Divide \[35{a^2} + 32a - 99\] by \[7a - 9\].

Hint: We can use long division method or factorisation method to solve this type sum.
Here we use Factorisation method, which is the process of reducing the bracket of a quadratic equation, instead of expanding the bracket and converting the equation to a product of factors which cannot be further.
While factorising a polynomial, these two points in mind:
1. Finding factors of the polynomial is just like doing any simple division, the only thing to be kept in mind is the accuracy of variables and coefficients.
2. Factorization by division method is the conventional method of finding factors of a polynomial equation.
Use, the above rule we divide the polynomial.

Complete step-by-step answer:
It is given that the equation we can write It as,
 \[\dfrac{{35{a^2} + 32a - 99}}{{7a - 9}}....\left( 1 \right)\]
Here, the given form \[35{a^2} + 32a - 99\] is considered as a polynomial to break the form
That is, \[35{a^2} + 32a - 99 = 0\]
To factorise, the given terms as
Multiply \[35\] by \[ - 99\], we get \[ - 3465\],
We have to break \[ - 3465\] by finding its divisors, its sum must be \[32\], the middle term.
We have to choose that type of break.
Write the divisors of \[ - 3465\] use prime factorisation method,
\[
  5\left| \!{\underline {\,
  {3465} \,}} \right. \\
  9\left| \!{\underline {\,
  {693} \,}} \right. \\
  11\left| \!{\underline {\,
  {77} \,}} \right. \\
  7\left| \!{\underline {\,
  7 \,}} \right. \\
 \]
Therefore \[3465 = 5 \times 9 \times 11 \times 7\]
\[3465 = 45 \times 77\]
Now, it is of the form, their difference \[(77 - 45)\] is \[32\]
Here the divisors of the given polynomial, their multiply is \[ - 3465\] , their sum is \[32\]
Hence the given polynomial will be written as \[35{a^2} - 45a + 77a - 99 = 0\]
Taking the common terms and we can write it as,
\[5a(7a - 9) + 11(7a - 9) = 0\]
Here we can write \[(7a - 9)\] commonly,
\[(7a - 9)(5a + 11) = 0\]
Now, the polynomial is broken into two factors, from now, the consideration of polynomials will fall.
That is, \[(7a - 9)(5a + 11)\] = \[35{a^2} + 32a - 99\]
Substitute the polynomial in \[\left( 1 \right)\] we get
\[\dfrac{{(7a - 9)(5a + 11)}}{{(7a - 9)}}\]
Cancel the same terms we get, \[(5a + 11)\]
\[\dfrac{{35{a^2} + 32a - 99}}{{7a - 9}} = (5a + 11)\]
Hence the required answer is \[(5a + 11)\].

Note: The alternative method is the normal division method. It is very easy to understand and solving will be easy, but we use the factorisation, when the factors can be found it is also very easy to solve.
Factorisation is the reverse function of multiplication. A form of disintegration, factorisation entails the gradual breakdown of a polynomial into its factors.