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. $5+2 \mathrm{x}$
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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