Question

find the remainder when
x³+3x²+3x+1 divided by

(i)x+1 (ii) $x -\frac{1}{2}$ (iii) $x$

$\red{\bf{please\: solve\: it\: as \:soon \:as \:possible}}$​

Answer 1

$\pink{\bf{(i)x+1}}$

• $\pink{\bf{put \:the\: value\: of \:x+1=0}}$

⠀⠀⠀⠀⠀⠀⠀⠀${\bf{x+1=0}}$

⠀⠀⠀⠀⠀⠀⠀⠀${\bf{x= -1}}$

• $\pink{\bf{now\: put}}$ $\pink{\bf{-1}}$$\pink{\bf{\: in\: place\: of\:x}}$

${\bf{P(0)=x³+3x²+3x+1}}$

${\bf{P(-1)=(-1)³+3(-1)²+3(-1)+1}}$

${\bf{= -1+3-3+1}}$

${\bf{= 4-4}}$

$\red{\bf{=0}\boxed{\bf{answer}}}$

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$\pink{\bf{(ii)x -\frac{1}{2}}}$

• $\pink{\bf{put\: the\: value\: of \: x -\frac{1}{2}=0}}$

⠀⠀⠀⠀⠀⠀⠀${\bf{x -\frac{1}{2}=0}}$

⠀⠀⠀⠀⠀⠀⠀${\bf{x= \frac{1}{2}}}$

• $\pink{\bf{now \:put \frac{1}{2}\:in \:the\: place\: of\:x}}$

${\bf{P(0)=x³+3x²+3x+1}}$

P(${\bf{\frac{1}{2}}}$)${\bf{= x( \frac{1}{2})³+3(\frac{1}{2} )²+3( \frac{1}{2}) +1}}$

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

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

$={\pink{\frac{27}{8} \pink{\bf{\boxed{answer}}}}}$

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$\pink{\bf{(iii)\:x}}$

• $\pink{\bf{put \:the \:value\: of x=0}}$

${\bf{x=0}}$

${\bf{x³+3x²+3x+1}}$

${\bf{ (0)³ + 3(0)²+3(0)+1}}$

$\red{\bf{=1}}\boxed{\pink{\bf{answer}}}$

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Answer 2

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x + 1 = 0

x = -1

P(x) = x³ + 3x² + 3x + 1

P(-1) = (-1)³ + 3(-1)² + 3(-1) + 1

P(-1) = -1 + 3 - 3 + 1

P(-1) = 0

Remainder = 0

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x - 1/2 = 0

x = 1/2

P(x) = x³ + 3x² + 3x + 1

P(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1

P(1/2) = 1/8 + 3/4 + 3/2 + 1

P(1/2) = 27/8

Remainder = 27/8

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x = 0

P(x) = x³ + 3x² + 3x + 1

P(0) = (0)³ + 3(0)² + 3(0) + 1

P(0) = 0 + 0 + 0 + 1

P(0) = 1

Remainder = 1

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