find the remainder when x^3 + 3x^2+ 3x + 1 is divided by
Q 1) x-1/2 ​

2 thoughts on “find the remainder when x^3 + 3x^2+ 3x + 1 is divided by <br />Q 1) x-1/2 ​”

1. Answer:

   27/8

   Step-by-step explanation:

   (x³+3x²+3x+1 ) ÷ (x-1/2)

   x-1/2=0

   x=1/2

   by putting x=1/2 in x³+3x²+3x+1

   (1/2)³ + [3(1/2)²] + 3(1/2) + 1

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

   = {1+(3×2) + (3×4) + (1×1)} / 8 (by l.c.m)

   = [1+6+12+1]/8

   = 27/8

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2. Given:–

   •x^3 + 3x^2 + 3x + 1 is divided by x -1/2.

   To Find:–

   •Find the remainder.

   Solution:–

   Firstly we have to solve this

   [tex] \: \: \sf \: x – \frac{1}{2} = 0 \\ \\ \: \: \sf \: x = \frac{1}{2} [/tex]

   Hence, the value of x is 1/2.

   Now we will find the remainder of x^3 + 3x^2 +3x+1.

   Now put the values

   [tex] \: \: \sf \: {x}^{3} + 3 {x}^{2} + 3x + 1 = 0 \\ \\ \: \: \sf \: {( \frac{1}{2}) }^{3} + 3 \times {( \frac{1}{2}) }^{2} + 3 \times \frac{1}{2} + 1 \\ \\ \: \: \sf \: \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1 \\ \\ \: \: \sf \: \frac{1 + 6 + 12}{8} + 1 \\ \\ \: \: \sf \: \frac{19}{8} + 1\\ \\ \: \: \sf \: \frac{19 + 8}{8} \\ \\ \: \: \sf \: \frac{27}{8} [/tex]

   Henceforth,value of x^3 + 3x^2 +3x+1 is 27/8.

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