Find the remainder when

Question:

Find the remainder when $x^{3}+3 x^{3}+3 x+1$ is divided by,

1. $x+1$

2. $x-1 / 2$

3. $x$

4. $x+\pi$

5. $5+2 \mathrm{x}$

 

Solution:

Here, $f(x)=x^{3}+3 x^{2}+3 x+1$

by remainder theorem

1 ⟹ x + 1 = 0

⟹ x = -1

substitute the value of x in f(x)

$f(-1)=(-1)^{3}+3(-1)^{2}+3(-1)+1$

= -1 + 3 - 3 + 1

= 0

2. x - 1/2

Here, $f(x)=x^{3}+3 x^{2}+3 x+1$

By remainder theorem

⟹ x - 1/2 = 0

⟹ x = 1/2

substitute the value of x in f(x)

$f(1 / 2)=(1 / 2)^{3}+3(1 / 2)^{2}+3(1 / 2)+1$

$=(1 / 2)^{3}+3(1 / 2)^{2}+3(1 / 2)+1$

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

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

= 27/8

3. x

Here, $f(x)=x^{3}+3 x^{2}+3 x+1$

by remainder theorem

⟹ x = 0

substitute the value of x in f(x)

$f(0)=0^{3}+3(0)^{2}+3(0)+1$

= 0 + 0 + 0 + 1

= 1

4. x + π

Here, $f(x)=x^{3}+3 x^{2}+3 x+1$

by remainder theorem

⟹ x + π = 0

⟹ x = - π

Substitute the value of x in f(x)

$f(-\pi)=(-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1$

5. 5 + 2x

Here, $f(x)=x^{3}+3 x^{2}+3 x+1$

by remainder theorem

5 + 2x = 0

2x = -5

x = -5/2

substitute the value of x in f(x)

$f(-5 / 2)=(-5 / 2)^{3}+3(-5 / 2)^{2}+3(-5 / 2)+1$

= −125/8 + 3(25/4) + 3(-5/2) + 1

= −125/8 + 75/4 - 15/2 + 1

Taking L.C.M

$=\frac{-125+150-50+8}{8}$

= −27/8

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