Lesson 6: Practice 1 Solutions

Find the quotient using synthetic division.

$(5 a^{3} + 14 a^{2} - 7 a + 9) \div (a + 4)$

$5 a^{2} - 6 a + 18 - \frac{63}{a + 4}$

2) $\frac{x^{2} - 8 x + 12}{x - 3}$

$x - 5 - \frac{3}{x - 3}$

3) $2 x - 1 4 x^{3} - 6 x^{2} + 10 x + 2$

$2 x^{2} - 2 x + 4 + \frac{3}{x - \frac{1}{2}} \frac{(3) (2)}{(x - \frac{1}{2}) (2)}$

Notes

Remember to divide every term by the coefficient of the divisor before using synthetic division. To write the final answer, multiply the remainder by this coefficient to clear fractions.

$2 x^{2} - 2 x + 4 + \frac{6}{2 x - 1}$

4) $(x^{3} - 3 x^{2} + 5 x - 6) \div (x - 2)$

Notes

Q: How do you know that the divisor is a factor of the polynomial?
A: The remainder is zero.

$x^{2} - x + 3$

5) $\frac{9 x^{4} + 6 x^{3} - 12 x^{2} - 8 x + 4}{3 x + 2}$

$\frac{(9 x^{4} + 6 x^{3} - 12 x^{2} - 8 x + 4) \div 3}{(3 x + 2) \div 3} = \frac{3 x^{4} + 2 x^{3} - 4 x^{2} - \frac{8}{3} x + \frac{4}{3}}{x + \frac{2}{3}}$

$3 x^{3} + 4 x + \frac{\frac{4}{3}}{x + \frac{2}{3}} \frac{(\frac{4}{3}) (3)}{(x + \frac{2}{3}) (3)}$

Notes

Q: What is the coefficient of the divisor?
A: 3

Q: When there is a 0 in a column, what does this mean for the quotient?

A: That degree term has a coefficient of zero and is not needed to be written as part of the final answer.

$3 x^{3} + 4 x + \frac{4}{3 x + 2}$

6) $(x^{4} - 13 x^{4} + 36) \div (x + 3)$

$x^{3} - 3 x^{2} - 4 x + 12$

7) $\frac{b^{3} - 8}{b - 2}$

Notes

Notice that this is the difference of cubes. You can use synthetic division or the polynomial identity for factoring a difference of cubes.

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

8) $x - 6 x^{4} - 1 2 x^{2} - 8 x - 7 6$

Notes

Q: What is the coefficient of the cubic term? Explain.
A: It is zero because this is not listed in the given polynomial. Zero multiplied by anything is zero.

$x^{3} + 6 x^{2} + 24 x + 136 + \frac{740}{x - 6}$

Notes

Problems 9–12
When the directions say to find the remainder, the quotient does not need to be written.

Use the Remainder Theorem to determine P(k).

9) $P (x) = 2 x^{3} + x^{2} - 4 x + 3; P (- 1)$

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$P (- 1) = 6$

10) $P (x) = x^{5} - 4 x^{3} + x^{2} - 5; P (5)$

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Notes

You can use a calculator to work with the large terms more quickly.

P(5) = 2645

11) $P (x) = 9 x^{3} + 13 x^{2} - 6 x + 8; P (- 2)$

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Notes

Q: How do you know that –2 is a root of the polynomial?
A: Because the remainder is zero.

Q: How do you write this as a factor?
A: x + 2

$P (- 2) = 0$

12) $P (x) = 3 x^{2} - 2 x - 1; P (- \frac{1}{3})$

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Notes

Q: Name the root and factor of this problem.
A: $- \frac{1}{3}$ is a root of the polynomial. $(x - \frac{1}{3}) o r (3 x - 1)$ is a factor.

$P (\frac{1}{3}) = 0$

Notes

Problems 13–14
Remember to check your work by substituting the value of n back into the polynomial.

Find the missing value.

13) $P (3) = 2; P (x) = 2 x^{3} - 3 x^{2} - 5 x + n$

$\begin{array}{rcl} n + 12 & = & 2 \end{array}$

$\begin{array}{rcl} n & = & - 10 \end{array}$

14) $P (- 1) = 5; P (x) = x^{3} + n x - 8$

$\begin{array}{rcl} - 8 - 1 (n + 1) & = & 5 \\ - n - 1 & = & 13 \\ - n & = & 14 \end{array}$

$\begin{array}{rcl} n & = & - 14 \end{array}$