# How do you find the roots of #12x^3+31x^2-17x-6=0#?

##### 1 Answer

#### Answer:

Use the rational roots theorem, then factor to find:

#x = 2/3" "# or#" "x = -1/4" "# or#" "x = -3#

#### Explanation:

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

That means that the only possible *rational* zeros are:

#+-1/12, +-1/6, +-1/4, +-1/3, +-1/2, +-2/3, +-3/4, +-1, +-3/2, +-2, +-3, +-6#

Notice that the signs of the coefficients of

So let's look among the positive possibilities:

#f(1) = 12+31-17-6 = 20#

#f(1/2) = 12(1/8)+31(1/4)-17(1/2)-6 = (6+31-34-24)/4 = -21/4#

#f(2/3) = 12(8/27)+31(4/9)-17(2/3)-6 = (32+124-102-54)/9 = 0#

So

#12x^3+31x^2-17x-6 = (3x-2)(4x^2+13x+3)#

To factor the remaining quadratic, use an AC method:

Look for a pair of factors of

The pair

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

#4x^2+13x+3 = (4x^2+12x)+(x+3)#

#color(white)(4x^2+13x+3) = 4x(x+3)+1(x+3)#

#color(white)(4x^2+13x+3) = (4x+1)(x+3)#

So the other two zeros are:

#x = -1/4" "# and#" "x = -3#