Question

find the remainder when x³+3x²+1 is divided by (i) x+1 (ii) x-½ (iii) x.​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \: \bull \: \sf \: A \: polynomial \: f(x) = {x}^{3} + 3 {x}^{2} + 1$

$\large\underline{\sf{To\:Find - }}$

Remainder when f(x) is divided by

$\: \: \: \: \: \: \: \: \: \: \: \bull \: \sf \: x \: + \: 1$

$\: \: \: \: \: \: \: \: \: \: \: \bull \: \sf \: x - \dfrac{1}{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \bull \: \sf \: x$

Concept Used :-

Remainder Theorem :-

This theorem states that if f(x) is a polynomial in x, then the remainder on dividing f(x) by x − a is f(a).

Let's solve the problem now!!

$\large\underline{\sf{Solution-(i)}}$

$\rm :\longmapsto\:Let \: x + 1 = 0$

$\rm :\longmapsto\:x = - 1$

Now, we have to find the remainder when f(x) is divided by x + 1.

Using Remainder Theorem,

• The remainder is given by f(-1)

So,

$\rm :\longmapsto\:f( - 1) = {( - 1)}^{3} + 3 {( - 1)}^{2} + 1$

$\rm :\longmapsto\:f( - 1) = - 1 + 3 + 1$

$\bf\implies \:f( - 1) = 3$

$\large\underline{\sf{Solution-(ii)}}$

$\rm :\longmapsto\:Let \: x \: - \: \dfrac{1}{2} = 0$

$\rm :\longmapsto\:x \: = \: \dfrac{1}{2}$

Now, we have to find the remainder when f(x) is divided by x + 1/2.

Using Remainder Theorem,

• The remainder is given by f(-1/2)

So,

$\rm :\longmapsto\:f\bigg( \dfrac{1}{2} \bigg) = {\bigg( \dfrac{1}{2} \bigg) }^{3} + 3 {\bigg( \dfrac{1}{2} \bigg) }^{2} + 1$

$\rm :\longmapsto\:f\bigg( \dfrac{1}{2} \bigg) = \dfrac{1}{8} + \dfrac{3}{4} + 1$

$\rm :\longmapsto\:f\bigg( \dfrac{1}{2} \bigg) = \dfrac{1 + 6 + 8}{8}$

$\bf\implies \:f\bigg( \dfrac{1}{2} \bigg) = \dfrac{15}{8}$

$\large\underline{\sf{Solution-(iii)}}$

$\rm :\longmapsto\:Let \: x \: = \: 0$

$\rm :\implies\:x = 0$

Now, we have to find the remainder when f(x) is divided by x

Using Remainder Theorem,

• The remainder is given by f(0)

So,

$\rm :\longmapsto\:f(0) = {(0)}^{3} + 3 \times {(0)}^{2} + 1$

$\bf\implies \:f(0) = 1$

Additional Information :-

Factor Theorem :-

This theorem states that if f(x) is a polynomial in x then the remainder on dividing f(x) by x − a is 0, then x - a is factor of f(x).