다항식 전체 문제

${x}^{3}-3{x}^{2}-2x-1-\frac{3}{x+3}$

This time we will show the division all in one step. We need to add two placeholders in order to divide.

To check, multiply $(2a+3)(4{a}^{2}-6a+9).$

The result should be $8{a}^{3}+27.$

${x}^{2}+4x+16$

$25{x}^{2}+10x+4$

Write the dividend with decreasing powers of $x.$
Write the coefficients of the terms as the first row of the synthetic division.
Write the divisor as $x-c$ and place c in the synthetic division in the divisor box.
Bring down the first coefficient to the third row.
Multiply that coefficient by the divisor and place the result in the second row under the second coefficient.
Add the second column, putting the result in the third row.
Multiply that result by the divisor and place the result in the second row under the third coefficient.
Add the third column, putting the result in the third row.
Multiply that result by the divisor and place the result in the third row under the third coefficient.
Add the final column, putting the result in the third row.
The quotient is $2{x}^{2}-1x+3$ and the remainder is 2.

The division is complete. The numbers in the third row give us the result. The $2\quad -1\quad 3$ are the coefficients of the quotient. The quotient is $2{x}^{2}-1x+3.$ The 2 in the box in the third row is the remainder.

Check:

$\quad \begin{array}{ccc} \\ \\ \text{(quotient)(divisor)}+\text{remainder} & = & \text{dividend} \\ (2{x}^{2}-1x+3)(x+2)+2 & \overset{?}{=} & 2{x}^{3}+3{x}^{2}+x+8 \\ 2{x}^{3}-{x}^{2}+3x+4{x}^{2}-2x+6+2 & \overset{?}{=} & 2{x}^{3}+3{x}^{2}+x+8 \\ 2{x}^{3}+3{x}^{2}+x+8 & = & 2{x}^{3}+3{x}^{2}+x+8\checkmark \end{array}$

$3{x}^{2}+4x-2;\,2$

$4{x}^{2}-3x+1;\,1$

The polynomial ${x}^{4}-16{x}^{2}+3x+12$ has its term in order with descending degree but we notice there is no ${x}^{3}$ term. We will add a 0 as a placeholder for the ${x}^{3}$ term. In $x-c$ form, the divisor is $x-(-4).$

We divided a 4^th degree polynomial by a 1^st degree polynomial so the quotient will be a 3^rd degree polynomial.

Reading from the third row, the quotient has the coefficients $1\quad -4\quad 0\quad 3,$ which is ${x}^{3}-4{x}^{2}+3.$ The remainder is 0.

${x}^{3}-4{x}^{2}+5;\,0$

${x}^{3}-3{x}^{2}+2;\,0$

ⓐ

ⓑ In part ⓐ we found $\left(\frac{f}{g}\right)(x)$ and now are asked to find $\left(\frac{f}{g}\right)(-4).$

ⓐ $\left(\frac{f}{g}\right)(x)=x-8$ ⓑ $\left(\frac{f}{g}\right)(-3)=-11$

ⓐ $\left(\frac{f}{g}\right)(x)=x-9$ ⓑ undefined

To use the Remainder Theorem, we must use the divisor in the $x-c$ form. We can write the divisor $x+2$ as $x-(-2).$ So, our $c$ is $-2.$

To find the remainder, we evaluate $f(c)$ which is $f(-2).$

To evaluate $f(-2),$ substitute $x=-2.$
Simplify.
	The remainder is 5 when $f(x)={x}^{3}+3x+19$ is divided by $x+2.$
Check: Use synthetic division to check.
The remainder is 5.

$-1$

$6$

The Factor Theorem tells us that $x-4$ is a factor of $f(x)={x}^{3}-64$ if $f(4)=0.$

Since $f(4)=0,$ $x-4$ is a factor of $f(x)={x}^{3}-64.$

yes

yes