What are the zeroes of #f(x) = x^3 – 3x^2 – 16x + 48#?

1 Answer
Jan 16, 2016

Answer:

#=color(red)((x-3)color(blue)((x-4)(x+4)#

Explanation:

#f(x) = x^3-3x^2-16x+48#

Whenever we come across such polynomials, the first step should be to see if you can factorize it. If not we can use the rational root theorem and use factor theorem.

#x^3-3x^2-16x+48=0#

Factor by grouping.

#color(red)((x^3-3x^2))+color(blue)((-16x+48)#
#=color(red)(x^2)(x-3)+color(blue)(-16)(x-3)#
#=color(red)((x-3)color(blue)((x^2-16))#
#=color(red)((x-3)color(blue)((x^2-4^2))#

#x^2-4^2# can be factored using difference of squares.

Difference of square rule : #a^2-b^2=(a-b)(a+b)#

#x^2-4^2=(x-4)(x+4)#

Continuing with our problem

#=color(red)((x-3)color(blue)((x-4)(x+4)#