What is the complete solution set of the function including multiplcity?

Question

f(x)=x⁴-256

and please explain too!

Answer 1

Notice that 256 = 16², so we can write

f(x) = x⁴ - 16² = (x² - 16)(x² + 16).

To find the solution set, we look for the roots of f. In other words, we solve

0 = f(x) = (x² - 16)(x² + 16).

But (x² - 16)(x² + 16) = 0 if and only if either:

(x² - 16) = 0 or (x² + 16) = 0.

Let's deal with these one at a time.

If (x² - 16) = 0, then x² = 16, and taking the square root of both sides, we find

x = ± 4.

So two of the four solutions (we know there are four solutions because this is a quartic polynomial) are 4 and -4, each with multiplicity 1.

Now the second option: if (x² + 16) = 0, then x² = -16. Taking the square root of both sides, we get

x = ± √ (-16) = ± 4 √(-1) = ± 4i

where i is the imaginary unit, equal to the square root of -1.

So the other two solutions are 4i and -4i, each with multiplicity 1.

So our full solution set is:

x = -4, 4, 4i, -4i.

Answer 2

Hi,

f(x)=x⁴-256

First, you need to know about this property a² - b²= ( a - b)( a + b) (1)

Second, you need to know about 256 that it is 4⁴

f(x)=x⁴-256 = x⁴- 4⁴= (x²)² - (4²)²= ( x² - 4² ) ( x² + 4²) base of above info (1)

one more time for this x² - 4² you can use (1) that it means x² - 4²= ( x - 2 ) ( x + 2)

so f(x) = ( x² - 4² ) ( x² + 4²) = ( x - 4 ) ( x + 4) ( x² + 4²)

f(x) = 0 it means ( x - 4 ) ( x + 4) ( x² + 4²) = 0 you know must be any parenthesis equals zero

x-4 = 0 ===> x = 4 & x+4 = 0 ====> x = -4 & x² + 4²= 0 no solution in real number but in complex number the answers will be x² + 4²= 0 ====> x² = - 4² after taking root of both sides x = ± 4i

{ 4 , -4 , 4i , -4i } is set of solution

I hope it is useful,

Minoo

Answer 3

I assume f(x) = 0

Factoring

(x²-16)(x²+16)

(x-4)(x+4)(x²+16)

(x-4)(x+4)(x-4i)(x+4i) =0

x-4 = 0 --> x=4

x+4 = 0 --> x=-4

x-4i = 0 --> x = 4i

x+4i=0 --> x = -4i

Solutions: x = 4, -4, 4i, -4i