How do you use synthetic division to divide #(2x^4 + 3x^2 - 1) / (x - 1/2)#?

1 Answer
Dec 22, 2016

Answer:

#2x^3+x^2+7/2x+7/4#
with a remainder of #(-1/8)#

Explanation:

I suspect the dividend should have had the term #3x^3# instead of #3x^2#;
however here is the answer to the question as it was asked.

#(color(red)2x^4+color(red)3x^2color(red)(-1)) div (x-color(blue)(1/2))#

#{: (,,(x^4),(x^3),(x^2),(x^1),(x^0)), (,"|",color(red)2,0,color(red)3,0,color(red)(-1)), (underline(" +"color(white)(1/2)),"|",underline(color(white)(1/2" ")),underline(color(blue)(1/2)xxcolor(green)2=1),underline(color(blue)(1/2)xxcolor(orange)(1)=1/2),underline(color(blue)(1/2)xxcolor(brown)(7/2)=7/4),underline(color(blue)(1/2)xxcolor(cyan)(7/4)=7/8)), (color(blue)(1/2),"|",color(white)(1/2)color(green)(2),color(white)(1/2)color(orange)(1),color(brown)(7/2),color(cyan)(7/4),color(magenta)(-1/8)), (,,(x^3),(x^2),(x^1),(x^0),"Remainder") :}#

#(color(red)2x^4+color(red)3x^2color(red)(-1)) div (x-color(blue)(1/2))= color(green)2x^3color(orange)(+1)x^2color(brown)(+7/2)xcolor(cyan)(+7/4)# with a remainder of #(color(magenta)(-1/8))#