How do you factor given that #f(-6)=0# and #f(x)=2x^3+7x^2-33x-18#?

Question

2953 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-08T07:23:33

# f(x)=(x+6)(x-3)(2x+1)#.

The given cubic polynomial #f# has #-6# as its zero.

#:. (x-(-6))=(x+6)# is a factor of #f(x)#.

We split the terms of #f(x)# in such a way that #(x+6)# may turn

out as a factor, as shown below :-

#f(x)=2x^3+7x^2-33x-18#,

#=ul(2x^3+12x^2)-ul(5x^2-30x)-ul(3x-18)#,

#=2x^2(x+6)-5x(x+6)-3(x+6)#,

#=(x+6){2x^2-5x-3}#,

#=(x+6){ul(2x^2-6x)+ul(x-3)}#,

#=(x+6){2x(x-3)+1(x-3)}#.

# rArr f(x)=(x+6)(x-3)(2x+1)#.