Question

How do you simplify \[\dfrac{{{x^2} + 5x + 6}}{{x + 1}}.\dfrac{{{x^2} - 1}}{{x + 3}}\] ?

Answer 1

Hint: Here we need to simplify the numerator and the denominators. Since we have a quadratic equation in the numerator we find the factors of it and we have \[{x^2} - 1\] in the numerator as well. To simplify this we use algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] . After done with the simplification of numerators we substitute in the given problem and we simplify it.

Complete step-by-step answer:
Given,
 \[\dfrac{{{x^2} + 5x + 6}}{{x + 1}}.\dfrac{{{x^2} - 1}}{{x + 3}}\] .
We have \[{x^2} + 5x + 6\] , we can find the factors by expanding the middle term.
 \[ \Rightarrow {x^2} + 3x + 2x + 6\]
Now taking x common in the first two terms and 2 common in the remaining two terms,
 \[ \Rightarrow x(x + 3) + 2(x + 3)\]
Again taking \[(x + 3)\] common we have
 \[ \Rightarrow (x + 3)(x + 2)\]
That is we have, \[ \Rightarrow {x^2} + 5x + 6 = (x + 3)(x + 2)\]
Now, we have \[{x^2} - 1\] which is same as \[{x^2} - {1^2}\]
We know the algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] , applying this we have,
 \[ \Rightarrow {x^2} - {1^2} = \left( {x - 1} \right)\left( {x + 1} \right)\]
Substituting these in the given problem we have,
 \[ \Rightarrow \dfrac{{\left( {x + 3} \right)\left( {x + 2} \right)}}{{x + 1}}.\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 3}}\]
Cancelling the terms we have,
 \[ \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right)\]
We can stop here, but we can simplify it further,
 \[ \Rightarrow x\left( {x - 1} \right) + 2\left( {x - 1} \right)\]
 \[ \Rightarrow {x^2} - x + 2x - 2\]
 \[ \Rightarrow {x^2} + x - 2\] .
Thus we have the simplified form of \[\dfrac{{{x^2} + 5x + 6}}{{x + 1}}.\dfrac{{{x^2} - 1}}{{x + 3}}\] is \[ \Rightarrow {x^2} + x - 2\] .
So, the correct answer is “ \[ {x^2} + x - 2\] ”.

Note: We can also find the solution of given problem by equating the obtained simplified form \[{x^2} + x - 2\] or \[\left( {x + 2} \right)\left( {x - 1} \right)\] to zero.
That is,
 \[\left( {x + 2} \right)\left( {x - 1} \right) = 0\]
By zero product principle we have,
 \[\left( {x + 2} \right) = 0\] or \[\left( {x - 1} \right) = 0\]
 \[ \Rightarrow x = - 2\] or \[x = 1\] . This is the solution to the given problem.