Question

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

Answer 1

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

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

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

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

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

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Answer 2

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