How do you factor the expression $ {x^2} - x - 90? $

Hint: Factorizing reduces the higher degree equation into its linear equation. In the above-given question, we need to reduce the quadratic equation into its simplest form in such a way that the addition of products of the factors of the first and last term should be equal to the middle term i.e. $ - x $

Complete step by step answer:
 $ a{x^2} + bx + c $ is a general way of writing quadratic equations where a, b and c are numbers.
In the above expression,
a=1, b=-1, c=-90
 $ {x^2} - x - 90 $
First step is by multiplying the coefficient of $ {x^2} $ and the constant term -90, we get $ - 90{x^2} $ .
After this, factors of $ 90{x^2} $ should be calculated in such a way that their addition should be equal to \[ - x\].
Factors of -90 can be 10 and 9 or 30 and 3. But $ 30 + 3 \ne - 1 $ ,so we will use 10 and 9.
where \[ - 10x + 9x = - x\].
So, further, we write the equation by equating it with zero and splitting the middle term according to the factors.
 $
  {x^2} - x - 90 = 0 \\
  {x^2} + 9x - 10x - 90 = 0 \\
  $
Now, by grouping the first two and last two terms we get common factors.
 $
   \Rightarrow x(x + 9) - 10(x + 9) = 0 \\
   \Rightarrow (x - 10)(x - 9) = 0 \\
  $
Taking x common from the first group and 2 commons from the second we get the above equation.
Therefore, by solving the above quadratic equation we get factors as 10 and 9.

Note:
 In quadratic equation, an alternative way of finding the factors is by directly solving the equation by using a formula which is given below:
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
By substituting the values of a=1, b=-1 and c=-90 we get the factors of x.
 $
  x = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(1)( - 90)} }}{{2(1)}} \\
    \\
  $
 $ x = 9 $ or $ x = 10 $ .The values are 9 and 10.