Question

How do you factor ${x^2} - 3x - 18$?

Answer 1

Hint:
Whenever we need to factorise any quadratic equation we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation. Now we will get the equation in form of the ${\left( {a + b} \right)^2} \pm {\text{c or }}{\left( {a - b} \right)^2} \pm {\text{c }}$ and now we can simplify it further as much as we can.

Complete step by step solution:
Here we are given the quadratic equation which we need to factorise. So we must know that any equation of degree $2$ is known as the quadratic equation. In order to find its factor we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation.
So here we can see in the above equation which is ${x^2} - 3x - 18$ that the coefficient of $x{\text{ is }} - 3$
So we need to add and subtract the value ${\left( {\dfrac{{ - 3}}{2}} \right)^2} = \dfrac{9}{4}$ in the above quadratic equation which is given. Hence we will get:
${x^2} - 3x - 18 + \dfrac{9}{4} - \dfrac{9}{4}$
Now we can write it in the form:
$
  \left( {{x^2} - 3x + \dfrac{9}{4}} \right) - 18 - \dfrac{9}{4} \\
  \left( {{x^2} - 3x + \dfrac{9}{4}} \right) - \left( {18 + \dfrac{9}{4}} \right) - - - - - - (1) \\
 $
Now we know that we can write the term in the first bracket of the equation (1) as:
$\left( {{x^2} - 3x + \dfrac{9}{4}} \right) = {\left( {x - \dfrac{3}{2}} \right)^2}$
And also we can solve $18 + \dfrac{9}{4}$ and we know that here we can take the LCM as $4$ because we know that LCM is the lowest number that is completely divisible by both the denominators which are ${\text{1 and 4}}$
So we will get:
$18 + \dfrac{9}{4} = \dfrac{{72 + 9}}{4} = \dfrac{{81}}{4}$
Now we can substitute these two calculated values in the equation (1) and we will get:
${\left( {x - \dfrac{3}{2}} \right)^2} - \dfrac{{81}}{4}$
Now we can write it as:
${\left( {x - \dfrac{3}{2}} \right)^2} - \dfrac{{81}}{4}$
${\left( {x - \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{9}{2}} \right)^2}$
Now we know that:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
So we can apply it over here and get:
${\left( {x - \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{9}{2}} \right)^2}$
$\left( {x - \dfrac{3}{2} + \dfrac{9}{2}} \right)\left( {x - \dfrac{3}{2} - \dfrac{9}{2}} \right)$
On simplifying it we will get:
$
  \left( {x + \dfrac{6}{2}} \right)\left( {x - \dfrac{{12}}{2}} \right) \\
  \left( {x + 3} \right)\left( {x - 6} \right) \\
 $

Hence we get the quadratic equation after factoring as $\left( {x + 3} \right)\left( {x - 6} \right)$

Note:
Here we can do it even by splitting the $ - 3x$ in the quadratic equation as $ - 6x + 3x$
Then we will get:
${x^2} - 3x - 18$
$
  {x^2} - 6x + 3x - 18 \\
  x\left( {x - 6} \right) + 3\left( {x - 6} \right) \\
  \left( {x + 3} \right)\left( {x - 6} \right) \\
 $
Hence this method can also be used over here but the method we used is applicable always when we do not know how to split the coefficient of $x$ term.