How do you factor #42x^3 + 51x^2 – 18x#?

1 Answer
Jan 31, 2016

Answer:

#42x^3+51x^2-18x = color(green)(3x(2x+3)(7x-2))#

Explanation:

Extracting the obvious common factor of #3x# from #42x^3+51x^2-18x# gives
#color(white)("XXX")3x(14x^2+17x-6)#

There are several ways of factoring #(14x^2+17x-6)#

One way is to look for factors #(color(blue)(a)*color(green)(b))#of #14# and #(color(cyan)(c)*color(orange)(d))# of #(-6)#
such that #color(blue)(a)color(orange)(d)-color(green)(b)color(cyan)(c)=17#
in order to generate the factors #(color(blue)(a)x+color(cyan)(c))(color(green)(b)x+color(orange)(d))#

Table of #color(blue)(a)color(orange)(d)-color(green)(b)color(cyan)(c)#

#{: ("factors of " 14 (axxb)":"," | ",(color(blue)(1) * color(green)(14))," | ",(color(blue)(2) * color(green)(7))," | ",(color(blue)(7) * color(green)(2))," | ",(color(blue)(14) * color(green)(1))), ("----------------------",,"--------------------",,"--------------------",,"--------------------",,"--------------------"), ("factors of "(-6) (cxxd)":"," | ", ," | ", ," | ",," | ", ), (color(white)("XX")((color(cyan)(-1)) * color(orange)(6))," | ",-14+6=-8," | ", -7+12=5," | ",-2+42=40," | ",-1+84=83), (color(white)("XX")((color(cyan)(-2)) * color(orange)(3))," | ", -28+3=-25," | ",-14+6=-8," | ",color(red)(-4+21=17)," | ", ". . ." ), (color(white)("XX")(color(cyan)(2) * (color(orange)(-3)) )," | ",". . ." ," | ", ". . ."," | ",". . ." ," | ", ". . ." ), (color(white)("XX")(color(cyan)(1) * (color(orange)(-6))) ," | ", ". . ." , " | ", ". . .", " | ",". . ." ," | ", ". . .") :}#

Which gives us the factoring #(ax+c)(bx+d)#
#color(white)("XXX")14x^2+17x-6 = (color(blue)(7)xcolor(cyan)(-2))(color(green)(2)xcolor(orange)(+3))#

#rArr42x^3+52x^2-18x = (3x)(7x-2)(2x+3)#