Question

How do you factor the trinomial: $\left( {{b^2} - b - 6} \right)$ ?

Answer 1

Hint: Given polynomial is of degree 2. Polynomials of degree 2 are known as Quadratic polynomials. Quadratic polynomials can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the polynomial remains unchanged.

Complete step-by-step answer:
 For factorising the given quadratic polynomial $\left( {{b^2} - b - 6} \right)$ , we can use the splitting method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
So, we have, $\left( {{b^2} - b - 6} \right)$
$ = $${b^2} - \left( {3 - 2} \right)b - 6$
We split the middle term $ - b$ into two terms $ - 3b$ and $ + 2b$ since the product of these two terms, $ - 6{b^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $ - b$.
$ = $${b^2} - 3b + 2b - 6$
Taking out $b$ common from first two brackets and $2$ common from last two brackets, we get,
$ = $$b\left( {b - 3} \right) + 2\left( {b - 3} \right)$
$ = $$\left( {b + 2} \right)\left( {b - 3} \right)$
So, the factored form of the quadratic and trinomial polynomial $\left( {{b^2} - b - 6} \right)$ is $\left( {b + 2} \right)\left( {b - 3} \right)$.

Note: Splitting of middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Prime factorization of the product of the coefficients of the ${x^2}$ term and constant term should be done in order to find the possible terms in which the middle term can be split.