# How do you find the zeros of the function `f(x) = 1/7x^3 - x ` algebraically?

mathsworkmusic | (Level 2) Educator

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We have the cubic function (a polynomial of order 3)

`f(x) = 1/7x^3 - x `

and wish to find algebraically the zeros, that is, the roots, where the function crosses the x axis.

To find the roots of the function, set it to zero and solve. We write, then, that

`f(x)=0 `

Implying that

`1/7x^3 - x =0 `

To find the roots of the equation we need to factor it. Each of the factors gives a root of the equation.

Begin by factoring out `x ` , giving

`x(1/7x^2 - 1) = 0 `

Incidentally, that `x ` is a factor means that 0 is a root, since setting the factor to zero and solving gives us a root. Directly, this gives `x=0 `.

As the equation is a cubic, there will be two other factors, making three, because 'cubic' means 'of order three', where the highest exponent of terms involving `x ` is 3, and where there are three roots to the equation.

The remaining expression, after we have factored out the term `x `, is a quadratic (second order) expression. An algebraic way to solve quadratics for their two roots is using the quadratic formula. This says that the roots `x_0 ` to the general quadratic equation

`ax^2 + bx + c `  , where `a,b, ` and `c ` are constants (numerical coefficients),

are given by

`x_0 = -b pm sqrt(b^2-4ac)/(2a) `

In the case of the quadratic expression we have here, `a=1/7 `, `b=0 ` and `c=-1 `. We find then that the two roots of this expression are given by

`x_0 = 0 pm sqrt(-4(1/7)(-1))/(2/7) = pm (7(sqrt(4/7)))/2 = pm (7(2/sqrt(7)))/2 = pm 7/sqrt(7) = pm sqrt(7) `

Therefore the three roots of `f(x) ` are  `0,sqrt(7),-sqrt(7) `

We should check that these are correct by substituting them back into `f(x) ` . The root `0 ` is clearly correct. Putting in the root `sqrt(7) `  we get

`f(sqrt(7)) = 1/7(sqrt(7))^3 - sqrt(7) = 1/7 7^(3/2) - sqrt(7) = 7^(1/2) - 7^(1/2) = 0 `

Similarly for the root `-sqrt(7) `  we get

`f(-sqrt(7)) = 1/7(-sqrt(7))^3 - (-sqrt(7)) = -1/7 7^(3/2) + sqrt(7) = -7^(1/2) + 7^(1/2) = 0 `

Therefore the zeros (or roots) of f(x) are x=0, `sqrt(7) ` and -`sqrt(7) ` .

Sources:

nick-teal | High School Teacher | (Level 3) Adjunct Educator

Posted on

First set your equation equal to 0.

`1/7x^3 + x = 0`

` `

Our goal is to find every value of x that will make the left side equal to 0.

There can be up to 3 solutions because this is a cubic function.  (The highest power of x is 3).` `

The first step you should take is factoring out an x.  This gives you

`x(1/7x^2 - 1) = 0`

In order to find values of x that equal 0, we will look at the two parts of this equation (the x, and the stuff inside the parenthesis) separately.  If one term equals zero, then the entire thing will equal zero, 0 times anything is 0.

Looking at the x term, all we have to do to make x equal to 0, is make it 0.  How easy!  So that is our first solution, x = 0.

Now when we look at the stuff inside the parenthesis, we want to figure out how to make `(1/7x^2 - 1)= 0`

` `

equal to zero.

We can solve this using the quadratic formula by setting a = 1/7, b = 0, c = -1.

After doing so we see that `x = -sqrt(7)`  and `x = +sqrt(7)`

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are also solutions to our problem.

Now we have found 3 possible x values, `x = 0, x = +-sqrt(7)`

And since we knew that the maximum number of solutions we could find is 3, we are finished.

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