Help me please Im stuck on this problem

Mathematics

1 answer:

satela [25.4K]3 years ago

5 0

Answer:

10$

Step-by-step explanation:

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Factor completely 81x^8-1

marysya [2.9K]

The complete factorized form for the given expression is $\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$

<u>Step-by-step explanation:</u>

Step 1: Given expression:

$81 x^{8}-1$

Step 2: Trying to factor as a Difference of Squares

Factoring $81 x^{8}-1$

As we know the theory that the difference of two perfect squares, $A^{2}-B^{2}$ can be factored into (A+B) (A-B)

from this, when analysing, 81 is the square of 9, $x^{8}$ is the square of $x^{4}$ . Hence, we can write the given expression as,

$\left(9 x^{4}\right)^{2}-1^{2}$

By using the theory, we get

$\left(9 x^{4}+1\right)\left(9 x^{4}-1\right)$

Again, we can further factorise the term $\left(9 x^{4}-1\right)$

$9 x^{4}$ is the square of $3 x^{2}$ . Therefore, it can be expressed as below

$\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$

Now, we can not factorise further the term $\left(3 x^{2}-1\right)$ . Because it will come as $\sqrt{3} x$ (3 is not a square term). Thereby concluding that the complete factorisation for the given expression is $\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$