + 1(8)}{8}$

$= \frac{1 + 6 + 12 + 8}{8}$

$ans \: = \frac{27}{8}$

$\: Thus, remainder = p( \frac{1}{2} ) = \frac{27}{8}$

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hope it was helpful to you

Answer 2

$\: \huge{{\mathfrak\red{⛄answer⛄}}}$

$x - \frac{1}{2}$

$x² + 3x² + 3x +1\: \: \: \div \: \: \: x - \frac{1}{2}$

➭ put divisors = 0

$x - \frac{1}{2} = 0$

$x = \frac{1}{2}$

➭ Let p(x)= x³ + 3x² + 3x + 1

$putting \: \: \: x = \frac{1}{2}$

$p=( \frac{1}{2})=( \frac{1}{2})^{3} +3( \frac{1}{2})^{2}+3( \frac{1}{2})+1$

$= \frac{1}{8} + 3( \frac{1}{4} ) + \frac{3}{2} + 1$

$= \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$

$= \frac{1 + 3(2) + 3(4) + 1(8)}{8}$

$= \frac{1 + 6 + 12 + 8}{8}$

$ans \: = \frac{27}{8}$

$\: Thus, remainder = p( \frac{1}{2} ) = \frac{27}{8}$

Thanks!