Questions & explanations
1. A student attempted to simplify the expression \(\frac{x^2 - 5x + 6}{x^2 - 4} \cdot \frac{x^2 - 2x - 8}{x^2 - 9x + 18}\) and obtained \(\frac{x-2}{x+2}\). Which of the following best describes the student's error?
- The student incorrectly factored \(x^2 - 5x + 6\) as \((x-2)(x-3)\) instead of \((x-3)(x-2)\) (order doesn't matter).
- The student incorrectly factored \(x^2 - 2x - 8\) as \((x-4)(x+2)\) instead of \((x-4)(x+2)\) (correct).
- The student incorrectly canceled factors that are not common, or made a sign error in the final simplification.
- The student incorrectly factored \(x^2 - 9x + 18\) as \((x-6)(x-3)\) instead of \((x-3)(x-6)\) (order doesn't matter).
- The student made no error; the simplified expression is indeed \(\frac{x-2}{x+2}\).
Answer: The student incorrectly canceled factors that are not common, or made a sign error in the final simplification.
First, factor each polynomial: \(x^2 - 5x + 6 = (x-2)(x-3)\); \(x^2 - 4 = (x-2)(x+2)\); \(x^2 - 2x - 8 = (x-4)(x+2)\); \(x^2 - 9x + 18 = (x-3)(x-6)\). Then the expression becomes \(\frac{(x-2)(x-3)}{(x-2)(x+2)} \cdot \frac{(x-4)(x+2)}{(x-3)(x-6)} = \frac{(x-2)(x-3)(x-4)(x+2)}{(x-2)(x+2)(x-3)(x-6)} = \frac{x-4}{x-6}\). The correct simplified form is \(\frac{x-4}{x-6}\), not \(\frac{x-2}{x+2}\). The student likely canceled \((x-2)\) and \((x+2)\) but also incorrectly canceled \((x-3)\) with something else or made a sign error. Option c correctly identifies that the student made an error in cancellation or simplification.
2. Simplify: \(\frac{x^2 - y^2}{x^2 + 2xy + y^2} \div \frac{x^2 - 2xy + y^2}{x^2 - y^2}\).
- 1
- \(\frac{(x-y)^2}{(x+y)^2}\)
- \(\frac{(x+y)^2}{(x-y)^2}\)
- \(\frac{x+y}{x-y}\)
- \(\frac{x-y}{x+y}\)
Answer: 1
Rewrite division as multiplication by reciprocal: \(\frac{x^2 - y^2}{x^2 + 2xy + y^2} \times \frac{x^2 - y^2}{x^2 - 2xy + y^2}\). Factor: \(x^2 - y^2 = (x-y)(x+y)\), \(x^2 + 2xy + y^2 = (x+y)^2\), \(x^2 - 2xy + y^2 = (x-y)^2\). Then expression becomes \(\frac{(x-y)(x+y)}{(x+y)^2} \times \frac{(x-y)(x+y)}{(x-y)^2} = \frac{x-y}{x+y} \times \frac{x+y}{x-y} = 1\).
3. Simplify the expression: \(3x^2y(2x^3y^2 - 4xy + 5y^3)\).
- \(6x^5y^3 - 12x^3y^2 + 15x^2y^4\)
- \(6x^6y^3 - 12x^3y^2 + 15x^2y^4\)
- \(6x^5y^3 - 12x^3y^2 + 15x^2y^3\)
- \(6x^5y^3 - 12x^2y^2 + 15x^2y^4\)
- \(6x^5y^3 - 12x^3y^2 + 15x^3y^4\)
Answer: \(6x^5y^3 - 12x^3y^2 + 15x^2y^4\)
Multiply the monomial \(3x^2y\) by each term of the polynomial:
- \(3x^2y \cdot 2x^3y^2 = 6x^{2+3}y^{1+2} = 6x^5y^3\)
- \(3x^2y \cdot (-4xy) = -12x^{2+1}y^{1+1} = -12x^3y^2\)
- \(3x^2y \cdot 5y^3 = 15x^2y^{1+3} = 15x^2y^4\)
Combine: \(6x^5y^3 - 12x^3y^2 + 15x^2y^4\).
4. Simplify: \(\frac{x^2 - y^2}{x^2 + 2xy + y^2}\)
- \(\frac{x - y}{x + y}\)
- \(\frac{x + y}{x - y}\)
- \(\frac{x - y}{x + y}\) for \(x \neq -y\)
- \(\frac{x + y}{x - y}\) for \(x \neq y\)
- \(\frac{x^2 - y^2}{(x+y)^2}\)
Answer: \(\frac{x - y}{x + y}\) for \(x \neq -y\)
Factor numerator: \(x^2 - y^2 = (x-y)(x+y)\). Denominator: \(x^2 + 2xy + y^2 = (x+y)^2\). Cancel common factor \((x+y)\) (provided \(x+y \neq 0\)): \(\frac{(x-y)(x+y)}{(x+y)^2} = \frac{x-y}{x+y}\). Option c includes the domain restriction.
5. A rectangular garden has length (x+5) meters and width (x-2) meters. A path of uniform width w surrounds the garden. The total area of the garden plus path is (x^2 + 10x + 16) square meters. If w is a positive integer, what is the value of w?
- 1
- 2
- 3
- 4
- 5
Answer: 2
Let w be path width. Total length = x+5+2w, total width = x-2+2w. Area = (x+5+2w)(x-2+2w) = x^2 + (3+4w)x + (5+2w)(-2+2w). Set equal to x^2+11x+18. Equate x coefficients: 3+4w=11 => w=2. Check constant: (5+4)(-2+4)=9*2=18. So w=2.
6. Simplify: \(\frac{2}{x-y} + \frac{3}{x+y} - \frac{5x}{x^2 - y^2}\).
- 0
- \(\frac{5}{x+y}\)
- \(\frac{5}{x-y}\)
- \(\frac{5x}{x^2 - y^2}\)
- \(\frac{5y}{x^2 - y^2}\)
Answer: 0
Note \(x^2 - y^2 = (x-y)(x+y)\). Write each term with denominator \((x-y)(x+y)\): \(\frac{2(x+y)}{(x-y)(x+y)} + \frac{3(x-y)}{(x-y)(x+y)} - \frac{5x}{(x-y)(x+y)} = \frac{2x+2y+3x-3y-5x}{(x-y)(x+y)} = \frac{0}{(x-y)(x+y)} = 0\).
7. Simplify: ((x^2 - 9)/(x^2 - 4x + 3)) * ((x^2 - 2x - 3)/(x^2 + 4x + 3)).
- (x+3)/(x+1)
- (x-3)/(x-1)
- (x+3)/(x-1)
- (x-3)/(x+1)
- 1
Answer: (x-3)/(x+1)
Factor: (x^2-9) = (x-3)(x+3); (x^2-4x+3) = (x-1)(x-3); (x^2-2x-3) = (x-3)(x+1); (x^2+4x+3) = (x+1)(x+3). Multiply: [(x-3)(x+3)(x-3)(x+1)] / [(x-1)(x-3)(x+1)(x+3)] = (x-3)/(x-1). Cancel common factors: (x-3), (x+3), (x+1).
8. Factor the polynomial: 6x^2 + 11x - 35
- (3x - 5)(2x + 7)
- (3x + 5)(2x - 7)
- (6x - 5)(x + 7)
- (6x + 5)(x - 7)
- (2x - 5)(3x + 7)
Answer: (3x - 5)(2x + 7)
To factor 6x^2 + 11x - 35, find factors of 6 and -35 that sum to 11. Factors of 6: (1,6), (2,3). Factors of -35: (1,-35), (5,-7), (7,-5), (35,-1). Try (3x - 5)(2x + 7) = 6x^2 + 21x - 10x - 35 = 6x^2 + 11x - 35. Correct.
9. Factor completely: \(2x^2 - 3xy - 2y^2\).
- \((2x+y)(x-2y)\)
- \((2x-y)(x+2y)\)
- \((2x+2y)(x-y)\)
- \((2x-2y)(x+y)\)
- \((2x+3y)(x-y)\)
Answer: \((2x+y)(x-2y)\)
Multiply \(a\cdot c = 2 \cdot (-2) = -4\). Find two numbers that multiply to -4 and add to -3: -4 and 1. Rewrite: \(2x^2 -4xy + xy -2y^2\). Factor by grouping: \(2x(x-2y) + y(x-2y) = (x-2y)(2x+y)\). So answer is a.
10. Simplify: ((x^2 - 9) / (x^2 + 6x + 9)) * ((x^2 + 3x) / (x^2 - 3x))
- 1
- x/(x+3)
- (x+3)/x
- x/(x-3)
- (x-3)/x
Answer: 1
Factor each expression: x^2 - 9 = (x-3)(x+3); x^2 + 6x + 9 = (x+3)^2; x^2 + 3x = x(x+3); x^2 - 3x = x(x-3). Then product = [(x-3)(x+3) / (x+3)^2] * [x(x+3) / x(x-3)] = Cancel (x-3), (x+3), x: = 1.
11. Simplify: ( (x^4 - 16) / (x^2 - 4x + 4) ) ÷ ( (x^2 + 4) / (x - 2) )
- x + 2
- x - 2
- (x+2)/(x-2)
- (x-2)/(x+2)
- 1
Answer: x + 2
Factor: x^4 - 16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4). x^2 - 4x + 4 = (x-2)^2. Division: multiply by reciprocal: [(x-2)(x+2)(x^2+4) / (x-2)^2] * [(x-2)/(x^2+4)] = Cancel (x-2), (x^2+4): = (x+2).
12. Factor the quadratic expression 2x² + 7x + 3. Which of the following is one of its factors?
- 2x + 1
- x + 3
- 2x + 3
- x + 1
- 2x + 2
Answer: 2x + 1
Factor 2x² + 7x + 3: multiply a and c: 2*3=6. Find two numbers that multiply to 6 and add to 7: 1 and 6. Rewrite: 2x² + x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3). So 2x+1 is a factor.