Question

How do you find all the zeros of `f(x)=4x^3-4x^2-7x-2`?

Answer 1

`f(x)` has zeros: `x=-1/2` with multiplicity `2` and `x=2`.

Explanation:

`f(x) = 4x^3-4x^2-7x-2`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-2` and `q` a divisor of the coefficient `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-1, +-2`

We can quickly see that `f(1)` and `f(-1)` cannot be zero, since `7` is odd and all the other coefficients are even.

We find:

`f(2) = 32-16-14-2 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`4x^3-4x^2-7x-2 = (x-2)(4x^2+4x+1)`

`color(white)()`
To find the remaining zeros I could just try the other rational values we identified and factor by the next one we find, but there's a shortcut:

Do you see the pattern `4,4,1` of the coefficients of the remaining quadratic?

Did you know that `441 = 21^2` ?

As a result, we can spot that `4x^2+4x+1` is a perfect square trinomial:

`4x^2+4x+1 = (2x+1)^2`

This shows us that `4x^2+4x+1` has a repeated zero `x=-1/2`

graph{4x^3-4x^2-7x-2 [-4.877, 5.123, -2.4, 2.6]}