Practice 1 Solutions

Note

Problems 1–4
When dividing by a monomial, be sure that your student simplifies the numerical coefficients as well as the variables. The variables should be simplified using exponent rules.

Simplify. Write the polynomial expression with positive exponents.

$\frac{20 a^{3} b^{2} - 15 a b^{3} + 10 a^{2} b}{5 a b}$

$\frac{20 a^{3} b^{2}}{5 a b} - \frac{15 a b^{3}}{5 a b} + \frac{10 a^{2} b}{5 a b}$

$4 a^{2} b - 3 b^{2} + 2 a$

$(13 x^{4} y^{5} + 39 x^{2} - 26) \div 13 x^{2} y$

$\frac{13 x^{4} y^{5} + 39 x^{2} - 26}{13 x^{2} y} \frac{13 x^{4} y^{5}}{13 x^{2} y} + \frac{39 x^{2}}{13 x^{2} y} - \frac{26}{13 x^{2} y}$

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

$\frac{2 p q^{4} + 12 p^{2} q^{2} - 9 p^{3} q + 8 p q}{- 3 p^{3} q}$

$\frac{2 p q^{4}}{- 3 p^{3} q} + \frac{12 p^{2} q^{2}}{- 3 p^{3} q} - \frac{9 p^{3} q}{- 3 p^{3} q} + \frac{8 p q}{- 3 p^{3} q}$

$- \frac{2 q^{3}}{3 p^{2}} - \frac{4 q}{p} + 3 - \frac{8}{3 p^{2}}$

$(4 x^{4} - 5 x^{3} + 8 x^{2} - 6 x + 2) {(4 x)}^{- 1}$

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

$x^{3} - \frac{5 x^{2}}{4} + 2 x - \frac{3}{2} + \frac{1}{2 x}$

Simplify. Write the remainder as a fraction if one exists.

$(3 y^{3} + 17 y^{2} + 22 y + 8) \div (y + 4)$

$y + 4 3 y^{2} + 5 y + 2 3 y^{3} + 17 y^{2} + 22 y + 8 - (3 y^{3} + 12 y^{2}) 5 y^{2} + 22 y - (5 y^{2} + 20 y) 2 y + 8 - (2 y + 8) 0$

$3 y^{2} + 5 y + 2$

Note

Q: How do you determine where to place the terms in the quotient?
A: By place-value according to the degree of the term.

Q: How can you determine if your solution is correct?
A: Find the product of the quotient and the divisor.

Q: What if your solution also contains a remainder?
A: You find the product first and then add in the remainder.

Q: When an expression is raised to the negative first power is this the dividend, divisor, quotient or remainder.
A: The divisor

$x^{2} - 3 x + 1 2 x^{3} - 3 x^{2} - 7 x + 3$

$x^{2} - 3 x + 1 2 x + 3 2 x^{3} - 3 x^{2} - 7 x + 3 - (2 x^{3} - 6 x^{2} + 2 x) 3 x^{2} - 9 x + 3 - (3 x^{2} - 9 x + 3) 0$

$2 x + 3$

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

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

$3 x + 5 + \frac{6}{x - 1}$

$(x^{4} - 3 x^{2} + x - 5) {(x + 1)}^{- 1}$

$x + 1 x^{3} - x^{2} - 2 x + 3 x^{4} + 0 x^{3} - 3 x^{2} + x - 5 - (x^{4} + x^{3}) - x^{3} - 3 x^{2} - (- x^{3} - x^{2}) - 2 x^{2} + x - (- 2 x^{2} - 2 x) 3 x - 5 - (3 x + 3) - 8 - \frac{8}{x + 1}$

$x^{3} - x^{2} - 2 x + 3 - \frac{8}{x + 1}$

$2 y - 1 4 y^{2} - 8 y + 3$

$2 y - 1 2 y - 3 4 y^{2} - 8 y + 3 - (4 y^{2} - 2 y) - 6 y + 3 - (- 6 y + 3) 0$

$2 y - 3$

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

$x + 2 3 x^{2} - 4 x + 8 3 x^{3} + 2 x^{2} + 0 x - 8 - (3 x^{3} + 6 x^{2}) - 4 x^{2} + 0 x - (- 4 x^{2} - 8 x) 8 x - 8 - (8 x + 16) - 24 - \frac{24}{x + 2}$

$3 x^{2} - 4 x + 8 - \frac{24}{x + 2}$

$\frac{3 x^{4} + 2 x^{2} + 16 x + 11}{x^{2} + 2 x + 1}$

$x^{2} + 2 x + 1 3 x^{2} - 6 x + 11 3 x^{4} + 0 x^{3} + 2 x^{2} + 16 x + 11 - (3 x^{4} + 6 x^{3} + 3 x^{2}) - 6 x^{3} - x^{2} + 16 x - (- 6 x^{3} - 12 x^{2} - 6 x) 11 x^{2} + 22 x + 11 - (11 x^{2} + 22 x + 11) 0$

$3 x^{2} - 6 x + 11$

$(2 x^{3} + 13 x^{2} - x - 110) {(x - \frac{5}{2})}^{- 1}$

$x - \frac{5}{2} 2 x^{2} + 18 x + 44 2 x^{3} + 13 x^{2} - x - 110 - (2 x^{3} - 5 x^{2}) 18 x^{2} - x - (18 x^{2} - 45 x) 44 x - 110 - (44 x - 110) 0$

$2 x^{2} + 18 x + 44$

$\frac{5 a^{3} - 30 a^{2} + 70}{5 a}$

$\frac{5 a^{3}}{5 a} - \frac{30 a^{2}}{5 a} + \frac{70}{5 a}$

$a^{2} - 6 a + \frac{14}{a}$

$(5 y^{4} + 3 y^{3} + 8) \div (y + 2)$

$y + 2 5 y^{3} - 7 y^{2} + 14 y - 28 5 y^{4} + 3 y^{3} + 0 y^{2} + 0 y + 8 - (5 y^{4} + 10 y^{3}) - 7 y^{3} + 0 y^{2} - (- 7 y^{3} - 14 y^{2}) 14 y^{2} + 0 y - (14 y^{2} + 28 y) - 28 y + 8 - (- 28 y - 56) 64 + \frac{64}{y + 2}$

$5 y^{3} - 7 y^{2} + 14 y - 28 + \frac{64}{y + 2}$

Note

Problems 15–16
Remind your student to use their Formula Sheet when working with figures.

The volume of a rectangle prism is $2 x^{3} - 4 x - 16 x + 42 {cm}^{3} .$ The area of the base is $2 x^{2} - 10 x + 14 {cm}^{2} .$ Find the height.

$V = B h 2 x^{3} - 4 x^{2} - 16 x + 42 = (2 x^{2} - 10 x + 14) h h = \frac{2 x^{3} - 4 x^{2} - 16 x + 42}{2 x^{2} - 10 x + 14}$

$2 x^{2} - 10 x + 14 x + 3 2 x^{3} - 4 x^{2} - 16 x + 42 - (2 x^{3} - 10 x^{2} + 14 x) 6 x^{2} - 30 x + 42 - (6 x^{2} - 30 x + 42) 0$

The height is x + 3 cm.

The area of a triangle is $x^{2} + 8 x + 7 m^{2} .$ The height is x + 1 meters. Determine the length of the base.

$A = \frac{1}{2} b h 2 A = b h 2 (x^{2} + 8 x + 7) = b (x + 1) b = \frac{2 x^{2} - 16 x + 14}{x + 1} x + 1 2 x + 14 2 x^{2} + 16 x + 14 - (2 x^{2} + 2 x) 14 x + 14 - (14 x + 14) 0$

The base is 2x + 14 meters.

Note

Q: How can you clear the fraction from the area of a triangle formula?
A: Multiply both sides of the formula by 2.

Your student may also factor this problem to find the length of the base.