Answer
Quotient $=5x^2-13$
Remainder $=x+27$
Work Step by Step
The given expression is
$(5x^4-3x^2+x+1)\div(x^2+2)$
Rewrite the expression as
$(5x^4+0x^3-3x^2+x+1)\div(x^2+0x+2)$
Perform long division to obtain:
$\begin{matrix}
& 5x^2 & -13 & & & & \leftarrow &\text{Quotient}\\
&-- &-- &--&--& \\
x^2+0x+2) &5x^4&+0x^3&-3x^2&+x&+1 & \\
& 5x^4 &+0x^3 &+10x^2 & && \leftarrow &5x^2(x^2+0x+2) \\
& -- & -- & --& && \leftarrow &\text{subtract} \\
& 0 & 0& -13x^2 &+x &+1 \\
& & & -13x^2 &0x &-26& \leftarrow & -13(x^2+0x+2) \\
& & & -- & -- & -- & \leftarrow & \text{subtract} \\
& & & 0&x &+27& \leftarrow & \text{Remainder}
\end{matrix}$
Checking:
(Quotient)(divisor)+ Remainder
$=(5x^2-13)(x^2+2)+x+27$
$=5x^4-13x^2+10x^2-26+x+27$
$=5x^4-3x^2+x+1$
$=$ Dividend
Hence, the Quotient is $5x^2-13$ and the remainder is $x+27$.
