# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given #f(x)=8x^3-6x^2-23x+6#?

##### 1 Answer

#### Answer:

We can tell that

The zeros, which are all rational are:

#### Explanation:

Given:

#f(x) = 8x^3-6x^2-23x+6#

**Descartes' Rule of Signs**

The coefficients of

The coefficients of

**Rational root theorem**

By the rational root theorem, any *rational* zeros of

That means that the only possible rational zeros are:

#+-1/8, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-6#

Note that the rational root theorem only tells us about possible *rational* zeros, not about all possible zeros which may be irrational or non-real complex.

**Actual zeros**

It is probably easiest to try small integers first rather than fractions.

We find:

#f(1) = 8-6-23+6 = -15#

#f(2) = 8(color(blue)(2))^3-6(color(blue)(2))^2-23(color(blue)(2))+6 = 64-24-46+6 = 0#

So

#8x^3-6x^2-23x+6 = (x-2)(8x^2+10x-3)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of

The pair

Use this pair to split the middle term and factor by grouping:

#8x^2+10x-3 = (8x^2+12x)-(2x+3)#

#color(white)(8x^2+10x-3) = 4x(2x+3)-1(2x+3)#

#color(white)(8x^2+10x-3) = (4x-1)(2x+3)#

Hence the other two zeros are

graph{(y-(8x^3-6x^2-23x+6))(60(x-2)^2+y^2-0.2)(60(x-1/4)^2+y^2-0.2)(60(x+3/2)^2+y^2-0.2) = 0 [-5, 5, -20, 20]}