How do you find all the zeros of #f(x)=12x^3+31x^2-17x-6#?

1 Answer
Aug 3, 2016

Answer:

#f(x)# has zeros: #2/3#, #-1/4#, #-3#

Explanation:

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

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 #-6# and #q# a divisor of the coefficient #12# of the leading term.

So the only possible rational zeros are:

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

That's rather a lot of possibilities to check, so let's search by approximately binary chop:

#f(0) = -6#

#f(1) = 12+31-17-7 = 19#

#f(1/2) = 3/2+31/4-17/2-6 = (6+31-34-24)/4 = -21/4#

#f(3/4) = 12(27/64)+31(9/16)-17(3/4)-6 = (81+279-204-96)/16 = 15/4#

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

So #x=2/3# is a zero and #(3x-2)# is a factor:

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

To factor #4x^2+13x+2# we can use an AC method: Look for a pair of factors of #AC = 4*3=12# with sum #B=13#.

The pair #12, 1# works. Use this pair to split the middle term and factor by grouping:

#4x^2+13x+3 = 4x^2+12x+x+3 = 4x(x+3)+1(x+3) = (4x+1)(x+3)#

So the remaining zeros are #x=-1/4# and #x=-3#