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

##### 1 Answer

#### Answer:

#### Explanation:

Given:

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

By the rational roots theorem, any rational zeros of

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

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

To factor the remaining quadratic note that

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

Putting it all together:

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