How do you factor completely: #10x^5 + 4x^4 + 8x^3#?

# 10x^5 + 4x^4 + 8x ^3 #

First we take out the obvious common factor of #x^3#. Let's take out the slightly less obvious factor of #2# as well.

# = 2 x^3 ( 5 x^2 + 2 x + 4) #

That's pretty good. If we can factor the quadratic equation we can make more progress. To check if we can, we can check if the discriminant #b^2-4ac# is a perfect square. Well #2^2-4(5)(4)=-76# is a negative number, which won't ever be a perfect square.

Depending on what grade we're in, we either stop here or factor using complex numbers. I'll continue.

Pro tip: The Shakespeare Quadratic Formula (#2b# or #-2b#) says #x^2-2bx+c# has zeros #x= b\pm sqrt{b^2-c}# and #ax^2+2bx+c# has zeros #x = \frac 1 a(-b \pm sqrt{b^2 - ac})#.

By the Shakespeare Quadratic Formula the quadratic has zeros

# x = 1/ 5 ( -1 \pm \sqrt{-19}) = 1/5 (-1 pm i \sqrt{19})#

so we can factor

#5 x^2 + 2 x + 4#

#= 5(x - 1/ 5 ( -1 + i\sqrt{19}))(x - 1/ 5 ( -1 - i\sqrt{19}))#

and the entire expression

# 10x^5 + 4x^4 + 8x ^3 #

# = 10 x^3 (x - 1/ 5 ( -1 + i\sqrt{19}))(x - 1/ 5 ( -1 - i\sqrt{19}))#