Practice 1 Solutions

State the restrictions on the denominator.

Note

Restrictions will be written in numerical order from least to greatest, and will use the ± symbol for answers that include the positive and negative value of a term.

$\frac{8 x^{2}}{x^{3} + 2 x^{2} + x}$

$x (x^{2} + 2 x + 1) x {(x + 1)}^{2} \begin{matrix} x \neq 0 & x + 1 \neq 0 \\ x \neq - 1 \end{matrix}$

$x \neq 0, - 1$

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

$(2 r^{3} + 3 r^{2}) - (2 r + 3) r^{2} (2 r + 3) - 1 (2 r + 3) (r^{2} - 1) (2 r + 3) (r + 1) (r - 1) (2 r + 3) \begin{matrix} r + 1 \neq 0 & r - 1 \neq 0 & 2 r + 3 \neq 0 \\ r \neq - 1 & r \neq 1 & r \neq - \frac{3}{2} \end{matrix}$

$r \neq - \frac{3}{2}, \pm 1$

$\frac{4 p + 5}{4 p^{2} - 16}$

$4 (p^{2} - 4) 4 (p + 2) (p - 2) \begin{matrix} p + 2 \neq 0 & p - 2 \neq 0 \\ p \neq - 2 & p \neq 2 \end{matrix}$

$p \neq \pm 2$

$\frac{x + 3}{x^{4} - 81}$

$(x^{2} + 9) (x^{2} - 9) (x^{2} + 9) (x + 3) (x - 3) \begin{matrix} x^{2} + 9 \neq 0 & x + 3 \neq 0 & x - 3 \neq 0 \\ Not Possible & x \neq - 3 & x \neq 3 \\ in real number \\ system \end{matrix}$

$x \neq \pm 3$

Note

You will learn how to solve $x^{2} + 9 = 0$ using the set of complex numbers in a later lesson.

Simplify. State the restrictions on the denominator.

$\frac{6 x^{2} + x - 1}{2 x^{2} - 13 x + 6} \cdot \frac{16 x^{2} + 40 x + 25}{12 x^{2} + 11 x - 5}$

$\frac{(2 x + 1) (3 x - 1)}{(2 x - 1) (x - 6)} \cdot \frac{(4 x + 5) (4 x + 5)}{(4 x + 5) (3 x - 1)} \frac{(2 x + 1) (3 x - 1)}{(2 x - 1) (x - 6)} \cdot \frac{(4 x + 5) (4 x + 5)}{(4 x + 5) (3 x - 1)}$

$\frac{(2 x + 1) (4 x + 5)}{(2 x - 1) (x - 6)}, x \neq - \frac{5}{4}, \frac{1}{3}, \frac{1}{2}, 6$

$\frac{x^{2} + 7 x + 12}{x^{2} - x - 12} \div \frac{x^{2} + 3 x + 2}{x^{2} - 9}$

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

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

Note

If asked, when expressions are divided, the numerator and denominator of the divisors need to be named as part of the restrictions.

Simplify.

Note

Problems 7–12
Solving for restrictions is not shown because they can be found using mental math.

$\frac{2 z^{2} + 13 z + 15}{z^{2} + 4 z + 4} \cdot \frac{2 z^{2} + z - 6}{z^{2} - 25}$

$\frac{(2 z + 3) (z + 5)}{(z + 2) (z + 2)} \cdot \frac{(2 z - 3) (z + 2)}{(z + 5) (z - 5)} \frac{(2 z + 3) (z + 5)}{(z + 2) (z + 2)} \cdot \frac{(2 z - 3) (z + 2)}{(z + 5) (z - 5)} z \neq \pm 5, - 2$

$\frac{(2 z + 3) (2 z - 3)}{(z + 2) (z - 5)}, z \neq \pm 5, - 2$

$\frac{b^{2} - 9 b + 20}{4 b^{2} + 6 b} \div \frac{2 b^{2} - 32}{4 b^{2} + 12 b + 9} \cdot \frac{8 b + 8}{b^{2} - 3 b - 4}$

$\frac{(b - 5) (b - 4)}{2 b (2 b + 3)} \div \frac{2 (b^{2} - 16)}{(2 b + 3) (2 b + 3)} \cdot \frac{8 (b + 1)}{(b - 4) (b + 1)} \frac{(b - 5) (b - 4)}{2 b (2 b + 3)} \cdot \frac{(2 b + 3) (2 b + 3)}{2 (b + 4) (b - 4)} \cdot \frac{8 (b + 1)}{(b - 4) (b + 1)} \frac{(b - 5) (b - 4)}{2 b (2 b + 3)} \cdot \frac{(2 b + 3) (2 b + 3)}{2 (b + 4) (b - 4)} \cdot \frac{{}^{2}8 (b + 1)}{(b - 4) (b + 1)} b \neq \pm 4, - \frac{3}{2}, - 1$

$\frac{2 (b - 5) (2 b + 3)}{b (b + 4) (b - 4)}, b \neq \pm 4, - \frac{3}{2}, - 1$

Note

Q: What is the divisor of this problem?
A: The middle expression $\frac{2 b^{2} - 32}{4 b^{2} + 12 b + 9}$

$\frac{g^{4} - 13 g^{2} + 36}{g^{2} - 10 g + 25} \div \frac{g^{2} - 5 g + 6}{g^{2} - g - 30} \cdot \frac{g^{2} - 25}{g^{2} + 7 g + 10}$

$\frac{(g^{2} - 4) (g^{2} - 9)}{(g - 5) (g - 5)} \div \frac{(g - 2) (g - 3)}{(g - 6) (g + 5)} \cdot \frac{(g + 5) (g - 5)}{(g + 5) (g + 2)} \frac{(g + 2) (g - 2) (g + 3) (g - 3)}{(g - 5) (g - 5)} \cdot \frac{(g - 6) (g + 5)}{(g - 2) (g - 3)} \cdot \frac{(g + 5) (g - 5)}{(g + 5) (g + 2)} \frac{(g + 2) (g - 2) (g + 3) (g - 3)}{(g - 5) (g - 5)} \cdot \frac{(g - 6) (g + 5)}{(g - 2) (g - 3)} \cdot \frac{(g + 5) (g - 5)}{(g + 5) (g + 2)} g \neq \pm 5, \pm 2, 3, 6$

$\frac{(g + 3) (g - 6) (g + 5)}{(g - 5)}, g \neq \pm 5, \pm 2, 3, 6$

$\frac{7 x}{x^{2} - 4 x - 21} \div \frac{x^{3} - 49 x}{x^{2} + 9 x + 18}$

$\frac{7 x}{(x - 7) (x + 3)} \div \frac{x (x^{2} - 49)}{(x + 6) (x + 3)} \frac{7 x}{(x - 7) (x + 3)} \div \frac{x (x + 7) (x - 7)}{(x + 6) (x + 3)} \frac{7 x}{(x - 7) (x + 3)} \cdot \frac{(x + 6) (x + 3)}{x (x + 7) (x - 7)} \frac{7 x}{(x - 7) (x + 3)} \cdot \frac{(x + 6) (x + 3)}{x (x + 7) (x - 7)} x \neq \pm 7, - 6, - 3, 0$

$\frac{7 (x + 6)}{{(x - 7)}^{2} (x + 7)}, x \neq \pm 7, - 6, - 3, 0$

$\frac{14 c^{4} - c^{3} - 3 c^{2}}{9 c^{3} - 25 c} \cdot \frac{9 c^{2} - 9 c - 10}{49 c^{2} + 42 c + 9}$

$\frac{c^{2} (7 c + 3) (2 c - 1)}{c (9 c^{2} - 25)} \cdot \frac{(3 c - 5) (3 c + 2)}{(7 c + 3) (7 c + 3)} \frac{c^{2} (7 c + 3) (2 c - 1)}{c (3 c + 5) (3 c - 5)} \cdot \frac{(3 c - 5) (3 c + 2)}{(7 c + 3) (7 c + 3)} \frac{c^{2} (7 c + 3) (2 c - 1)}{c (3 c + 5) (3 c - 5)} \cdot \frac{(3 c - 5) (3 c + 2)}{(7 c + 3) (7 c + 3)} c \neq \pm \frac{5}{3}, - \frac{3}{7}, 0$

$\frac{c (2 c - 1) (3 c + 2)}{(3 c + 5) (7 c + 3)}, c \neq \pm \frac{5}{3}, - \frac{3}{7}, 0$

Note

Q: Why does the term “3c” not simplify out of the expression?
A: Because 3c is part of two different binomial expressions and only identical binomial expressions simplify out.

$\frac{4 y^{2} + 4 y - 35}{3 y^{2} - 20 y - 7} \cdot \frac{y^{2} - 11 y + 28}{6 y^{2} + 19 y - 7} \div \frac{y^{2} - 16}{9 y^{2} - 1}$

$\frac{(2 y - 5) (2 y + 7)}{(3 y + 1) (y - 7)} \cdot \frac{(y - 4) (y - 7)}{(3 y - 1) (2 y + 7)} \div \frac{(y + 4) (y - 4)}{(3 y + 1) (3 y - 1)} \frac{(2 y - 5) (2 y + 7)}{(3 y + 1) (y - 7)} \cdot \frac{(y - 4) (y - 7)}{(3 y - 1) (2 y + 7)} \cdot \frac{(3 y + 1) (3 y - 1)}{(y + 4) (y - 4)} \frac{(2 y - 5) (2 y + 7)}{(3 y + 1) (y - 7)} \cdot \frac{(y - 4) (y - 7)}{(3 y - 1) (2 y + 7)} \cdot \frac{(3 y + 1) (3 y - 1)}{(y + 4) (y - 4)} y \neq - \frac{7}{2}, \pm \frac{1}{3}, \pm 4, 7$

$\frac{2 y - 5}{y + 4}, y \neq - \frac{7}{2}, \pm \frac{1}{3}, \pm 4, 7$

Find the missing side.

The area of a triangle is $\frac{2 x + 1}{5 x + 2} m^{2}$ and the area of a rectangle is $\frac{4 x^{2} + 2 x}{5} m^{2}$ . Find the probability of the triangle. $P (triangle) = \frac{area of triangle}{area of rectangle}$ What is the probability of the triangle if x = 1?

$\frac{\frac{2 x + 1}{5 x + 2}}{\frac{4 x^{2} + 2 x}{5}} = \frac{2 x + 1}{5 x + 2} \div \frac{4 x^{2} + 2 x}{5} \frac{2 x + 1}{5 x + 2} \cdot \frac{5}{4 x^{2} + 2 x} \frac{2 x + 1}{5 x + 2} \cdot \frac{5}{2 x (2 x + 1)} \frac{2 x + 1}{5 x + 2} \cdot \frac{5}{2 x (2 x + 1)} P (triangle, x = 1) = \frac{5}{2 (1) (5 (1) + 2)} = \frac{5}{2 (7)} = \frac{5}{14}$

$P (triangle) = \frac{5}{2 x (5 x + 2)} P (triangle, x = 1) = \frac{5}{14}$

Note

Remember that a fraction bar means division, or $\frac{n}{d} = n \div d$ . This means you can write the fraction vertically or horizontally.
Remember to simplify and then substitute in the value for x. More problems related to probability will be in later lessons.

Find the height of a trapezoid when the sum of the bases is $\frac{4 x + 5}{x^{2} + 9 x - 10}$ inches and the area is $\frac{12 x^{2} + 19 x + 5}{x^{2} - 100} {inches}^{2}$ . Then explain if x = 1 is a possible solution.

$\begin{array}{rcl} A & = & \frac{1}{2} h (b_{1} + b_{2}) \\ 2 A & = & h (b_{1} + b_{2}) \\ \frac{2 A}{(b_{1} + b_{2})} & = & h \\ h & = & \frac{\frac{2 (12 x^{2} + 19 x + 5)}{x^{2} - 100}}{\frac{4 x + 5}{x^{2} + 9 x - 10}} = \frac{2 (12 x^{2} + 19 x + 5)}{x^{2} - 100} \div \frac{4 x + 5}{x^{2} + 9 x - 10} \\ h & = & \frac{2 (12 x^{2} + 19 x + 5)}{x^{2} - 100} \cdot \frac{x^{2} + 9 x - 10}{4 x + 5} \\ h & = & \frac{2 (4 x + 5) (3 x + 1)}{(x + 10) (x - 10)} \cdot \frac{(x + 10) (x - 1)}{(4 x + 5)} \\ h & \neq & \pm 10, - \frac{5}{4}, 1 \end{array}$

$h = \frac{2 (3 x + 1) (x - 1)}{(x - 10)} inches$

It is not possible for x = 1 because it is a restriction on the denominator (or excluded value).

All numbers that make the denominator of a rational expression equal to zero must be excluded as possible values.

A rational expression can also be described as the quotient of two polynomials, since the fraction bar represents a division problem.