Question

Can you help me find the zeros step by step?

Answer 1

To find the zeros in a function you have to find those values that make that:

`f(x)=0`

In this case you have an cubic equation so first we are going to factorize it:

`f(x)=x^3-x^2-5x+125^{}`

Knowing that 125 is equal to:

`125=5^3`

this function can be expressed like:

`f(x)=(x^3+5^3)-x^2-5x`

The sum of cubes is:

`a^3+b^3=(a+b)(a^2-ab+b^2)`

We can use this to the first part of the equation:

`(x^3+5^3)=(x+5)(x^2-5x+25)`

Now we can factorize the other part as follow:

`-x^2-5x=-x(x+5)`

So the equation now is:

`f(x)=\text{ }(x+5)(x^2-5x+25)-x(x+5)`

Now we can factorize the (x+5) as a common term

`(x+5)(x^2-5x+25-x)`

And if we organice this one we get:

`f(x)=(x+5)(x^2-6x+25)`using he first part ( x + 5) we can find one zero, as follow:
`x+5=0`
`x=-5`

Finally we can use the quadratic equation in the second part we can find the other zeros, as follow:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`
`x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(25)}}{2(1)}`
`x=\frac{6\pm\sqrt[]{36-100}}{2}=\frac{6\pm\sqrt[]{-64}}{2}`

As we get a root for a negative munber we can use the imaginary number:

`\sqrt[]{-64}=8i`

So:

`x=(6\pm8i)/(2)`

The zeros are now calculated with the two solutions ( + and -)

`x_1=(6+8i)/(2)=3+4i`
`x_2=(6-8i)/(2)=3-4i`

Then so, the zeros of the function

`f(x)=x^3-x^2-5x+125^{}`

are:

- 5

3 + 4i

3 - 4i

option B