factor completely
x^4-16=0
Assistance needed.
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I would insist that by 12th grade you would recognize the difference of squares pattern.
Problem x4 - 16 = 0
FIRST. Make [x4 - 16] a perfect square.
Dive -16 by two and square that number, add to both sides.
x4 -16 + 64 = 0 + 64
SECOND. Solve the equation.
(Simplify) → (x2 - 8)^2 this equal the left side of the equation.
so we have
(x2 -8)^2 = 64
Then
Remove the square by taking the square root of 64.
x2 - 8 = positive or negative 8
And then find your answers.
(x-2)(x+2)(x^2+4)= 0
Giving the solutions
x= +2;-2;-2i;2i
(i being the imaginary unit)
To factor the expression completely, x^4 - 16 = 0, we can start by recognizing that it is a difference of squares. This means we can rewrite the expression as (x^2)^2 - 4^2 = 0.
Now, we have the form a^2 - b^2, which factors into (a - b)(a + b). Applying this formula, we get [(x^2 - 4)(x^2 + 4)] = 0.
Next, we notice that x^2 - 4 is the difference of squares once again, so we can further factor it as (x - 2)(x + 2)(x^2 + 4) = 0.
So, the completely factored form of x^4 - 16 = 0 is (x - 2)(x + 2)(x^2 + 4) = 0.