Given: #x^4+2x^3-8x-16#
First observation is that #2xx8=16# so we have a possible connection there.
Lets have a look at the direct grouping:
#[color(white)(2/2)x^4+2x^3color(white)(2/2)]+[color(white)(2/2)-8x-16color(white)(2/2)]#
Consider the first brackets. If we factor out #x^3# we end up with:
#x^3[color(white)(2/2)x+2color(white)(2/2)]+[color(white)(2/2)-8x-16color(white)(2/2)]#
We can make the second brackets the same as the first if we factor out #(-8)# giving:
#x^3[color(white)(2/2)x+2color(white)(2/2)]-8[color(white)(2/2)x+2color(white)(2/2)]#
We now factor out the #(x+2)# giving:
#(x+2)(x^3-8)#
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#color(blue)("Trying to factor the part "(x^3-8)#
This is the same as #(x^3-2^3)#
Can we manipulate a quadratic out of this? Lets have a play with:
#(x-2)("something")=x^3-8#
#color(blue)("Dealing with the "x^3" term")#
To obtain #x^3# we try: #(x-2)(x^2+?)=x^3-2x^2+?#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Dealing with the "x^2" term")#
but there is no #-2x^2# term in #x^3-8# so we need to get rid of it.
Try: #(x-2)(x^2+x+?)=x^3+x^2-2x^2-2x+? color(red)(" Fail")#
Try: #(x-2)(x^2+2x+?)=x^3+2x^2-2x^2-4x+? color(green)(" Works!")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Dealing with the "-4x" term")#
#(x-2)(x^2+2x+?)=x^3-4x+?#
but there is no #x# term so we need to get rid of it.
Try: #(x-2)(x^2+2x+2)=x^3-4x+4xcolor(red)(-4) larr color(red)(" Fail")#
We got rid of the #x^2# term but ended up with the wrong constant.
Try: #(x-2)(x^2+2x+4)=x^3-4x+4x+8 larrcolor(green)(" Works")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all back together")#
#(x+2)(x^3-8) color(white)("d")->color(white)("d")(x+2)(x-2)(x^2+2x+4) #
/////////////////////////////////////////////////////////////////////////////////////////////////////
#color(blue)("Bottom line comment")#
If you end up with #(a^3-b^3)# this becomes:
#(a-b)(a^2+ab+b^2)#
Where #color(white)("d")-(-b)a=+ab#
