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# How do you solve $2{{x}^{2}}-x-6=0$ by factoring?

Last updated date: 23rd Feb 2024
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Hint: Factorisation is a method of expressing or writing a given number in the form of the product of other numbers. The numbers present in the product are called as factors of the original number.

Complete step by step solution:
Let us first understand what is meant by the term factorisation.Factorisation is a method of expressing or writing a given number in the form of the product of other numbers. The numbers present in the product are called as factors of the original number.For example, the number 6 can be written in the form of the product of the numbers 2 and 3. i.e. $6=3\times 2$. Here, 3 and 2 are the factors of 6. The method of factorisation can also be used for expressions containing variables.

Here, the expression is given as $2{{x}^{2}}-x-6$ …. (i).
We can write the expression (i) as $2{{x}^{2}}-x-6=2{{x}^{2}}-4x+3x-6$.
Now, we can take the term ‘2x’ as common from the first two terms and 3 as common from the second two terms as shown below.
$2{{x}^{2}}-x-6=2x(x-2)+3(x-2)$.
Now, take the term (x - 2) as a common term. Then,
$2{{x}^{2}}-x-6=(x-2)(2x+3)$
Therefore, we factored the expression.
But, it is given that $2{{x}^{2}}-x-6=0$.
Then this means that $(x-2)(2x+3)=0$
This means that either $x-2=0$ or $2x+3=0$
If we further simplify, then we get that either $x=2$ or $x=-\dfrac{3}{2}$.

Therefore, the roots or the solutions of the given quadratic equation are $x=2$ or $x=-\dfrac{3}{2}$.

Note: Here is a trick to solve a given quadratic equation with factorisation method.When we use factorisation method, we must split the middle term of the expression in two terms such that the sum of the two terms is equal to the middle terms and product of the two is equal to product of the other terms.