How do i factor x^8 - 16?

a^2 - b^2 ≡ (a + b)(a - b)

x^8 - 16
= (x^4)^2 - 4^2
= (x^4 + 4)(x^4 - 4)
= (x^4 + 4)[(x^2)^2 - 2^2]
= (x^4 + 4)(x^2 + 2)(x^2 - 2)

(x^4-4)(x^4+4)=0
(x^2-2)(x^2+2)(x^4+4)=0
(x+root2)(x-root2)(x^2+2)(x^4+4)=0

x^8 - 16 is the difference of two squares. Take the sq rt of x^8 which is x^4 * x^4 and 16 which is 4 * 4. Separate one by a + and the other by a -.

(x^4 + 4)(x^4 - 4). Now x^4 + 4 can't be factored any further but (x^4 - 4) can. It also is a difference of two squares (x^2 + 2)(x^2 - 2).

Answer: (x^4 + 4)(x^2 + 2)(x^2 - 2)

Notice that this is the difference of two perfect squares.
x^8 = x^4 * x^4
16 = 4 * 4
x^8 - 16 = (x^4)² - (4)²

Remember that you can factor the difference of two perfect squares.
(a - b)(a + b) = a² - b²

Given: x^8 - 16 = (x^4)² - (4)²
Means: a = x^4, b = 4

Apply the factoring formula.
x^8 - 16 = (x^4)² - (4)² =
(x^4 - 4)(x^4 + 4) =

Notice that x^4 - 4 is also the difference of two squares.
x^4 = (x²)²
4 = 2 * 2 = (2)²
x^4 - 4 = (x²)² - (2)²

Remember that you can factor the difference of two perfect squares.
(a - b)(a + b) = a² - b²

Given: x^4 - 4 = (x²)² - (2)²
Means: a = x², b = 2

Apply the factoring formula.
(x^4 - 4)(x^4 + 4) =
(x² - 2)(x² + 2)(x^4 + 4)

x² - 2 is also the difference of two squares.
x² = (x)²
2 = (√2)²
x² - 2 = (x)² - (√2)²

Remember that you can factor the difference of two perfect squares.
(a - b)(a + b) = a² - b²

Given: x² - 2 = (x)² - (√2)²
Means: a = x, b = √2

Apply the factoring formula.
(x² - 2)(x² + 2)(x^4 + 4)
(x - √2)(x + √2)(x² + 2)(x^4 + 4) <====== factored

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Sometimes you'll come across polynomials that match a particular pattern.

HINT: Memorize these commonly occurring factoring formulas

Perfect square binomials:
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

Difference of two perfect squares:
(a - b)(a + b) = a² - b²

Sum of two perfect cubes:
a³ + b³ = (a + b)(a² - ab + b²)

Difference of two perfect cubes:
a³ - b³ = (a - b)(a² + ab + b²)

Notice that the formulas for the perfect cubes are almost the same, but the signs are different. You can use the SOAP mnemonic to remember the signs.

S = The first sign is always the SAME as the sign between the 2 cubes
O = The second sign is always the OPPOSITE as the sign between the 2 cubes
AP = The last sign is ALWAYS POSITIVE

Through the entire answer, you're going to be using the identity a^2-b^2=(a+b)(a-b)

x^8-16
=(x^4+4)(x^4-4) [(x^4)^2=a^2=x^8]
=(x^4+1)(x^2+2)(x^2-2)
=(x^4+1)(x^2+2)(x+root2)(x-root2)

(x^4 + 4)(x^4 - 4)

Notice how the middle terms (-8x^4 + 8x^4) cancel out?

You can factor this even further by factoring out the second term with the same method, getting:

(x^4 + 4)(x^2 - 2)(x^2 + 2)

:)

=(x^4-4)(x^4+4)=(x^4+4)(x^2-2)(x^2+2)=
=(x^4+4)(x^2+2)(x-\/'''2'''')(x+\/''''2'''')
God bless you.

First recognize that this is the difference of two squares: x^2 - y^2

Watch closely: ( x^4 )^2 - 4^2

It's more apparent if you make a small substitution: Let, say m = x^4

=> m^2 - 4^2

See it now?

So go ahead and finish ----- m^2 - 4^2 = ( m + 4 )( m - 4 )

Now, put back in for m and you will be done.

---> m = x^4

=> m^2 - 4^2 = ( x^4 + 4 )( x^4- 4 )

Make sense?

:)