# Divide:(i) $x+2x^2+3x^4-x^5$ by $2x$(ii) $y^4-3y^3+\frac{1}{2y^2}$ by $3y$(iii) $-4a^3+4a^2+a$ by $2a$

Given:

The given expressions are:

(i) $x+2x^2+3x^4-x^5$ by $2x$

(ii) $y^4-3y^3+\frac{1}{2y^2}$ by $3y$

(iii) $-4a^3+4a^2+a$ by $2a$

To do:

We have to divide the given expressions.

Solution:

We have to divide the given polynomials by monomials using the formula $x^a \div x^b=a^{a-b}$

Polynomials:

Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.

Monomial:

A monomial is an expression that contains a single term composed of a product of constants and variables with non-negative integer exponents.

Therefore,

(i) The given expression is $x+2x^2+3x^4-x^5$ by $2x$.

$x+2x^2+3x^4-x^5 \div 2x=\frac{x}{2x}+\frac{2x^2}{2x}+\frac{3x^4}{2x}-\frac{x^5}{2x}$

$x+2x^2+3x^4-x^5 \div 2x=\frac{1}{2}+x^{2-1}+\frac{3}{2}x^{4-1}-\frac{1}{2}x^{5-1}$

$x+2x^2+3x^4-x^5 \div 2x=\frac{1}{2}+x^{1}+\frac{3}{2}x^{3}-\frac{1}{2}x^{4}$

$x+2x^2+3x^4-x^5 \div 2x=\frac{1}{2}+x+\frac{3}{2}x^{3}-\frac{1}{2}x^{4}$

Hence, $x+2x^2+3x^4-x^5$ divided by $2x$ is $\frac{1}{2}+x+\frac{3}{2}x^{3}-\frac{1}{2}x^{4}$.

(ii) The given expression is $y^4-3y^3+\frac{1}{2y^2}$ by $3y$.

$y^4-3y^3+\frac{1}{2y^2} \div 3y=\frac{y^4}{3y}-\frac{3y^3}{3y}+\frac{\frac{1}{2}y^2}{3y}$

$y^4-3y^3+\frac{1}{2y^2} \div 3y=\frac{1}{3}y^{4-1}-y^{3-1}+\frac{1}{2\times3}y^{2-1}$

$y^4-3y^3+\frac{1}{2y^2} \div 3y=\frac{1}{3}y^{3}-y^{2}+\frac{1}{6}y$

Hence, $y^4-3y^3+\frac{1}{2y^2}$ divided by $3y$ is $\frac{1}{3}y^{3}-y^{2}+\frac{1}{6}y$.

(iii) The given expression is $-4a^3+4a^2+a$ by $2a$.

$-4a^3+4a^2+a \div 2a=\frac{-4a^3}{2a}+\frac{4a^2}{2a}+\frac{a}{2a}$

$-4a^3+4a^2+a \div 2a=-2a^{3-1}+2a^{2-1}+\frac{1}{2}a^{1-1}$

$-4a^3+4a^2+a \div 2a=-2a^2+2a^1+\frac{1}{2}a^0$

$-4a^3+4a^2+a \div 2a=-2a^2+2a+\frac{1}{2}$             [Since $a^0=1$]

Hence, $-4a^3+4a^2+a$ divided by $2a$ is $-2a^2+2a+\frac{1}{2}$.

Updated on: 13-Apr-2023

37 Views

Get certified by completing the course