Question

How do you factor and solve \[{x^2} - x - 12?\]

Answer 1

Hint: Factorising 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 addition of products of the factors of first and last term should be equal to the middle term \[ - x\] .

Complete step-by-step answer:
 \[a{x^2} + bx + c\] is a general way of writing quadratic equations where a, b and c are the numbers.
In the above expression,
a=1, b=-1, c=12
 \[{x^2} - x - 12\]
First step is by multiplying the term \[{x^2}\] and the constant term -12, we get \[ - 12{x^2}\] .
 After this, factors of \[ - 12{x^2}\] should be calculated in such a way that their addition should be equal to -x .
Factors of 8 can be 3 and 4
where \[ - 4x + 3x = - x\] .
So, further we write the equation by equating it with zero and splitting the middle term according to the
factors.
 \[
   \Rightarrow {x^2} - x - 12 = 0 \\
   \Rightarrow {x^2} - 4x + 3x - 12 = 0 \;
 \]
Now, by grouping the first two and last two terms we get common factors.
 \[
   \Rightarrow x\left( {x - 4} \right) + 3\left( {x - 4} \right) = 0 \;
 \]
Taking x common from the first group and 1 common from the second we get the above equation.
We can further solve it we get,
 \[
   \Rightarrow \left( {x + 3} \right)(x - 4) = 0 \;
  x = - 3 \;
 \]
Or
 \[x = 4\]
Therefore, we get the above value for x.
So, the correct answer is “x=-3 or x=4”.

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=-12 we get the factors of x.
 \[x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 12} \right)} }}{{2\left( 1 \right)}}\] .Therefore, the value we get is 4 and 3.