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 $-2a^2+2a+\frac{1}{2}$.
- Related Articles
- 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 $3a$
- Solve the following pair of linear equations by the elimination method and the substitution(i) $x + y = 5$ and $2x – 3y = 4$(ii) $3x + 4y = 10$ and $2x – 2y = 2$(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$.
- Find the following products:(i) $(x + 4) (x + 7)$(ii) $(x - 11) (x + 4)$(iii) $(x + 7) (x - 5)$(iv) $(x - 3) (x - 2)$(v) $(y^2 - 4) (y^2 - 3)$(vi) $(x + \frac{4}{3}) (x + \frac{3}{4})$(vii) $(3x + 5) (3x + 11)$(viii) $(2x^2 - 3) (2x^2 + 5)$(ix) $(z^2 + 2) (z^2 - 3)$(x) $(3x - 4y) (2x - 4y)$(xi) $(3x^2 - 4xy) (3x^2 - 3xy)$(xii) $(x + \frac{1}{5}) (x + 5)$(xiii) $(z + \frac{3}{4}) (z + \frac{4}{3})$(xiv) $(x^2 + 4) (x^2 + 9)$(xv) $(y^2 + 12) (y^2 + 6)$(xvi) $(y^2 + \frac{5}{7}) (y^2 - \frac{14}{5})$(xvii) $(p^2 + 16) (p^2 - \frac{1}{4})$
- Simplify:(i) $2x^2 (x^3 - x) - 3x (x^4 + 2x) -2(x^4 - 3x^2)$(ii) $x^3y (x^2 - 2x) + 2xy (x^3 - x^4)$(iii) $3a^2 + 2 (a + 2) - 3a (2a + 1)$(iv) $x (x + 4) + 3x (2x^2 - 1) + 4x^2 + 4$(v) $a (b-c) - b (c - a) - c (a - b)$(vi) $a (b - c) + b (c - a) + c (a - b)$(vii) $4ab (a - b) - 6a^2 (b - b^2) -3b^2 (2a^2 - a) + 2ab (b-a)$(viii) $x^2 (x^2 + 1) - x^3 (x + 1) - x (x^3 - x)$(ix) $2a^2 + 3a (1 - 2a^3) + a (a + 1)$(x) $a^2 (2a - 1) + 3a + a^3 - 8$(xi) $\frac{3}{2}-x^2 (x^2 - 1) + \frac{1}{4}-x^2 (x^2 + x) - \frac{3}{4}x (x^3 - 1)$(xii) $a^2b (a - b^2) + ab^2 (4ab - 2a^2) - a^3b (1 - 2b)$(xiii) $a^2b (a^3 - a + 1) - ab (a^4 - 2a^2 + 2a) - b (a^3- a^2 -1)$.
- Simplify:$(x^3 - 2x^2 + 3x - 4) (x - 1) - (2x - 3) (x^2 - x + 1)$
- Multiply:$(3x^2 +y^2)$ by $(2x^2 + 3y^2)$
- Divide $2x^3-5x^2+8x-5 \ by \ 2x^2-3x+5$
- Which of the following expressions are not polynomials?:
(i) $x^2+2x^{-2}$
(ii) $\sqrt{ax}+x^2-x^3$
(iii) $3y^3-\sqrt{5}y+9$
(iv) $ax^{\frac{1}{2}}y^7+ax+9x^2+4$
(v) $3x^{-3}+2x^{-1}+4x+5$
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- Solve the following quadratic equation by factorization: $3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5, x≠\frac{1}{3}, \frac{-3}{2}$
- Multiply $-\frac{3}{2}x^2y^3$ by $(2x-y)$ and verify the answer for $x = 1$ and $y = 2$.
- Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:(i) $t^2-3, 2t^4 + t^3 - 2t^2 - 9t - 12$(ii) $x^2 + 3x + 1, 3x^4+5x^3-7x^2+2x + 2$(iii) $x^3 -3x + 1, x^5 - 4x^3 + x^2 + 3x + l$
- Subtract:(i) $-5xy$ from $12xy$(ii) $2a^2$ from $-7a^2$(iii) \( 2 a-b \) from \( 3 a-5 b \)(iv) \( 2 x^{3}-4 x^{2}+3 x+5 \) from \( 4 x^{3}+x^{2}+x+6 \)(v) \( \frac{2}{3} y^{3}-\frac{2}{7} y^{2}-5 \) from \( \frac{1}{3} y^{3}+\frac{5}{7} y^{2}+y-2 \)(vi) \( \frac{3}{2} x-\frac{5}{4} y-\frac{7}{2} z \) from \( \frac{2}{3} x+\frac{3}{2} y-\frac{4}{3} z \)(vii) \( x^{2} y-\frac{4}{5} x y^{2}+\frac{4}{3} x y \) from \( \frac{2}{3} x^{2} y+\frac{3}{2} x y^{2}- \) \( \frac{1}{3} x y \)(viii) \( \frac{a b}{7}-\frac{35}{3} b c+\frac{6}{5} a c \) from \( \frac{3}{5} b c-\frac{4}{5} a c \)
- Divide $x^5+3x^4-5x^3+14x^2+30x-16 \ by \ x^3-x^2+x+8$
- Solve the following quadratic equation by factorization: $\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}, x ≠ 1, -2, 2$
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies