Question

How do you factor ${x^3} + 3{x^2} + 7x + 21$?

Answer 1

Hint: The given question asks us to solve the above cubic equation by factoring the equation. The cubic equation is one whose degree is $3$since the term with highest power is the first term with the power of $3$the given equation is cubic. The first step to solve these question is to find a zero of the equation by hit or trial method by putting values such as $0,1,2,3, - 1, - 2, - 3$ but since this method is cumbersome because we have to check each and every value, we see that the ratio of the first two parts and last two parts is equal in the given equation. Thus we try to factor the equation by using this information.

Complete step by step solution:
The first two parts ration is $\dfrac{1}{3}$and then next two parts ratio is also the same so we will use this information to solve the given question,
${x^3} + 3{x^2} + 7x + 21 = {x^2}(x + 3) + 7(x + 3)$
${x^3} + 3{x^2} + 7x + 21 = ({x^2} + 7)(x + 3)$
Thus the given equation is factored the another important thing to remember here is that the quadratic factor in this question could have been solved further but that would have made the factors of the complex nature therefore it is left as it is,
So, the correct answer is “$ ({x^2} + 7)(x + 3)$”.

Note: The given equation is in the form of a cubic equation this type of question are best solved by hit and trial method but that method is very cumbersome since we saw that the ratios are same we applied that method in case it was not possible then we would have used that method and by hit and trial achieved the value of $ - 3$.