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

##### 1 Answer

#### Explanation:

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

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

So the only possible *rational* zeros of

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

Since the coefficients of

Note that:

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

So there is a Real root somewhat closer to

#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

#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

#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: