# How do you find the rational roots of #x^4+8x^3+7x^2-40x-60=0#?

##### 1 Answer

#### Answer:

The rational roots are:

It also has irrational roots:

#### Explanation:

Given:

#x^4+8x^3+7x^2-40x-60=0#

Let:

#f(x) = x^4+8x^3+7x^2-40x-60#

**Rational roots theorem**

By the rational roots theorem, any rational zeros of this quartic must be expressible in the form

That means that the only possible rational roots are:

#+-1, +-2, +-3, +-4, +-5, +-6, +-10, +-12, +-15, +-20, +-30, +-60#

**Descartes' Rule of Signs**

The pattern of signs of the coefficients of

The pattern of signs of the coefficients of

Given that there is at least one and possibly three negative real zeros, let us try the possible negative rational zeros first:

#f(-1) = 1-8+7+40-60 = -20#

#f(-2) = 16-64+28+80-60 = 0#

So

#x^4+8x^3+7x^2-40x-60 = (x+2)(x^3+6x^2-5x-30)#

Note that the ratio between the first and second terms of the remaining cubic is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#x^3+6x^2-5x-30 = (x^3+6x^2)-(5x+30)#

#color(white)(x^3+6x^2-5x-30) = x^2(x+6)-5(x+6)#

#color(white)(x^3+6x^2-5x-30) = (x^2-5)(x+6)#

#color(white)(x^3+6x^2-5x-30) = (x^2-(sqrt(5))^2)(x+6)#

#color(white)(x^3+6x^2-5x-30) = (x-sqrt(5))(x+sqrt(5))(x+6)#

Hence the remaining rational zero is