Determine the completely factored form of #f(x) = 12x^3 - 44x^2 + 49x - 15#?

1 Answer
Oct 21, 2016

Answer:

#12x^3-44x^2+49x-15 = (2x-3)(3x-5)(2x-1)#

Explanation:

#f(x) = 12x^3-44x^2+49x-15#

By the rational roots theorem, any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# and #q# a divisor of the coefficient #12# of the leading term.

In addition, note that the pattern of signs of coefficients of #f(x)# is #+ - + -# with #3# sign changes, while those of #f(-x)# are in the pattern #- - - -# with #0# sign changes. So by Descartes' Rule of Signs, #f(x)# has #3# or #1# positive Real zero and no negative Real zeros.

Hence the only possible rational zeros are:

#1/12, 1/6, 1/4, 1/3, 5/12, 1/2, 3/4, 5/6, 1, 5/4, 3/2, 5/3, 5/2, 3, 5, 15#

We could simply try each of these in turn, but if you are allowed, we can speed up the process of finding the zeros as follows:

Look at the derivative and find out where it is #0#, indicating a local maximum or minimum...

#f'(x) = 36x^2-88x+49#

#color(white)(f'(x)) = (6x-22/3)^2-484/9+49#

#color(white)(f'(x)) = (6x-22/3)^2-(sqrt(43)/3)^2#

#color(white)(f'(x)) = (6x-22/3-sqrt(43)/3)(6x-22/3+sqrt(43)/3)#

Hence zero at #x = 1/6(22/3+-sqrt(43)/3) = 1/18(22+-sqrt(43))#

#sqrt(43) ~~ 6.5#

So the local maximum and minimum are approximately at:

#1/18(22-6.5) = 31/36 ~~ 5/6# and #1/18(22+6.5) = 19/12 ~~ 3/2#

#f(5/6) = 20/9#

#f(3/2) = 0#

Hmmm - a zero near where we expect the minimum. Let's look at the other rational possibility nearby...

#f(5/3) = 0#

So #x=3/2# and #x=5/3# are zeros with corresponding factors #(2x-3)# and #(3x-5)#.

Looking at the coefficient of the leading term and constant term, the remaining factor (which must be rational) must be #(2x-1)#

So:

#12x^3-44x^2+49x-15 = (2x-3)(3x-5)(2x-1)#

graph{12x^3-44x^2+49x-15 [-0.448, 2.052, -1.26, 2.49]}