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I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?

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- The number of real roots of $1+x/1!+x^2/2!+x^3/3! + \cdots + x^6/6! =0$

As mentioned, long division works fine. But general modular techniques also work quickly, and require no specialized knowledge. For example, here modulo $\rm\:4\:x^2 + 5\:,\:$ we have $\rm\ {\color{red}{4\:x^2}} \equiv -5\ $ so $$\rm\:2\ f(x)\ =\ (-{\color{red}{4\:x^2}}-6)\ x+2\cdot {\color{red}{\:4\: x^2}}+10\ \equiv\ (5-6)\ x -10 + 10\ \equiv\:\: -x$$

Therefore $\rm\ 2\ f(x)\ =\:\: -x + (a\:x+b)\ (4\:x^2+5)\ $ so $\rm\ a = -1,\ b = 2\ $ by comparing coef’s of $\rm\:x^3,\ x^0\:.$

**Note** $\ $ I multipled $\rm f(x)$ by $2$ only to simplify the arithmetic (eliminate fractions).

The answer in no way depends on whether $a$ and $b$ are more than $1$ or less; the only fact about those that you need is that the polynomial you’re dividing by is not zero. You have this:

$$

\begin{array}{ccccccccccccccccc}

\\

4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\

& &

\end{array}

$$

So ask what you need to multiply $4$ by to get $-2$. In other words, divide $-2$ by $4$. You get $-2/4=-1/2$. So write

$$

\begin{array}{ccccccccccccccccc}

& & \frac{-1}{2} x \\

\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\

& &

\end{array}

$$

Then multiply:

$$

\begin{array}{ccccccccccccccccc}

& & \frac{-1}{2} x \\

\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\

& & -2x^3 & & & -& \frac 52 x

\end{array}

$$

Then subtract:

$$

\begin{array}{ccccccccccccccccc}

& & \frac{-1}{2} x \\

\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\

& & -2x^3 & & & -& \frac 52 x \\ \hline

& & & & 4x^2 & – & \frac 12 x & + 5 \\ \hline

\end{array}

$$

The ask what you have to multiply $4x^2$ by to get $4x^2$, then multiply, then subtract…..

You can do long division with polynomials almost the same way you would for integers. For example, $4x^2+5$ can be multiplied by $-\frac{1}{2}x$ to get a leading term of $-2x^3$, so we might say that $4x^2+5$ goes into $-2x^3 + 4x^2 – 3x + 5$ about $-\frac{1}{2}x$ times. As with long division, we then multiply $4x^2+5$ by $-\frac{1}{2}x$ and subtract the result from $-2x^3 + 4x^2 – 3x + 5$; then repeat the process until we can’t anymore. If there’s a remainder at the end, we divide it by the divisor and add that to the end of the result (again, similar to long division).

In this case you should verify that the result is $-\frac{1}{2}x + 1 + \frac{-x}{8x^2+10}$.

Nobody seems to have posted the *synthetic division* route; I’ll include it here for completeness.

We first monicize the divisor (i.e., set things up such that the divisor has leading coefficient $1$); thus, consider

$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}$$

and set up the array

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times& & &\times\\

-\frac54&\times&\times& & \\\hline

& & & &

\end{array}$$

Note that the coefficients in the leftmost column are the negated coefficients of the terms with degree less than the degree of the polynomial. The second dividing line (can’t do dashed lines in an `array`

environment, sorry) indicates that we expect the quotient to be linear ($3-2=1$). The $\times$ entries indicate the sections of the array that shouldn’t be filled.

This generalized synthetic division proceeds like this:

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times& & &\times\\

-\frac54&\times&\times& & \\\hline

&-\frac12& & &

\end{array}$$

(bring down leading coefficient)

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times&0& &\times\\

-\frac54&\times&\times&\frac58& \\\hline

&-\frac12& & &

\end{array}$$

(multiply the bottom of the first column by the coefficients in the leftmost column, and fill the array diagonally)

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times&0& &\times\\

-\frac54&\times&\times&\frac58& \\\hline

&-\frac12&1& &

\end{array}$$

(add numbers in second column)

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times&0&0&\times\\

-\frac54&\times&\times&\frac58&-\frac54\\\hline

&-\frac12&1& &

\end{array}$$

(multiply bottom of the second column by the coefficients in the leftmost column and fill diagonally)

$$\begin{array}{r|cc|cc}

&-\frac12&1&-\frac34&\frac54\\\hline

0&\times&0&0&\times\\

-\frac54&\times&\times&\frac58&-\frac54\\\hline

&-\frac12&1&-\frac18&0

\end{array}$$

(add numbers in third and fourth columns)

We thus have the result

$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac18x+0}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac12x}{4x^2+5}$$

I will try this way:

Since you are dividing a 3rd degree polynomial by a 2nd degree polynomial, WLOG, we may assume

$$-2x^3+4x^2-3x+5=(4x^2+5)(ax+b)+cx+d\quad(1)$$

Now, comparing the coefficients of $x^3$ and $x^2$ readily give $a=-\frac{1}{2}$ and $b=1$. Comparing coefficients of $x$, we have $5a+c=-3\Rightarrow c=-\frac{1}{2}$. Finally comparing the constant term, we get $5b+d=5\Rightarrow d=0$.

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