Question

Factorise the given expression ${y^{16}} - 63{y^8} - 64$
A.$({y^8} - 1)({y^4} + 8)({y^4} - 8)$
B.${({y^4} + 8)^2}({y^8} + 1)$
C.${({y^4} - 8)^2}({y^8} - 1)$
D.$({y^8} + 1)({y^4} + 8)({y^4} - 8)$

Answer 1

Hint: We consider y $^8 = t$ then solve or factorise the quadratic equation which is formed. Then we will solve further.

Complete step-by-step answer:
Consider, ${y^8} = t$
Rewrite the given equation in the form as shown: ${\left( {{y^8}} \right)^2} - 63{y^8} - 64$
Therefore, ${y^{16}} - 63{y^8} - 64$ becomes,
${y^{16}} - 63{y^8} - 64$ $ = {t^2} - 63t - 64$
Now, factorise${t^2} - 63t - 64$. We have to find or think two numbers whose sum is -63 and product is 64 .
$\therefore $ Numbers are -64 and 1. Such that Sum of numbers $ = - 64 + 1 = - 63$ and product of numbers $ = - 64 \times 1 = - 64$
$\therefore {t^2} - 63t - 64 = {t^2} - 64t + t - 64$
$ = t(t - 64) + 1(t - 64) = (t + 1)(t - 64)$
Put $t = {y^8}$,
$\therefore (t + 1)(t - 64) = \left( {{y^8} + 1} \right)\left( {{y^8} - 64} \right)$ - (A)
Now we will factorise $\left( {{y^8} - 64} \right)$ using formula ${a^2} - {b^2} = (a + b)(a - b)$
$ \Rightarrow \;\;\;{\kern 1pt} {y^8} - 64 = {\left( {{y^4}} \right)^2} - {(8)^2} = \left( {{y^4} + 8} \right)\left( {{y^4} - 8} \right)$
$\therefore $ (A) Becomes $\left( {{y^8} + 1} \right)\left( {{y^4} + 8} \right)\left( {{y^4} - 8} \right)$
 Therefore, factors of ${y^{16}} - 63{y^8} - 64$ are $\left( {{y^8} + 1} \right)\left( {{y^4} + 3} \right)\left( {{y^4} - 8} \right)$
Hence, Option (A) is correct.

Additional Information: Factoring means finding what to multiply together to get an expression.
When trying to factor, follow these steps:
"Factor out" any common terms see if it fits any of the identities, plus any more you may know Keep going till you can't factor any more
Factoring is a process of changing an expression from a sum or difference of terms to a product of factors.
A polynomial can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it.
The process involved in breaking a polynomial into the product of its factors is known as the factorization of polynomials


Note: While factoring the big powers first look at it, may be most of the times square of one power is given as another power which helps to make the factors of given function.