Question

How do you find all the zeros of `f(x)=x^3+7x^2-6x-72`?

Answer 1

Use the rational root theorem to help find the first zero, then complete the square to find the other two, giving:

`3`, `-4`, `-6`

Explanation:

`f(x) = x^3+7x^2-6x-72`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` where `p` is a divisor of the constant term `-72` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`, `+-3`, `+-4`, `+-6`, `+-12`, `+-18`, `+-24`, `+-36`, `+-72`

Trying each in turn, the first one that works is:

`f(3) = 3^3+(7*3^2)-(6*3)-72 = 27+63-18-72 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`x^3+7x^2-6x-72 = (x-3)(x^2+10x+24)`

One way of factoring the remaining quadratic factor `x^2+10x+24` is by completing the square:

`x^2+10x+24`

`=(x+5)^2-25+24`

`=(x+5)^2-1^2`

`=((x+5)-1)((x+5)+1)`

`=(x+4)(x+6)`

Hence zeros:

`x = -4` and `x=-6`