Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.4.2, AII.5.1, AII.6.4
An architech wants the front of his building to be twice as long as it is high. If the budget uses a formula to pay for materials to cover the front of the building, how tall should the front be? This type of question uses division to solve.
Factoring is unmultiplying, but it only works when it comes out exactly. To unmultiply polynomials when it is not factorable, use polynomial division.
Polynomial long division uses the same algorithm as long division with numbers.
The long division algorithm is
For example, consider dividing 137 by 3.
$$ \require{enclose} \begin{array}{rll} 45 \\ 3 \enclose{longdiv}{137} && \\ \underline{−12}\downarrow && \hbox{(\(3 × 4 = 12\))} \\ 17 && \hbox{(\(13 - 12 = 1\)), Bring down the 7} \\ \underline{-15} && \hbox{(\(3 × 5 = 15\))} \\ 2 && \hbox{(\(17 - 15 = 2\))} \\ && \hbox{Answer: 45 R 2 or \(45 \tfrac{2}{3}\)} \\ \end{array}$$
Division of polynomials is similar to long division of whole numbers. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. For example, divide 3x3 − 2x2 + 5x + 1 by x + 1 using the long division algorithm, it would look like this:
$$ \require{enclose} \begin{array}{rll} x + 1 \enclose{longdiv}{3x^3 - 2x^2 + 5x + 1} && \hbox{(Set up the division)} \\ \\ \\ \color{blue}{3x^2} \phantom{ + 10x + 4} && \hbox{(\(3x^3\) divided by \(x\))} \\ \color{blue}{x} + 1 \enclose{longdiv}{\color{blue}{3x^3} - 2x^2 + 5x + 1} && \\ \\ \\ \color{blue}{3x^2} \phantom{ + 10x + 4} && \\ \color{blue}{x + 1} \enclose{longdiv}{3x^3 - 2x^2 + 5x + 1} && \\ \color{blue}{3x^3 + 3x^2} \phantom{ + 10x + 4} && \hbox{(\(3x^2\) multiplied by \(x + 1\))} \\ \\ \\ 3x^2 \phantom{ + 10x + 10} && \\ x + 1 \enclose{longdiv}{\color{blue}{3x^3 - 2x^2} + 5x + \phantom{0}1} && \\ \underline{\color{blue}{-(3x^3 + 3x^2)}} \phantom{+} \downarrow \phantom{x + 4} && \\ \color{blue}{-5x^2 + 5x} \phantom{0 + 4} && \hbox{(Subtract and bring down next term)} \\ \\ \\ 3x^2 \color{blue}{- 5x} \phantom{ + 100} && \hbox{(\(-5x^2\) divided by \(x\))} \\ \color{blue}{x} + 1 \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ \color{blue}{-5x^2} + 5x \phantom{0 + 4} && \\ \\ \\ 3x^2 \color{blue}{- 5x} \phantom{ + 100} && \\ \color{blue}{x + 1} \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ -5x^2 + 5x \phantom{0 + 4} && \\ \color{blue}{-5x^2 - 5x} \phantom{0 + 4} && \hbox{(\(-5x\) multiplied by \(x + 1\))} \\ \\ \\ 3x^2 - 5x \phantom{ + 100} && \\ x + 1 \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ \color{blue}{-5x^2 + 5x} \phantom{0 + 4} && \\ \underline{\color{blue}{-(-5x^2 - 5x)}} \phantom{10} \downarrow && \\ \color{blue}{10x + 1} && \hbox{(Subtract and bring down next term)} \\ \\ \\ 3x^2 - 5x \color{blue}{ + 10} && \hbox{(\(10x\) divided by \(x\))} \\ \color{blue}{x} + 1 \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ -5x^2 + 5x \phantom{0 + 4} && \\ \underline{-(-5x^2 - 5x)} \phantom{10} \downarrow && \\ \color{blue}{10x} + 1 && \\ \\ \\ 3x^2 - 5x \color{blue}{ + 10} && \\ \color{blue}{x + 1} \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ -5x^2 + 5x \phantom{0 + 4} && \\ \underline{-(-5x^2 - 5x)} \phantom{10} \downarrow && \\ 10x + \phantom{0}1 && \\ \color{blue}{10x + 10} && \hbox{(\(10\) multiplied by \(x + 1\))} \\ \\ \\ \color{red}{3x^2 - 5x + 10} && \\ \color{purple}{x + 1} \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ -5x^2 + 5x \phantom{0 + 4} && \\ \underline{-(-5x^2 - 5x)} \phantom{10} \downarrow && \\ \color{blue}{10x + \phantom{0}1} && \\ \underline{\color{blue}{-(10x + 10)}} && \\ \color{blue}{-9} && \hbox{(Subtract)} \\ \end{array}$$
Answer: \(\color{red}{3x^2 - 5x + 10} \color{blue}{-} \frac{\color{blue}{9}}{\color{purple}{x + 1}}\)
To divide two polynomials using long division,
Divide 4x2 + 7x − 30 by x − 2.
$$ \require{enclose} \begin{array}{rll} x - 2 \enclose{longdiv}{4x^2 + 7x - 30} && \hbox{(Set up the division)} \\ \\ \\ \color{blue}{4x} \phantom{ + 100} && \hbox{(\(4x^2\) divided by \(x\))} \\ \color{blue}{x} - 2 \enclose{longdiv}{\color{blue}{4x^2} + 7x - 30} && \\ \\ \\ \color{blue}{4x} \phantom{ + 100} && \\ \color{blue}{x - 2} \enclose{longdiv}{4x^2 + 7x - 30} && \\ \color{blue}{4x^2 - 8x} \phantom{ + 100} && \hbox{(\(4x\) multiplied by \(x - 2\))} \\ \\ \\ 4x \phantom{ + 100} && \\ x - 2 \enclose{longdiv}{\color{blue}{4x^2 + 7x} - 30} && \\ \underline{\color{blue}{-(4x^2 - 8x)}} \phantom{+0} \downarrow && \\ \color{blue}{15x - 30} && \hbox{(Subtract and bring down next term)} \\ \\ \\ 4x \color{blue}{+ 15} && \hbox{(\(15x\) divided by \(x\))} \\ \color{blue}{x} - 2 \enclose{longdiv}{4x^2 + 7x - 30} && \\ \underline{-(4x^2 - 8x)} \phantom{+0} \downarrow && \\ \color{blue}{15x} - 30 && \\ \\ \\ 4x \color{blue}{+ 15} && \\ \color{blue}{x - 2} \enclose{longdiv}{4x^2 + 7x - 30} && \\ \underline{-(4x^2 - 8x)} \phantom{+0} \downarrow && \\ 15x - 30 && \\ \color{blue}{15x - 30} && \hbox{(\(15\) multiplied by \(x - 2\))} \\ \\ \\ \color{red}{4x + 15} && \\ \color{purple}{x - 2} \enclose{longdiv}{4x^2 + 7x - 30} && \\ \underline{-(4x^2 - 8x)} \phantom{+0} \downarrow && \\ \color{blue}{15x - 30} && \\ \underline{\color{blue}{-(15x - 30)}} && \\ \color{blue}{0} && \hbox{(Subtract)} \\ \\ \\ \end{array}$$
The quotient is 4x + 15. The remainder is 0. Write the result as
$$ \frac{4x^2 + 7x - 30}{x - 2} = \mathbf{4x + 15} $$
Divide 6x3 − 8x2 + 11x − 5 by 3x − 1.
$$ \require{enclose} \begin{array}{rll} 2x^2 - 2x + 3 && (6x^3 \text{ divided by } 3x \text{ is } 2x^2)\\ 3x - 1 \enclose{longdiv}{6x^3 - 8x^2 + 11x - 5} && \\ \underline{-(6x^3 - 2x^2)} \phantom{+} \downarrow \phantom{ 00000} && (\text{Multiply } 3x - 1 \text{ by } 2x^2)\\ -6x^2 + 11x \phantom{1 + 4} && (\text{Subtract. Bring down the next term. } -6x^2 \text{ divided by } 3x \text{ is } -2x)\\ \underline{-(-6x^2 + 2x)} \phantom{10} \downarrow && (\text{Multiply } 3x - 1 \text{ by } -2x)\\ 9x - 5 \phantom{)} && (\text{Subtract. Bring down the next term. } 9x \text{ divided by } 3x \text{ is } 3)\\ \underline{-(9x - 3)} && (\text{Multiply } 3x - 1 \text{ by } 3)\\ -2 && (\text{Subtract. The remainder is } -2) \\ \end{array}$$
The is remainder is −2. The result can be written as:
$$ \frac{6x^3-8x^2+11x-5}{3x - 1}=\mathbf{2x^2 - 2x + 3 + \frac{-2}{3x - 1}} $$
Divide (2x3 + 7x2 − x − 2) ÷ (x2 + 3x − 2).
$$ \require{enclose} \begin{array}{rll} 2x + 1 && (2x^3 \text{ divided by } x^2 \text{ is } 2x)\\ x^2 + 3x - 2 \enclose{longdiv}{2x^3 + 7x^2 - 2x - 4\phantom{)}} && \\ \underline{-(2x^3 + 6x^2 - 4x)} \phantom{1} \downarrow \phantom{)} && (\text{Multiply } x^2 + 3x - 2 \text{ by } 2x)\\ x^2 + 2x - 4\phantom{)} && (\text{Subtract. Bring down the next term. } x^2 \text{ divided by } x^2 \text{ is } 1)\\ \underline{-(x^2 + 3x - 2)} && (\text{Multiply } x^2 + 3x - 2 \text{ by } 1)\\ -x - 2 && (\text{Subtract. The remainder is } -x - 2) \\ \end{array}$$
The is remainder is −x − 2. The result can be written as:
$$ \frac{2x^3 + 7x^2 - 2x - 4}{x^2 + 3x - 2} = \mathbf{2x + 1 + \frac{-x - 2}{x^2 + 3x - 2}} $$
Divide (4x3 + 16x2 + 7x − 15) ÷ (2x + 3).
\(2x^2 + 5x - 4 + \frac{-3}{2x + 3}\)
As is evident from the previous examples, long division can be long and cumbersome. Synthetic Division is a shorthand methods of dividing polynomials by a binomial, x - k. Notice the leading coefficient of the binomial is 1.
To illustrate the process, recall the example at the beginning of the section.
Divide 3x3 − 2x2 + 5x + 1 by x + 1 using long division. The final form of the process looked like this:
$$ \require{enclose} \begin{array}{rll} 3x^2 - 5x + 10 && \\ x + 1 \enclose{longdiv}{3x^3 - 2x^2 + 5x + \phantom{0}1} && \\ \underline{-(3x^3 + 3x^2)} \phantom{+} \downarrow \phantom{x + 4} && \\ -5x^2 + 5x \phantom{0 + 4} && \\ \underline{-(-5x^2 - 5x)} \phantom{10} \downarrow && \\ 10x + \phantom{0}1 && \\ \underline{-(10x + 10)} && \\ -9 && \\ \end{array}$$
There is a lot of repetition in the table. To simplify it, don’t write the variables, but line up their coefficients in columns under the division sign. Also, eliminate the partial products to create a simpler version of the entire problem.
$$ \require{enclose} \begin{array}{rll} 3 \quad -5 \quad\quad 10 && \\ 1 \enclose{longdiv}{3 \quad -2 \quad\quad 5 \quad\quad \phantom{0}1} && \\ \underline{-3 \quad -3}\quad \phantom{+} \downarrow \phantom{000}\quad && \\ \underline{5 \quad\quad 5} \phantom{10} \quad \downarrow && \\ \underline{-10 \; -10} && \\ -9 && \\ \end{array}$$
Synthetic division further simplifies even more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 1, as is done in division of whole numbers, then multiplying and subtracting the middle product, change the sign of the “divisor” to –1, multiply and add. The process starts by bringing down the leading coefficient.
$$ \begin{array}{rrrrr} \underline{-1}| & 3 & -2 & 5 & 1 \\ & & -3 & 5 & -10 \\ \hline & 3 & -5 & 10 & |\underline{-9} \end{array} $$
Then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 3x2 − 5x + 10 and the remainder is −9. The process will be made more clear in example 4.
Synthetic division is a shortcut that can be used when the divisor is a binomial in the form x − k. In synthetic division, only the coefficients are used in the division process.
To divide a polynomial by x − k,
Use synthetic division to divide 4x2 – 13x + 15 by x – 2.
Begin by setting up the synthetic division. The binomial divisor is x – 2, so k is 2. Write k and the coefficients.
$$ \begin{array}{rrrr} \underline{2}| & 4 & -13 & 15 \\ \end{array} $$
Bring down the leading coefficient. Multiply the leading coefficient by k, and write the result in the next column.
$$ \begin{array}{rrrr} \underline{2}| & 4 & -13 & 15 \\ & & 8 & & \\ \hline & 4 & & & \end{array} $$
Next, add 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.
$$ \begin{array}{rrrr} \underline{2}| & 4 & -13 & 15 \\ & & 8 & -10 \\ \hline & 4 & -5 & |\underline{5} \end{array} $$
The quotient is 4x − 5. The remainder is 5. This is written as 4x – 5 + \(\mathbf{\frac{5}{x-2}}\).
Analysis: Just as with long division, check the work by multiplying the quotient by the divisor and adding the remainder.
(x – 2)(4x − 5) + 5 = 4x2 – 13x + 15
Use synthetic division to divide 2x3 − 5x2 − 8x + 15 by x − 3.
The binomial divisor is x − 3 so k = 3. Add each column, multiply the result by 3, and repeat until the last column is reached.
$$ \begin{array}{rrrrr} \underline{3}| & 2 & -5 & -8 & 15 \\ & & 6 & 3 & -15 \\ \hline & 2 & 1 & -5 & |\underline{\phantom{00}0} \end{array} $$
The quotient is 2x2 + x − 15. The remainder is 0, so x − 3 is a factor of 2x3 − 5x2 − 8x + 15.
Analysis: The graph of the polynomial function f(x) = 2x3 − 5x2 − 8x + 15 in figure 2 shows a zero at x = k = 3 because there is an x-intercept at x = 3. This confirms thatx − 3 is a factor of 2x3 − 5x2 − 8x + 15.
Use synthetic division to divide 2x4 + 3x3 − 2x2 + 4 by x + 2.
Notice there is no x-term. Use a zero as the coefficient for that term. All powers of x and the constant must be represented in the coefficients for synthetic division.
The binomial divisor is x + 2 so k = −2. Add each column, multiply the result by −2, and repeat until the last column is reached.
$$ \begin{array}{rrrrr} \underline{−2}| & 2 & 3 & -2 & 0 & 4 \\ & & -4 & 2 & 0 & 0 \\ \hline & 2 & -1 & 0 & 0 & |\underline{4} \end{array} $$
The remainder is 4, so the quotient is \(\mathbf{2x^3 - x^2 + \frac{4}{x + 2}}\).
Use synthetic division to divide x4 + 5x3 + 7x – 20 by x + 5.
\(x^3 + 7 + \frac{-55}{x + 5}\)
Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. An example of this was at the beginning of this section.
The area of a rectangle is given by x2 − 2x − 3. The width is x + 1. Find the length of the rectangle.
A sketch of the problem is in figure 3.
Write an equation by substituting the known values into the formula for the area of a rectangle.
A = l · w
x2 − 2x − 3 = l · (x + 1)
Now solve for h using synthetic division.
$$ l = \frac{x^2 - 2x - 3}{x + 1} $$
$$ \begin{array}{rrrr} \underline{−1}| & 1 & -2 & -3 \\ & & -1 & 3 \\ \hline & 1 & -3 & |\underline{\phantom{0}0} \end{array} $$
The quotient is x − 3 and the remainder is 0. The length of the rectangle is x − 3.
The area of a rectangle is 2x3 + 13x2 + 11x − 20. The width of the rectangle is x + 5. Find an expression for the length of the rectangle.
2x2 + 3x − 4
173 #1, 3, 5, 7, 9, 11, 13, 15, 21, 31, Mixed Review = 15
Mixed Review