Factorize the given equation: - $5{x^2} - 16x - 21$

Hint: We were given with a quadratic equation and we have to factorize it. Let us suppose a quadratic equation
 $ \Rightarrow a{x^2} + bx + c = 0$
Here, in this equation first multiply a and c and find all the combinations of the numbers that give the product ac. Now, take a set of numbers from these factors so that their addition or subtraction gives b. After this split the x term and write it in the form of addition or subtraction of these chosen factors (the set of numbers which we selected above).

Complete step-by-step answer:
We have a quadratic equation i.e.
$ \Rightarrow 5{x^2} - 16x - 21$ …….(1)
First of all, multiply 5 and 21 i.e.
$ \Rightarrow 5 \times 21 = 105$
Now, find all the combination of numbers that gives the product 105 i.e.
$ \Rightarrow 1 \times 105$, $5 \times 21$, $7 \times 15$, $3 \times 35$
After this take a set of numbers from these factors so that their addition or subtraction gives 16 i.e. the difference of 5 and 21 gives 16.
$ \Rightarrow 21 - 5 = 16$
Now, split the x term and write it in the form of subtraction in the equation 1 we get,
$ \Rightarrow 5{x^2} - 16x - 21 = 5{x^2} - (21 - 5)x - 21$
$ \Rightarrow 5{x^2} - 16x - 21 = 5{x^2} - 21x + 5x - 21$
Taking the first two terms and the last two terms separately. Take x common from first two terms and 1 from last two terms we get,
$ \Rightarrow 5{x^2} - 16x - 21 = x(5x - 21) + 1(5x - 21)$
Now take $5x - 21$common from the both terms we get,
$ \Rightarrow 5{x^2} - 16x - 21 = (5x - 21)(x + 1)$
Hence by factoring $5{x^2} - 16x - 21$we get $(5x - 21)(x + 1)$.

Note: The common mistake done by students is to take wrong combinations but we have to take care of the numbers chosen should give 105 when multiplied and give 16 while subtracting. Secondly, while splitting x term students generally make mistakes in sign of the term. So, to avoid these confusions take the addition or subtraction of numbers in brackets after the sign of x term.
You can find the roots of the equation by equating each factor with zero we get,
$ \Rightarrow (5x - 21)(x + 1) = 0$
$ \Rightarrow 5x - 21 = 0$ and $x + 1 = 0$
$ \Rightarrow x = \dfrac{{21}}{5}$and $x = - 1$
You can also find the roots by using the following formula i.e.
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$