IBEW Aptitude Test Polynomial Factoring Methods Questions and Answers 3 — Questions and Answers

Question 1: Solve: x² + 3x - 10 = 0

x = 2 and x = -5 (Correct answer)
x = -2 and x = 5
x = -2 and x = -5
x = 2 and x = 5

Correct answer: x = 2 and x = -5

Factor: (x+5)(x-2)=0. x=-5 or x=2.

Need product = -10, sum = +3: numbers are +5 and -2 (5×-2=-10, 5+(-2)=3). Factor: (x+5)(x-2)=0. Solutions: x+5=0 → x=-5; x-2=0 → x=2. Check: (-5)²+3(-5)-10=25-15-10=0 ✓, (2)²+3(2)-10=4+6-10=0 ✓.

Question 2: Factor: 4x² - 12x + 9

(2x - 3)² (Correct answer)
(4x - 3)(x - 3)
(2x - 9)(2x - 1)
(2x + 3)²

Correct answer: (2x - 3)²

This is a perfect square: (2x)² - 2(2x)(3) + 3² = (2x-3)².

4x² - 12x + 9: a=2x (since (2x)²=4x²), b=3 (since 3²=9). Check middle term: 2ab = 2(2x)(3) = 12x ✓. Pattern: a²-2ab+b² = (a-b)². Factored: (2x-3)². Verify: (2x-3)² = 4x²-12x+9 ✓.

Question 3: What is the GCF of 12x³y², 18x²y³, and 24xy⁴?

6xy² (Correct answer)
6x²y²
12xy
6xy³

Correct answer: 6xy²

GCF of 12, 18, 24 is 6. Lowest powers: x¹, y². GCF = 6xy².

Coefficient GCF: GCF(12,18,24)=6. Variable x: powers 3,2,1 → lowest=1. Variable y: powers 2,3,4 → lowest=2. GCF = 6x¹y² = 6xy². Factoring multi-variable polynomials by GCF is a skill tested on IBEW aptitude exams.

Question 4: Factor: x² - x - 12

(x + 3)(x - 4) (Correct answer)
(x - 3)(x + 4)
(x + 6)(x - 2)
(x - 6)(x + 2)

Correct answer: (x + 3)(x - 4)

Find two numbers with product=-12, sum=-1: -4 and +3. Factor: (x+3)(x-4).

Product = -12, sum = -1. Try: -4 and +3: (-4)(3)=-12, -4+3=-1 ✓. Factor: (x-4)(x+3). Verify: x²+3x-4x-12 = x²-x-12 ✓. When product is negative, the two numbers have opposite signs. The larger absolute value gets the sign of the sum.

Question 5: Simplify: (x² - 9)/(x + 3)

x - 3 (Correct answer)
x + 3
x - 9
(x-3)(x+3)

Correct answer: x - 3

x²-9 = (x-3)(x+3). Divide by (x+3): x-3.

Numerator: x²-9 = (x-3)(x+3). Denominator: (x+3). Cancel (x+3): (x-3)(x+3)/(x+3) = x-3, provided x ≠ -3. Simplifying rational expressions by factoring is a key algebra technique that applies to circuit analysis formulas.

Question 6: Use the quadratic formula to solve: x² - 4x - 5 = 0

x = 5 and x = -1 (Correct answer)
x = -5 and x = 1
x = 4 and x = -1
x = 5 and x = 1

Correct answer: x = 5 and x = -1

a=1, b=-4, c=-5. Discriminant=16+20=36. x=(4±6)/2. x=5 or x=-1.

Quadratic formula: x = (-b ± √(b²-4ac))/2a. a=1, b=-4, c=-5. b²-4ac = 16+20 = 36. √36 = 6. x = (4±6)/2. x = (4+6)/2 = 5 or x = (4-6)/2 = -1. Check by factoring: (x-5)(x+1) = x²-4x-5 ✓. The quadratic formula works for all quadratics.

IBEW Aptitude Test Polynomial Factoring Methods Questions and Answers 3

Solve: x² + 3x - 10 = 0