Q.1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1 (ii) x - $\frac{1}{2}$ (iii) x (iv) x + π (v) 5 + 2x

Solution:

The remainder can be obtained by either long division method or by remainder theorem.

(i) Dividing x3 + 3x2 + 3x + 1 by x + 1

p(x) = x3 + 3x2 + 3x + 1

g(x) = x + 1

Put g(x) = 0,

i.e., x+1 = 0 ⇒ x = -1

p(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1

= -1 + 3 - 3 + 1 = 0

Therefore, the remainder is 0.

• The same can be verified by long division method.

(ii) Dividing x3 + 3x2 + 3x + 1 by x

p(x) = x3 + 3x2 + 3x + 1

g(x) = x

Put g(x) = 0,

i.e., x = 0 ⇒ x = $\frac{1}{2}$

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

= + $\frac{3}{4}$ +$\frac{3}{2}$ + 1

= = $\frac{27}{8}$

Therefore, the remainder is $\frac{27}{8}$.

• The same can be verified by long division method.

(iii) Dividing x3 + 3x2 + 3x + 1 by x

p(x) = x3 + 3x2 + 3x + 1

g(x) = x

Put g(x) = 0, i.e., x = 0

p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 1

Therefore, the remainder is 1.

(iv) Dividing x3 + 3x2 + 3x + 1 by x + π

Now p(x) = x3 + 3x2 + 3x + 1

g(x) = x + π

Put g(x) = 0, i.e., x = -π

p(π) = (-π)3 + 3(π)2 + 3(π) + 1

= -π3 +3 π2 -3 π + 1

Therefore, the remainder is -π3 +3 π2 -3 π + 1.

(v) Dividing x3 + 3x2 + 3x + 1 by 5 + 2x

Now p(x) = x3 + 3x2 + 3x + 1

g(x) = 5 + 2x

Put g(x) = 0, i.e., x = - $\frac{5}{2}$

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

= $-\frac{125}{8}$ + $\frac{75}{4}$ - $\frac{15}{2}$ + 1

= $\frac{-125+150-60+8}{8}$ = $-\frac{27}{8}$

Therefore, the remainder is $-\frac{27}{8}$.