How do you factor #x^3-12x^2+41x-42#?

1 Answer
Apr 3, 2018

Answer:

#x^3-12x^2+41x-42 = (x-2)(x-3)(x-7)#

Explanation:

Given:

#f(x) = x^3-12x^2+41x-42#

By the rational roots 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 #-42# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1, +-2, +-3, +-6, +-7, +-14, +-21, +-42#

In addition, note that the pattern of signs of the coefficients of #f(x)# is #+ - + -#. With #3# changes of signs, Descartes' Rule of Signs tells us that #f(x)# has #3# or #1# positive real zeros. Also, the signs of the coefficients of #f(-x)# are in the pattern #- - - -#. With no changes of sign, Descartes' Rule of Signs tells us that #f(x)# has no negative real zeros.

So the only possible rational zeros are:

#1, 2, 3, 6, 7, 14, 21, 42#

We find:

#f(2) = (color(blue)(2))^3-12(color(blue)(2))^2+41(color(blue)(2))-42#

#color(white)(f(2)) = 8-48+82-42 = 0#

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

#x^3-12x^2+41x-42 = (x-2)(x^2-10x+21)#

To factor the remaining quadratic note that #3+7 = 10# and #3*7 = 21#. So:

#x^2-10x+21 = (x-3)(x-7)#

Putting it all together:

#x^3-12x^2+41x-42 = (x-2)(x-3)(x-7)#