Factor the expression, if possible. x ^4 - 81?

x ^4 - 81 can be written as:
(x² - 9)(x² + 9)

a^2 - b^2 = (a + b)(a - b)

x^4 - 81
= (x^2 + 9)(x^2 - 9)
= (x^2 + 9)(x + 3)(x - 3)

x=3 and X=-3

(x² - 9)(x² + 9)
(x - 3)(x + 3)(x² + 9)

note that a^2-b^2=(a+b)(a-b) this is called the difference of squares rule
in this case a=a^2 and b=9
=>x^4-81=(a^2+9)(a^2-9)
again a^2-9 can be simplified using the difference of squares rule. so a^2=(a+3)(a-3)
putting these together:
x^4-81=(a^2+9)(a+3)(a-3)
it's easy! even i can do it!

difference of two squares has factors of the sum and difference of two terms.

simply get the principal root of the first term and the second term

x ^4 = x^2
81 = 9

then you will have (x^2 + 9)(x^2-9)
(x^2-9) is also difference of 2 terms

thus, you will have
(x-3)(x+3)

therefore, x ^4 - 81 = (x-3)(x+3)(x^2 + 9)


answer: (x-3)(x+3)(x^2 + 9)

hope that helps

factor out the terms

(x^2+9) (x^2-9) (then evalluate if the terms can still be
factor.then it will become)

(x^2+9) ( (X+9) (X-9) ) ----------then it is the final answer






just use your COMMON SENSE


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