Question

What are all the possible rational zeros for `f(x)=10x^3-15x^2-16x+12` and how do you find all zeros?

Answer 1

The "possible" rational zeros are:

`+-1/10, +-1/5, +-3/10, +-2/5, +-1/2, +-3/5, +-4/5, +-1, +-6/5, +-3/2, +-2, +-12/5, +-3, +-4, +-6, +-12`

The actual zeros are:

`2" "` and `" "-1/4+-sqrt(265)/20`

Explanation:

Given:

`f(x) = 10x^3-15x^2-16x+12`

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 `12` and `q` a divisor of the coefficient `10` of the leading term.

That means that the only possible rational zeros are:

`+-1/10, +-1/5, +-3/10, +-2/5, +-1/2, +-3/5, +-4/5, +-1, +-6/5, +-3/2, +-2, +-12/5, +-3, +-4, +-6, +-12`

Trying a few of the simpler values, we eventually find:

`f(2) = 10(color(blue)(2))^3-15(color(blue)(2))^2-16(color(blue)(2))+12`

`color(white)(f(2)) = 80-60-32+12`

`color(white)(f(2)) = 0`

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

`10x^3-15x^2-16x+12 = (x-2)(10x^2+5x-6)`

The remaining quadratic factor is in the standard form `ax^2+bx+c` with `a=10`, `b=5` and `c=-6`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 5^2-4(10)(-6) = 25+240 = 265`

This is positive but not a perfect square. Hence we can deduce that the remaining zeros are real but irrational. We can find the zeros using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (-5+-sqrt(265))/(2*10)`

`color(white)(x) = (-5+-sqrt(265))/20`

`color(white)(x) = -1/4+-sqrt(265)/20`