5.4 Dividing polynomials  (Page 3/6)

Begin by setting up the synthetic division. Write $k$ and the coefficients.

Bring down the lead coefficient. Multiply the lead coefficient by $k.$

Continue by adding the numbers in the second column. Multiply the resulting number by $k.$ Write the result in the next column. Then add the numbers in the third column.

The result is $5x+12.$ The remainder is 0. So $x-3$ is a factor of the original polynomial.

The binomial divisor is $x+2$ so $k=\mathrm{-2.}$ Add each column, multiply the result by –2, and repeat until the last column is reached.

The result is $4x^2+2x-10.$ The remainder is 0. Thus, $x+2$ is a factor of $4x^3+10x^2-6x-20.$

Notice there is no x -term. We will use a zero as the coefficient for that term.

The result is $-9x^3+x^2+8x+8+\frac{2}{x-1}.$

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in [link] .

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

To solve for $h,$ first divide both sides by $3x.$

Now solve for $h$ using synthetic division.

The quotient is $x^2+x-9$ and the remainder is 0. The height of the solid is $x^2+x-9.$