Question

Divide the polynomial by the given monomial:
i)$\left( 5{{x}^{2}}-6x \right)\div 3x$
ii) $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}$
iii) $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{x}^{2}}$
iv)$\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x$
v) $\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}$

Answer 1

Hint: First make the expression from $a\div b$ to the form of $\dfrac{a}{b}$ then separate the terms, that is in the form of $\left( \dfrac{a}{c}-\dfrac{b}{c} \right)$. Then cancel the like terms and get the desired result.


Complete step-by-step answer:
Now we can write $a\div b$ as, $\dfrac{a}{b}.$

_{Now consider first expression,}

i)$\left( 5{{x}^{2}}-6x \right)\div 3x$
Then we can write $\left( 5{{x}^{2}}-6x \right)\div 3x$ as, $\dfrac{5{{x}^{2}}-6x}{3x}.$
If $\left( \dfrac{a-b}{c} \right)$ is the form then we can write as $\left( \dfrac{a}{c}-\dfrac{b}{c} \right).$
So, $\left( \dfrac{5{{x}^{2}}-6}{3x} \right)$ can be written as,
$\left( \dfrac{5{{x}^{2}}}{3x}-\dfrac{6x}{3x} \right)$
Cancelling the like terms, the above expression can further simplified as,
$\left( \dfrac{5x}{3}-2 \right)$
Therefore, $\left( 5{{x}^{2}}-6x \right)\div 3x=\left( \dfrac{5x}{3}-2 \right).$

Now consider the second expression,

ii) $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}$
As we can write $a\div b$ as $\dfrac{a}{b}.$
So, we can write $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}$ as $\dfrac{3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}}{{{y}^{2}}}.$
If $\left( \dfrac{q-b+c}{d} \right)$ can be written as $\left( \dfrac{a}{d}-\dfrac{b}{d}+\dfrac{c}{d} \right).$
So, $\dfrac{3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}}{{{y}^{2}}}$ can be written as,
$\left( \dfrac{3{{y}^{8}}}{{{y}^{4}}}-\dfrac{4{{y}^{6}}}{{{y}^{4}}}+\dfrac{5{{y}^{4}}}{{{y}^{4}}} \right)$
Cancelling the like terms, the above expression can be further simplified as,
$\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right)$
Therefore, $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}=\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right).$

Now consider the third expression,

iii) $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{x}^{2}}$
As we can write $a\div b$ as $\dfrac{a}{b}.$
So we can write $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{x}^{2}}$ as,
$\dfrac{8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)}{4{{x}^{2}}{{y}^{2}}{{x}^{2}}}$
If $e\left( \dfrac{a+b+c}{d} \right)$, then it can be written as $\dfrac{ea}{d}+\dfrac{eb}{d}+\dfrac{ec}{d}.$
So, $\dfrac{8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)}{4{{x}^{2}}{{y}^{2}}{{x}^{2}}}$ can be written as,
$\dfrac{8{{x}^{3}}{{y}^{2}}{{z}^{2}}}{4{{x}^{2}}{{y}^{2}}{{x}^{2}}}+\dfrac{8{{x}^{2}}{{y}^{3}}{{z}^{2}}}{4{{x}^{2}}{{y}^{2}}{{x}^{2}}}+\dfrac{8{{x}^{2}}{{y}^{2}}{{z}^{3}}}{4{{x}^{2}}{{y}^{2}}{{x}^{2}}}$
Cancelling the like terms, the above expression can be simplified as,
(2x + 2y + 2z)
Therefore, $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{x}^{2}}=2\left( x+y+z \right)$.

Now consider the fourth expression,

iv)$\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x$
As we can write $a\div b$ as $\dfrac{a}{b}.$
So, we can write $\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x$ as,
 $\dfrac{{{x}^{3}}+2{{x}^{2}}+3x}{2x}$
If $\left( \dfrac{a+b+c}{d} \right)$, then it can be written as $\dfrac{a}{b}+\dfrac{b}{d}+\dfrac{c}{d}.$
So, $\left( \dfrac{{{x}^{3}}+2{{x}^{2}}+3x}{2x} \right)$ can be written as,
 $\dfrac{{{x}^{3}}}{2x}+\dfrac{2{{x}^{2}}}{2x}+\dfrac{3x}{2x}$
Cancelling the like terms, the above expression can be simplified as,
$\left( \dfrac{{{x}^{2}}}{2}+x+\dfrac{3}{2} \right)$
Therefore, \[\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x=\dfrac{{{x}^{2}}}{2}+x+\dfrac{3}{4}.\]

Now consider the last expression,

v) $\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}$

As we can write $a\div b$ as $\dfrac{a}{b}.$
If $\left( \dfrac{a-b}{c} \right)$, then it can be written as $\left( \dfrac{a}{c}-\dfrac{b}{c} \right).$
Then we can write $\dfrac{{{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}}}{{{p}^{3}}{{q}^{3}}}$ as,
\[\left( \dfrac{{{p}^{3}}{{q}^{6}}}{{{p}^{3}}{{q}^{3}}}-\dfrac{{{p}^{6}}{{q}^{3}}}{{{p}^{3}}{{q}^{3}}} \right)\]
Cancelling the like terms, the above expression can be written as,
${{q}^{2}}-{{p}^{3}}$
Therefore, $\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}={{q}^{2}}-{{p}^{3}}$

Note: Students must be careful while dividing separately and cancelling the common power between numerator and denominator. They should also do calculations carefully to avoid any mistakes. Another approach is to use a long division method of polynomials.