1.
Item 2. Simplify the rational expression and state the non-permissible values. \(\frac{-64x^3}{14y^3}\)
Solution:\(\frac{-64x^3}{14y^3}\)Simplify the coefficient by dividing numerator and denominator by \(2\):\(\frac{-64x^3}{14y^3}=\frac{2(-32x^3)}{2(7y^3)}\)\(=\frac{-32x^3}{7y^3}\)The denominator is \(7y^3\), so \(y\neq0\). There is no \(x\) in the denominator, so \(x\) has no restriction.Answer: A
2.
Item 5. Simplify. \(\frac{4(a+8)}{4(7a-8)}\times\frac{7a(7a-8)}{a+8}\)
Solution:\(\frac{4(a+8)}{4(7a-8)}\times\frac{7a(7a-8)}{a+8}\)Cancel the common factor \(4\):\(\frac{a+8}{7a-8}\times\frac{7a(7a-8)}{a+8}\)Cancel \((a+8)\) and \((7a-8)\):\(=1\times7a\)\(=7a\)Answer: C
3.
Item 1. Which of the following are rational expressions?<br>A) \(\frac{-46x-x^2}{4xy}\)<br>B) \(\frac{\sqrt{x}}{\sqrt{2}}\)<br>C) \(\frac{1}{x}\)<br>D) \(\frac{3x^2-8x+9}{9}\)
Solution:A) \(\frac{-46x-x^2}{4xy}\) is a rational expression because it is a quotient of polynomials.B) \(\frac{\sqrt{x}}{\sqrt{2}}\) is not a rational expression because the variable \(x\) is inside a radical.C) \(\frac{1}{x}\) is a rational expression because it is a quotient of polynomials.D) \(\frac{3x^2-8x+9}{9}\) is a rational expression because it is a quotient of polynomials.Therefore, A, C, and D are rational expressions.Answer: E
4.
Item 4. Simplify the rational expression and state the restrictions on the variable(s). \(\frac{2x-16}{x^2-7x-8}\)
Solution:\(\frac{2x-16}{x^2-7x-8}\)Factor the numerator:\(2x-16=2(x-8)\)Factor the denominator:\(x^2-7x-8=(x-8)(x+1)\)So:\(\frac{2x-16}{x^2-7x-8}=\frac{2(x-8)}{(x-8)(x+1)}\)Cancel \((x-8)\):\(=\frac{2}{x+1}\)Restrictions come from the original denominator: \((x-8)(x+1)\neq0\), so \(x\neq8\) and \(x\neq-1\).Answer: A
5.
Item 6. Simplify. \(\frac{4b^2-16b}{3b^2-15b}\times\frac{b^2-b-20}{4b^2+16b}\)
Solution:\(\frac{4b^2-16b}{3b^2-15b}\times\frac{b^2-b-20}{4b^2+16b}\)Factor each part:\(4b^2-16b=4b(b-4)\)\(3b^2-15b=3b(b-5)\)\(b^2-b-20=(b-5)(b+4)\)\(4b^2+16b=4b(b+4)\)So:\(\frac{4b(b-4)}{3b(b-5)}\times\frac{(b-5)(b+4)}{4b(b+4)}\)Cancel \(4\), \((b-5)\), and \((b+4)\):\(=\frac{b-4}{3b}\)This answer is not listed in options A-D.Answer: E
6.
Item 3. State the quantity you have to multiply or divide the first expression by to get the second expression. \(\frac{x^2-9}{2x-6},\; \frac{x+3}{2}\)
Solution:First expression: \(\frac{x^2-9}{2x-6}\)Factor the numerator and denominator:\(x^2-9=(x-3)(x+3)\)\(2x-6=2(x-3)\)So:\(\frac{x^2-9}{2x-6}=\frac{(x-3)(x+3)}{2(x-3)}\)The second expression is \(\frac{x+3}{2}\).To get the second expression from the first expression, divide by \((x-3)\).Answer: A
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