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 $-2a^2+2a+\frac{1}{2}$.
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