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

Dec 22, 2016

$2 {x}^{3} + {x}^{2} + \frac{7}{2} x + \frac{7}{4}$
with a remainder of $\left(- \frac{1}{8}\right)$

#### Explanation:

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

$\left(\textcolor{red}{2} {x}^{4} + \textcolor{red}{3} {x}^{2} \textcolor{red}{- 1}\right) \div \left(x - \textcolor{b l u e}{\frac{1}{2}}\right)$

{: (,,(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") :}

$\left(\textcolor{red}{2} {x}^{4} + \textcolor{red}{3} {x}^{2} \textcolor{red}{- 1}\right) \div \left(x - \textcolor{b l u e}{\frac{1}{2}}\right) = \textcolor{g r e e n}{2} {x}^{3} \textcolor{\mathmr{and} a n \ge}{+ 1} {x}^{2} \textcolor{b r o w n}{+ \frac{7}{2}} x \textcolor{c y a n}{+ \frac{7}{4}}$ with a remainder of $\left(\textcolor{m a \ge n t a}{- \frac{1}{8}}\right)$