Question

Factorise the expression and divide them as directed
I. $ \left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right) $
II. $ \left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right) $
III. $ \left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right) $
IV. $ 4yz\left( {{z^2} + 6 - 16} \right) \div 2y\left( {z + 8} \right) $
V. $ 5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right) $
VI. $ 12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right) $
VII. $ 39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right) $

Answer 1

Hint: If there are common terms then divide them first and after that if the equation is the quadratic form try to factorize then divide if there are common terms. Again, divide the common terms by using the division method if there are common terms left after factorization.

Complete step-by-step answer:
The given equation is $ {y^2} + 7y + 10 $ .
We will take $ 10 $ as $ 5 $ and $ 2 $ . Since, the sum of $ 5 $ and $ 2 $ is $ 7 $ .
So, on rearranging the above expression, we get,
 $ \begin{array}{l}
{y^2} + 5y + 2y + 10\\
y\left( {y + 5} \right) + 2\left( {y + 5} \right)
\end{array} $
In the above expression, $ y + 5 $ is common we get,
  $ \left( {y + 2} \right)\left( {y + 5} \right) $
So, if we divide the above with $ \left( {y + 5} \right) $ we get,
 $ \left( {y + 2} \right) $
Therefore, on dividing we get $ \left( {y + 2} \right) $ .

(ii)
The given equation is $ {m^2} - 14m - 32 $ .
We will take $ 32 $ as $ 16 $ and $ 2 $ .Since, sum of $ 16 $ and $ - 2 $ is $ 14 $ .
So, on rearranging the above expression, we get,
 $ \begin{array}{l}
{m^2} - 16m + 2m - 32\\
m\left( {m - 16} \right) + 2\left( {m - 16} \right)
\end{array} $
In the above expression $ m - 16 $ is common we get,
 $ \left( {m + 2} \right)\left( {m - 16} \right) $
So if we divide the above with $ \left( {m + 2} \right) $ we get,
 $ m - 16 $
Therefore, on dividing we get $ m - 16 $ .

(iii)
The given equation is $ 5{p^2} - 25p + 20 $ .
Since, $ 5 $ is common in the equation, so we can rearrange it as,
 $ 5\left( {{p^2} - 5p + 4} \right) $
We will take $ 4 $ as multiplication of $ 4 $ and $ 1 $ . Since, sum of $ 1 $ and $ 4 $ is $ 5 $ .
So, on rearranging the above expression, we get,
 $ \begin{array}{l}
5\left( {{p^2} - 5p + 4} \right)\\
5\left( {{p^2} - 4p - p + 4} \right)
\end{array} $
In the above expression $ p - 1 $ is common, we get,
 $ \left( {p - 4} \right)\left( {p - 1} \right) $
So if we divide the above with $ p - 1 $ we get,
 $ 5\left( {p - 4} \right) $
Therefore, on dividing we get $ 5\left( {p - 4} \right) $ .

(iv)
The given equation is $ 4yz\left( {{z^2} + 6z - 16} \right) $ .
Since, $ 2y $ is common in the numerator and denominator let us divide the equation by $ 2y $ .
Then we will have the equation,
 $ 2z\left( {{z^2} + 6z - 16} \right) $
We will take $ 16 $ as $ 8 $ and $ - 2 $ . Since, sum of $ 8 $ and $ - 2 $ is $ 6 $ .
So on rearranging the above expression, we get,
 $ \begin{array}{l}
2z\left( {{z^2} + 8z - 2z - 16} \right)\\
2z\left( {z\left( {z + 8} \right) - 2\left( {z + 8} \right)} \right)
\end{array} $
In the above expression $ z + 8 $ is common we get,
 $ 2z\left( {z - 2} \right)\left( {z + 8} \right) $
So if we divide the above with $ \left( {z + 8} \right) $ we get,
 $ 2z\left( {z - 2} \right) $
Therefore, on dividing we get $ 2z\left( {z - 2} \right) $ .

(v)
The given equation is $ 5pq\left( {{p^2} - {q^2}} \right) $ .
Since, $ p $ is common in the numerator and denominator let us divide $ p $ .
Then we will have the equation,
 $ 5q\left( {{p^2} - {q^2}} \right) $
Since, we can write $ \left( {{p^2} - {q^2}} \right) $ as:
 $ {p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right) $
Now, our equation will become,
 $ 5\left( {p + q} \right)\left( {p - q} \right) \div 2\left( {p + q} \right) $
So if we divide the above with $ \left( {p + q} \right) $ , we get,
 $ \dfrac{5}{2}q\left( {p - q} \right) $

Therefore, on dividing we get $ \dfrac{5}{2}q\left( {p - q} \right) $ .

(vi)
The given equation is $ 12xy\left( {9{x^2} - 16{y^2}} \right) $ .
Since $ 4xy $ is common in the numerator and denominator let us divide $ 4xy $ .
We have, $ 3\left( {9{x^2} - 16{y^2}} \right) \div 2\left( {3x + 4y} \right) $
Since, $ {p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right) $ , so we can write $ 9{x^2} - 16{y^2} $ as: $ 9{x^2} - 16{y^2} = \left( {3x - 4y} \right)\left( {3x + 4y} \right) $
So if we divide the above with $ \left( {3x + 4y} \right) $ ,
 $ 3\left( {3x - 4y} \right) $
Therefore, on dividing we get $ 3\left( {3x - 4y} \right) $ .

 (vii)
The given equation is $ 39{y^3}\left( {50{y^2} - 98} \right) $ .
Since $ 26{y^2} $ is common in the numerator and denominator let us divide $ 26{y^2} $ .
We have $ 3y\left( {25{y^2} - 49} \right) \div \left( {5y + 7} \right) $ . Since, $ {p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right) $ , so we can write it as: $ 25{y^2} - 49 = \left( {5y - 7} \right)\left( {5y + 7} \right) $
So if we divide the above with $ 5y + 7 $ ,
 $ 3\left( {5y - 7} \right) $
Therefore, on dividing we get $ 3\left( {5y - 7} \right) $ .

Note: In this problem, only if the numerator and denominator have the same factor then only we can divide otherwise we need to go with the normal division method. For some equations we cannot factorise for such types with normal division. For example, we cannot factorize
 $ {x^2} + x + 1 $ .