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# Find the remainder when $$x^3 + 3x^2 + 3x + 1$$ is divided by i)$$x + 1$$ii)$$x - \frac{1}{2}$$iii)$$x$$iv)$$x + \pi$$v)$$5 + 2x$$

Answer :

Let $$p(x) = x^3 + 3x^2 + 3x + 1$$

i)The zero of $$x + 1$$ is -1
[$$\because x+1= 0 \Rightarrow x = -1$$ ]

$$\therefore p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$$
$$\Rightarrow p(-1) = -1 + 3 - 3 +1$$
$$\Rightarrow p(-1) = 0$$

Hence, By remainder theorem, required remainder = 0.

ii)The zero of $$x - \frac{1}{2}$$ is x = $$\frac{1}{2}$$
[$$\because (x - \frac{1}{2} = 0 )$$]

$$\therefore p(\frac{1}{2} ) = (\frac{1}{2} )^3 + 3(\frac{1}{2} )^2 + 3(\frac{1}{2} )+ 1$$
$$\Rightarrow p(\frac{1}{2}) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} +1$$
$$\Rightarrow p(\frac{1}{2} ) = \frac{27}{8}$$

Hence, By remainder theorem, required remainder = $$\frac{27}{8}$$ .

iii)The zero of $$x$$ is x = 0 [$$\because (x = 0)$$ ]

$$\therefore p(0) = (0)^3 + 3(0)^2 + 3(0) + 1$$
$$\Rightarrow p(-1) = 1$$

Hence, By remainder theorem, required remainder = 1.

iv) The zero of $$x + \pi$$ is $$x = -\pi$$ [$$\because (x + \pi = 0)$$]

$$\therefore p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1$$

$$\Rightarrow p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1$$

Hence, By remainder theorem, required remainder$$= -\pi^3 + 3\pi^2 - 3\pi + 1$$.

v) The zero of $$5 + 2x$$ is x = $$\frac{-5}{2}$$ [$$\because (5 + 2x = 0)$$]

$$\therefore p(\frac{-5}{2} ) = (\frac{-5}{2})^3 + 3(\frac{-5}{2})^2 + 3(\frac{-5}{2} )+ 1$$
$$\Rightarrow p(\frac{-5}{2} ) = \frac{-125}{8} + \frac{75}{4} - \frac{15}{2} + 1$$
$$\Rightarrow p(\frac{-5}{2} ) = \frac{-27}{8}$$

Hence, By remainder theorem, required remainder = $$\frac{-27}{8}$$.