Question

How do you factor $6{x^2} - 216$ ?

Answer 1

Hint: In this question, we need to find the factors of the given polynomial. The given problem is simple to solve. Firstly, we find out the common term in the given polynomial and factor it out. Then we obtain an expression where the both terms are perfect squares. We factor this using the difference of squares formula given by ${a^2} - {b^2} = (a + b)(a - b)$. Then substitute the values of a and b, then simplify the given expression and obtain the result.

Complete step by step answer:
Given the polynomial of the form $6{x^2} - 216$ …… (1)
We are asked to find the factors of the above polynomial given by the equation (1).
Note that above polynomial has a degree 2, so it is a quadratic polynomial.
Now let us consider the polynomial given by the equation (1).
We can write 216 as $216 = 6 \times 36$.
Then the equation becomes,
$ \Rightarrow 6{x^2} - 6 \times 36$
Now we factor out the common term which is 6, we get,
$ \Rightarrow 6({x^2} - 36)$
We know that 36 is a perfect square of 6. Hence we rewrite 36 as, $36 = {6^2}$.
So we have,
$ \Rightarrow 6({x^2} - {6^2})$
Note that the both terms inside the parenthesis are perfect squares. So we factor using the difference of squares formula which is given by,
${a^2} - {b^2} = (a + b)(a - b)$
Here we have $a = x$ and $b = 6$.
Hence we get the expression as,
$ \Rightarrow 6(x + 6)(x - 6)$

Therefore, the factorization of $6{x^2} - 216$ is given by $6(x + 6)(x - 6)$.

Note: It is important to remember the basic algebraic formulas . They are given by,
(1) ${(a + b)^2} = {a^2} + 2ab + {b^2}$
(2) ${(a - b)^2} = {a^2} - 2ab + {b^2}$
(3) ${a^2} - {b^2} = (a + b)(a - b)$
Also note that when we multiply the above obtained factors, we need to get the given equation as a result. If we don’t get the given equation as a result, then our solution is not correct. Also we must know the values of squares of numbers as its simplifies our work.