Find all of the Zeros:

Question

Equation: f(x)=x^3+64

How many zeros are there?

Answer 1

Here you go:

We want to find the zeros of the function f(x) = x³ + 64. What does this mean?

The zeros of the function f(x) = x³ + 64 are the values of x which result in x³ + 64 = 0. That is:

If x³ + 64 = 0, then x is a zero of the function f(x) = x³ + 64. First, we need to factor out x³ + 64:

In case you haven't learned this already, there is a formula called the Sum of Cubes Formula, which states the following:

if a and b are two real numbers,

a³ + b³ = (a + b) (a² - ab + b²)

How can we apply this formula to x³ + 64? Let x = a, and let 64 = b³. So 64 is some number b multiplied by itself three times. What number multiplied by itself three times equals 64? 4*4*4 = 16*4 = 64, so b = 4.

Now, plug in x for a and 4 for b into the sum of cubes formula from above, and we obtain:

x³ + 64 = (x+4)(x² -4x + 4²) = (x+4)(x² -4x + 16).

Since f(x) = x³ + 64, we can see now that f(x) = (x+4)(x² -4x + 16), since x³ + 64 = (x+4)(x² -4x + 16).

Moving forward with f(x) = (x+4)(x² -4x + 16): When does f(x) = 0? First, plug in -4 for x:

f(-4) = (-4+4) ((-4)² -4(-4) + 16) = (0)(16+16+16) = (0)(48) = 0.

So when x = -4, f(x) = 0. So we see that x = -4 is one zero of the function f(x) = x³ + 64.

However, we need to solve for the rest of the zeros.

Going back to f(x) = x³ + 64 = (x+4)(x² -4x + 16), it was easy to see that when we plug in x = -4, (x+4) = 0, so

(x+4)(x² -4x + 16) = 0. But there are also values for x in which (x² -4x + 16) = 0. How do we solve for these?

By using the quadratic formula. When we use the quadratic formula and obtain a result, the result we obtain is the values for x in which (x² -4x + 16) = 0. These are known as the roots of the polynomial (x² -4x + 16).

The quadratic formula is stated as follows:

If we have a polynomial ax² + bx + c, then the roots of the polynomial are found using the formula:

x = ( -b ± √(b² - 4(a)(c) ) ⁄ 2(a) . Here one root will be -b + the square root, and the other root will be -b - the square root.

Our polynomial is: x² - 4x + 16. Let's put this into the form ax² + bx + c. Then:

x² - 4x + 16 = ax² + bx + c , which means that a = 1, b = -4, and c = 16. Plugging these values into the quadratic formula:

x = ( -b ± √(b² - 4(a)(c) ) ⁄ 2(a)

x = ( -(-4) ± √((-4)² - 4(1)(16) ) ⁄ 2(1)

x = (4 ± √(16 - 64)) ⁄ 2

So our first root will be x₁, and we will compute it as follows:

x₁ = (4 + √(16 - 64)) ⁄ 2 = (4 + √-48)) ⁄ 2 = (4 + (√(-1) √(48)) ) ⁄ 2 = (4 + (√(-1) √(16) √(3)) ) ⁄ 2

= (4 + ( 4 i √(3) ) ⁄ 2 (since √16 = 4 and √(-1) = i .)

= (2 + ( 2 i √(3) ) )

Similarly, the second root, x₂, is solved for as follows:

x₂ = (4 - √(16 - 64)) ⁄ 2 = (4 - √-48)) ⁄ 2 = (4 - (√(-1) √(48)) ) ⁄ 2 = (4 - (√(-1) √(16) √(3)) ) ⁄ 2

= (4 - ( 4 i √(3) ) ⁄ 2 (since √16 = 4 and √(-1) = i .)

= (2 - ( 2 i √(3) ) ) .

You can just plug x² - 4x + 16 into an online quadratic formula calculator for efficiency purposes and it will give you the roots of the polynomial, but I wanted to show you an example in case you are presented a question like this in an exam.

So, x₁ = (2 + ( 2 i √(3) ) ) and x₂ = (2 - ( 2 i √(3) ) ). When either x₁ or x₂ are substituted for x in x² - 4x + 16 and evaluated, then x² - 4x + 16 will be equal to zero. Therefore, if x = (2 + ( 2 i √(3) ) ) or x = (2 - ( 2 i √(3) ) ), then:

f(x) = x³ + 64 = (x+4)(x² -4x + 16) = (x+4)(0) = 0. So (2 + ( 2 i √(3) ) ) and (2 - ( 2 i √(3) ) ) are zeros of the equation f(x) = x³ + 64.

In summary, the zeros for f(x) = x³ + 64 occur when x = -4, x = (2 + ( 2 i √(3) ) ), or x = (2 - ( 2 i √(3) ) ). These three values for x are the solutions to your problem.

So there are 3 zeros for the equation f(x) = x³ + 64.

Note that (2 + ( 2 i √(3) ) ) and (2 - ( 2 i √(3) ) ) are complex numbers, meaning they have an imaginary component (note the i symbol in each solution). If your professor only wants the real zeros, then the answer is x = -4.

Hope this all helps. Please feel free to leave a rating. I know this solution is long and comprehensive but this is how it's done, so I wanted to give you as good of an explanation as possible. Please feel free to comment with any questions you have for me regarding this problem, happy to help.

Answer 2

There are 3, here is the reason why.

Anything of the form x³+a³ can be factored into the following polynomials: (x+a)(x²-ax+a^2). Since you have x³+64 = x³+4³, this factors into (x+4)(x²-4x+4²) = (x+4)(x²-4x+16). The first zero is when x+4 = 0. The second and third zeros are from the when the polynomial x²-4x+16 = 0.