Question

Factorize the given equation: ${x^2} + 3x - \left( {{a^2} + a - 2} \right)$.

Answer 1

Hint: In this question, we are given an equation which further has another equation in itself. So, we will start by factoring the smaller equation in the given equation. After we have found its factors, we will compare the resultant equation with ${x^2} - \left( {a + b} \right)x + ab$ and will find out the values of a and b and then, we will form the equation in this format. Next, we will take x common and find the factors just like we find of any other equation. Then, we will simplify and we will have our answer.

Complete step-by-step solution:
We are given an equation and we have been asked to factorize the given equation.
If we look on the given equation closely, we will notice that there is another quadratic equation in the already given quadratic equation. So, we will start factorizing that equation first.
$ \Rightarrow {a^2} + a - 2$
Factorize this equation using splitting the middle term. Think of any two factors of 2- positive or negative, such that when they are multiplied they will give -2 and when they are added, they will give 1. Two such factors of 2 are 2 and -1.
$ \Rightarrow {a^2} + 2a - a - 2$
Simplifying,
$ \Rightarrow a\left( {a + 2} \right) - 1\left( {a + 2} \right)$
$ \Rightarrow \left( {a - 1} \right)\left( {a + 2} \right)$
Putting this in the given equation as ${a^2} + a - 2 = \left( {a - 1} \right)\left( {a + 2} \right)$.
$ \Rightarrow {x^2} + 3x - \left( {a - 1} \right)\left( {a + 2} \right)$
If we compare this equation with the formula - ${x^2} - \left( {a + b} \right)x + ab$, we can see that $ab = \left( {a - 1} \right)\left( {a + 2} \right)$ and, $a = \left( {a - 1} \right),b = \left( {a + 2} \right)$. Now, as we can see that $a + b = 3$, we will place a and b in such a way that it gives us 3.
Using them to factorize the given equation,
$ \Rightarrow {x^2} + \left[ {\left( {a + 2} \right) - \left( {a - 1} \right)} \right]x - \left( {a - 1} \right)\left( {a + 2} \right)$
Simplifying further,
$ \Rightarrow {x^2} + \left( {a + 2} \right)x - \left( {a - 1} \right)x - \left( {a - 1} \right)\left( {a + 2} \right)$
Taking x common to form factors,
$ \Rightarrow x\left( {x + a + 2} \right) - \left( {a - 1} \right)\left( {x + a + 2} \right)$
$ \Rightarrow \left( {x + a + 2} \right)\left( {x - a + 1} \right)$

$\therefore $ On factoring ${x^2} + 3x - \left( {{a^2} + a - 2} \right)$, we will get $\left( {x + a + 2} \right)\left( {x - a + 1} \right)$.

Note: The step where we place a and b in such a way that it gives us 3 has not been explained in the question deeply as it is not required. But it is explained below for understanding:
We know that $a = \left( {a - 1} \right),b = \left( {a + 2} \right)$ and $a + b = 3$. So, let us place them in such a way that they give us 3 when added or subtracted.
$ \Rightarrow \left( {a + 2} \right) + \left[ { - \left( {a - 1} \right)} \right] = 3$
Hence, on simplification we will get,
$ \Rightarrow \left( {a + 2} \right) - \left( {a - 1} \right) = 3$
Therefore, this has been used in the question.