Question

Divide as directed.
(i) \[5\left( 2x+1 \right)\left( 3x+5 \right)\div \left( 2x+1 \right)\]
(ii) \[26xy\left( x+5 \right)\left( y-4 \right)\div 13x\left( y-4 \right)\]
(iii) \[52pqr\left( p+q \right)\left( q+r \right)\left( r+p \right)\div 104pq\left( q+r \right)\left( r+p \right)\]
(iv) \[20\left( y+4 \right)\left( {{y}^{2}}+5y+3 \right)\div 5y\left( y+4 \right)\]
(v) \[x\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)\div x\left( x+1 \right)\]

Answer 1

Hint: To solve this question we will consider every option separately and then divide them using the fact that \[\dfrac{ab}{a}=b\]. Similarly, \[\dfrac{\left( x+1 \right)\left( x-1 \right)}{\left( x+1 \right)}=\left( x-1 \right)\]. So, we will apply this to solve all given parts of the question.

Complete step-by-step solution:
Consider
(i) \[5\left( 2x+1 \right)\left( 3x+5 \right)\div \left( 2x+1 \right)\]
\[\dfrac{5\left( 2x+1 \right)\left( 3x+5 \right)}{\left( 2x+1 \right)}\]
Now as \[\left( 2x+1 \right)\] is common in both numerator and denominator \[\Rightarrow \] we can cancel it.
Cancelling \[\left( 2x+1 \right)\] from both numerator and denominator we get,
\[5\left( 3x+5 \right)\]
So the answer of (i) is \[5\left( 3x+5 \right)\].
Consider option (ii) \[26xy\left( x+5 \right)\left( y-4 \right)\div 13x\left( y-4 \right)\].
\[\dfrac{26xy\left( x+5 \right)\left( y-4 \right)}{13x\left( y-4 \right)}\]
As \[x\left( y-4 \right)\] is common in both numerator and denominator \[\Rightarrow \] cancelling \[\left( x \right)\left( y-4 \right)\] from numerator and denominator we have,
\[=\dfrac{26y\left( x+5 \right)}{13}\]
Now as \[13\times 2=26\]
\[\Rightarrow \dfrac{26}{13}=2\]
\[2y\left( x+5 \right)\]
Answer of (ii) is \[2y\left( x+5 \right)\].
Consider option (iii) \[52pqr\left( p+q \right)\left( q+r \right)\left( r+p \right)\div 104pq\left( q+r \right)\left( r+p \right)\]
\[=\dfrac{52pqr\left( p+q \right)\left( q+r \right)\left( r+p \right)}{104pq\left( q+r \right)\left( r+p \right)}\]
Now as \[pq\left( q+r \right)\left( r+p \right)\] is common so we can cancel it from the numerator and denominator.
\[\Rightarrow \dfrac{52r\left( p+q \right)}{104}\]
As \[\dfrac{104}{52}=2\]
\[\Rightarrow \dfrac{52}{104}=\dfrac{1}{2}\]
Applying this we get
\[\dfrac{r}{2}\left( p+q \right)\].
Consider (iv) = \[20\left( y+4 \right)\left( {{y}^{2}}+5y+3 \right)\div 5y\left( y+4 \right)\]
(iv) = \[\dfrac{20\left( y+4 \right)\left( {{y}^{2}}+5y+3 \right)}{5y\left( y+4 \right)}\]
Now as \[\left( y+4 \right)\] is common so cancelling \[\left( y+4 \right)\] from both numerator and denominator we get,
(iv) = \[\dfrac{20\left( {{y}^{2}}+5y+3 \right)}{5y}\] and \[\dfrac{20}{5}=4\].
\[\Rightarrow \] (iv) = \[\dfrac{4\left( {{y}^{2}}+5y+3 \right)}{y}\]
So answer of (iv) is \[\dfrac{4{{y}^{2}}+20y+12}{y}\] or \[4y+20+\dfrac{12}{y}\].
Consider (v) \[x\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)\div x\left( x+1 \right)\]
(v) = \[\dfrac{x\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)}{x\left( x+1 \right)}\]
Now as \[x\left( x+1 \right)\] is common so cancelling it from both numerator and denominator we get,
(v) = \[\left( x+2 \right)\left( x+3 \right)\], which is the answer of (v).

Note: The key point here to note is that while dividing ab by a.
\[\dfrac{ab}{a}=b\], whenever \[a\ne 0\].
If the value of a in any case = 0 then, \[\dfrac{ab}{a}\ne b\].
Here we were not having any condition on x so all were valid.