Question

Factorize: \[{{\left( {{x}^{2}}+x \right)}^{2}}+4\left( {{x}^{2}}+x \right)-12\]

Answer 1

Hint: To find the value of \[x\] , we will substitute the \[x\] terms in terms of another variable and then use the quadratic formula to find the value of \[x\] . To find the roots of the quadratic equation, we use the quadratic formula on the question, by using the long method formula:
The roots of the quadratic equation is \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
where \[a,\text{ }b,\text{ }c\] are referred from the equation written in the form of \[a{{x}^{2}}+bx+c=0\] .

Complete step-by-step answer:
To solve the question, we first equate the value of \[x\] terms from the quadratic equation \[{{\left( {{x}^{2}}+x \right)}^{2}}+4\left( {{x}^{2}}+x \right)-12\] where we take the value of \[{{x}^{2}}+x\] as \[t\] . Using the value of \[t\] , we form a new quadratic equation where we find the equation as:
 \[\Rightarrow {{\left( {{x}^{2}}+x \right)}^{2}}+4\left( {{x}^{2}}+x \right)-12\]
 \[\Rightarrow {{\left( t \right)}^{2}}+4\left( t \right)-12\]
Now let us find the roots of the above equation, by using the quadratic formula:
 \[\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
With \[a=1\] , \[b=4\] and \[c=-12\] we get the value of the roots as:
 \[\Rightarrow \dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 1\times -12}}{2\times 1}\]
 \[\Rightarrow \dfrac{-4\pm \sqrt{16+48}}{2}\]
 \[\Rightarrow \dfrac{-4\pm \sqrt{64}}{2}\]
 \[\Rightarrow \dfrac{-4+8}{2},\dfrac{-4-8}{2}\]
 \[\Rightarrow \dfrac{4}{2},\dfrac{-12}{2}\]
Hence, the value of the roots of the quadratic equation in terms of \[t\] is given as:
 \[\Rightarrow 2,-6\]
Now we equate the value of the \[t\] in terms of \[{{x}^{2}}+x\] , we get the value of \[x\] as:
For \[2\] : \[c=2\] or \[{{x}^{2}}+x=2\]
and
For \[-6\] : \[c=-6\] or \[{{x}^{2}}+x=-6\]
Taking the two roots we are equating to the value of \[c\] with \[{{x}^{2}}+x\] and the two roots we get are:
 \[{{x}^{2}}+x=2\] , \[{{x}^{2}}+x=-6\] or \[\left( {{x}^{2}}+x-2 \right)\] , \[\left( {{x}^{2}}+x+6 \right)\]
Therefore, the factorization of \[{{\left( {{x}^{2}}+x \right)}^{2}}+4\left( {{x}^{2}}+x \right)-12\] is \[\left( {{x}^{2}}+x-2 \right)\] , \[\left( {{x}^{2}}+x+6 \right)\] .
So, the correct answer is “ \[\left( {{x}^{2}}+x-2 \right)\] , \[\left( {{x}^{2}}+x+6 \right)\]”.

Note: Students may get wrong if they started solving the questions from the get go by expanding the equation as it is. The process gives the correct answer but the equation tends to stretch long with multiple power variables of \[x\] and finding common or shortcuts for such a long equation tends to get problematic at the end as the roots of the equation are quadratic themselves.