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

1 Answer
Aug 5, 2016

Answer:

#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#