# Divide:(i) $-x^6+2x^4+4x^3+2x^2$ by $\sqrt2x^2$(ii) $-4a^3+4a^2+a$ by $2a$(iii) $\sqrt3a^4+2\sqrt3a^3+3a^2-6a$ by $3a$

Given:

The given expressions are:

(i) $-x^6+2x^4+4x^3+2x^2$ by $\sqrt2x^2$

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

(iii) $\sqrt3a^4+2\sqrt3a^3+3a^2-6a$ by $3a$

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 $-x^6+2x^4+4x^3+2x^2$ by $\sqrt2x^2$.

$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-x^6}{\sqrt2x^2}+\frac{2x^4}{\sqrt2x^2}+\frac{4x^3}{\sqrt2x^2}+\frac{2x^2}{\sqrt2x^2}$

$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{6-2}+\frac{\sqrt2 \times \sqrt2}{\sqrt2}x^{4-2}+\frac{2\sqrt2 \times \sqrt2}{\sqrt2}x^{3-2}+\frac{\sqrt2 \times \sqrt2}{\sqrt2}x^{2-2}$

$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{4}+\frac{\sqrt2}{1}x^{2}+\frac{2\sqrt2}{1}x^{1}+\frac{\sqrt2}{1}x^{0}$

$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{4}+\sqrt2x^{2}+2\sqrt2x+\sqrt2$             [Since $x^0=1$]

Hence, $-x^6+2x^4+4x^3+2x^2$ divided by $\sqrt2x^2$ is $\frac{-1}{\sqrt2}x^{4}+\sqrt2x^{2}+2\sqrt2x+\sqrt2$.

(ii) 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 $x^0=1$]

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

(iii) The given expression is $\sqrt3a^4+2\sqrt3a^3+3a^2-6a$ by $3a$.

$\sqrt3a^4+2\sqrt3a^3+3a^2-6a \div 3a=\frac{\sqrt3a^4}{3a}+\frac{2\sqrt3a^3}{3a}+\frac{3a^2}{3a}-\frac{6a}{3a}$

$\sqrt3a^4+2\sqrt3a^3+3a^2-6a \div 3a=\frac{\sqrt3}{\sqrt3 \times \sqrt3}a^{4-1}+\frac{2\sqrt3}{\sqrt3 \times \sqrt3}a^{3-1}+\frac{3}{3}a^{2-1}-2a^{1-1}$

$\sqrt3a^4+2\sqrt3a^3+3a^2-6a \div 3a=\frac{1}{\sqrt3}a^{3}+\frac{2}{\sqrt3}a^{2}+a^{1}-2a^{0}$

$\sqrt3a^4+2\sqrt3a^3+3a^2-6a \div 3a=\frac{1}{\sqrt3}a^{3}+\frac{2}{\sqrt3}a^{2}+a-2$             [Since $a^0=1$]

Hence, $\sqrt3a^4+2\sqrt3a^3+3a^2-6a$ divided by $3a$ is $\frac{1}{\sqrt3}a^{3}+\frac{2}{\sqrt3}a^{2}+a-2$.

Updated on: 13-Apr-2023

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