How do you find all the real and complex roots of #f(x)= 3x^3 + 36x^2 + 156x + 240#?

1 Answer
Dec 11, 2016

Answer:

#f(x)# has zeros #-4# and #-4+-2i#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this later, with #a=(x+4)# and #b=2i#

#f(x) = 3x^3+36x^2+156x+240#

#color(white)(f(x)) = 3(x^3+12x^2+52x+80)#

We can start to simplify this cubic using a linear substitution to eliminate the square term:

#x^3+12x^2+52x+80 = (x^3+12x^2+48x+64) + (4x+16)#

#color(white)(x^3+12x^2+52x+80) = (x+4)^3 + 4(x+4)#

We could put #t = x+4# at this stage, but there is no constant term, so let's leave #(x+4)# as #(x+4)# and just factor it from here:

#color(white)(x^3+12x^2+52x+80) = (x+4)((x+4)^2+4)#

#color(white)(x^3+12x^2+52x+80) = (x+4)((x+4)^2-(2i)^2)#

Having made the quadratic factor into a difference of squares, we can apply the difference of squares identity to find:

#color(white)(x^3+12x^2+52x+80) = (x+4)((x+4)-2i)((x+4)+2i)#

#color(white)(x^3+12x^2+52x+80) = (x+4)(x+4-2i)(x+4+2i)#

Hence the zeros of #f(x)# are #-4# and #-4+-2i#