Question

How do you find all the zeros of `f(x)= 6x^3 +25x^2 +3x -4`?

Answer 1

`f(x)` has zeros: `1/3`, `-1/2` and `-4`

Explanation:

`f(x) = 6x^3+25x^2+3x-4`

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 `-4` and `q` a divisor of the coefficient `6` of the leading term.

So the only possible rational zeros of `f(x)` are:

`+-1/6, +-1/3, +-1/2, +-2/3, +-1, +-4/3, +-2, +-4`

Since the coefficients of `f(x)` have one change of sign, we can use Descartes' rule of signs to deduce that it has exactly one positive Real zero.

Note that:

`f(0) = -4` and `f(1) = 6+25+3-4 = 30`

So there is a Real root somewhat closer to `0` than to `1`.

`f(1/6) = 6(1/216)+25(1/36)+3(1/6)-4`

`=(1+25+18-144)/36 = -100/36 = -25/9`

`f(1/3) = 6(1/27)+25(1/9)+3(1/3)-4`

`=(2+25+9-36)/9 = 0`

So `x=1/3` is a zero and `(3x-1)` a factor:

`6x^3+25x^2+3x-4 = (3x-1)(2x^2+9x+4)`

We can factor the remaining quadratic using an AC method:

FInd a pair of factors of `AC=2*4=8` with sum `B=9`. The pair `8, 1` works. Use this pair to split the middle term and factor by grouping:

`2x^2+9x+4`

`=2x^2+8x+x+4`

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

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

Hence the other two zeros are: `x=-1/2` and `x=-4`