Question

1. Find the remainder when x³+3x²+3x+1 is divided by
(i)x+1
(ii)x-1/2
(iii)x
(iv)x+π
(v)5+2x
​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

We have to use the concept given below:

p(x) divided by (ax + b) leaves remainder p(-b/a).

 

Now,

p(x) = x³ + 3x² + 3x + 1 = 0

(i)  Divisor is (x + 1).

$ax+b=x+1 \\ \\ a=1\ \ \ ; \ \ \ b=1 \\ \\ \\ \displaystyle -\frac{b}{a}=\ -\frac{1}{1}=\ -1$

∴ p(x) divided by (x + 1) leaves remainder p(-1).

$p(-1)=(-1)^3+3(-1)^2+3(-1)+1 \\ \\ p(-1)=-1+3-3+1 \\ \\ p(-1)=\bold{0}$

Thus the remainder is 0.

[Here's no need to find p(-1), because p(x) = x³ + 3x² + 3x + 1 = (x + 1)³]

(ii)  Divisor is (x - 1/2).

$ax+b=x-\displaystyle \frac{1}{2} \\ \\ \\ a=1\ \ \ ; \ \ \ b=-\frac{1}{2} \\ \\ \\ -\frac{b}{a}=\ -\frac{-\frac{1}{2}}{1}=\ \frac{1}{2}$

∴ p(x) divided by (x - 1/2) leaves remainder p(1/2).

$\displaystyle p\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)+1 \\ \\ \\ p\left(\frac{1}{2}\right)=\frac{1}{8}+3 \times \frac{1}{4}+3 \times \frac{1}{2}+1 \\ \\ \\ p\left(\frac{1}{2}\right)=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1 \\ \\ \\ p\left(\frac{1}{2}\right)=\frac{1}{8}+\frac{6}{8}+\frac{12}{8}+\frac{8}{8} \\ \\ \\ p\left(\frac{1}{2}\right)=\frac{1+6+12+8}{8} \\ \\ \\ p\left(\frac{1}{2}\right)=\bold{\frac{27}{8}}$

$\displaystyle p\left(\frac{1}{2}\right)=\bold{\left(\frac{3}{2}\right)^3}$

Thus the remainder is (3/2)³.

(iii)  Divisor is x.

$a=1\ \ \ ; \ \ \ b=0 \\ \\ \\ \displaystyle -\frac{b}{a}=\ -\frac{0}{1}\ = 0$

∴ p(x) divided by x leaves remainder p(0).

$p(0)=(0)^3+3(0)^2+3(0)+1 \\ \\ p(0)=0+0+0+1 \\ \\ p(0)=\bold{1}$

Thus the remainder is 1.

[p(x) = x³ + 3x² + 3x + 1 = x(x² + 3x + 3) + 1]

(iv)  Divisor is (x + π).

$a=1\ \ \ ; \ \ \ b=\pi \\ \\ \\ \displaystyle -\frac{b}{a}=\ -\frac{\pi}{1}\ =-\pi$

∴ p(x) divided by (x + π) leaves remainder p(-π).

$p(-\pi)=(-\pi)^3+3(-\pi)^2+3(-\pi)+1 \\ \\ p(-\pi)=(-\pi)^3+3(-\pi)^2 \cdot 1+3(-\pi) \cdot 1^2+1^3 \\ \\ p(-\pi)=(-\pi+1)^3 \\ \\ p(-\pi)=\bold{(1-\pi)^3}$

Thus the remainder is (1 - π)³.

[p(x) = (x + 1)³   ;   p(-π) = (-π + 1)³ = (1 - π)³]

(v) Divisor is (5 + 2x) = (2x + 5).

$a=2\ \ \ ; \ \ \ b=5 \\ \\ \\ -\displaystyle \frac{b}{a}=\ -\frac{5}{2}$

∴ p(x) divided by (5 + 2x) leaves remainder p(-5/2).

$\displaystyle p\left(-\frac{5}{2}\right)=\left(-\frac{5}{2}\right)^3+3\left(-\frac{5}{2}\right)^2+3\left(-\frac{5}{2}\right)+1 \\ \\ \\ p\left(-\frac{5}{2}\right)=-\frac{125}{8}+3 \cdot \frac{25}{4}-3 \cdot \frac{5}{2}+1 \\ \\ \\ p\left(-\frac{5}{2}\right)=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1 \\ \\ \\ p\left(-\frac{5}{2}\right)=-\frac{125}{8}+\frac{150}{8}-\frac{60}{8}+\frac{8}{8} \\ \\ \\ p\left(-\frac{5}{2}\right)=\frac{-125+150-60+8}{8} \\ \\ \\ p\left(-\frac{5}{2}\right)=-\frac{27}{8}$

$\displaystyle p\left(-\frac{5}{2}\right)=\bold{\left(-\frac{3}{2}\right)^3}$

Thus the remainder is (-3/2)³.