What is the maximum number of distinct real rational roots of x⁵ − 3x³ + 2 = 0? — 5.A degree-5 polynomial can have at most 5 real roots (by the Fundamental Theorem of Algebra). The Rational Root Theorem limits candidates, but the maximum count is 5.
When dividing p(x) by (2x − 4), the Remainder Theorem tells us to evaluate: — p(2).(2x − 4) = 2(x − 2), so the root is x = 2. The Remainder Theorem evaluates p at c = 2.
If p(x) has degree 6 and you perform synthetic division by (x − 2) three times in a row, what degree is the remaining polynomial? — 3.Each synthetic division reduces the degree by 1. Starting at degree 6 and dividing 3 times: 6 − 3 = 3.
The division algorithm states p(x) = d(x)q(x) + r(x). What must be true about degree of r(x)? — deg r < deg d.In polynomial division, the remainder r(x) must have degree strictly less than the degree of the divisor d(x). This is what terminates the algorithm.
How many candidates does the Rational Root Theorem give for 6x³ + x² − 5x + 2? — 12.Candidates = ±(factors of 2)/(factors of 6) = ±{1,2}/{1,2,3,6} = ±{1, 2, 1/2, 1/3, 1/6, 2/3} — 12 candidates.
What is the factorization of x⁴ − 16? — (x² + 4)(x + 2)(x − 2).x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2). Factor completely: x² − 4 continues to factor.
If (x + 2) is a factor of x³ + 5x² + 8x + 4, which is NOT also a factor? — (x + 4).Dividing by (x+2): x³+5x²+8x+4 = (x+2)(x²+3x+2) = (x+2)(x+1)(x+2) = (x+2)²(x+1). The factors are (x+2) and (x+1), NOT (x+4).
What pattern does x⁶ − 64 follow? — Both difference of squares and difference of cubes.x⁶ − 64 = (x³)² − 8² (difference of squares) AND (x²)³ − 4³ (difference of cubes). Both patterns apply.
What is special about the trinomial factor in sum/difference of cubes? — It is always irreducible over the reals.The trinomial a²∓ab+b² from sum/difference of cubes always has a negative discriminant (disc = −3b²/a²-related), making it irreducible over the reals.
What is the equation of the oblique asymptote of f(x) = (x² + x + 1)/(x − 1)? — y = x + 2.Degree of numerator (2) > degree of denominator (1) → oblique asymptote. Divide: x² + x + 1 ÷ (x−1) = x + 2 remainder 3. Oblique asymptote: y = x + 2.
Which value is excluded from the range of f(x) = (x + 1)/(x − 3)? — y = 1.The horizontal asymptote y = 1 (equal degrees, leading coefficients both 1) is never actually reached. So y = 1 is excluded from the range.
For the function f(x) = (x² + x − 2)/(x − 1), what is the y-value of the hole? — 3.Factor: (x+2)(x−1)/(x−1) → hole at x=1. The y-value at the hole: substitute x=1 into the simplified form x+2 = 1+2 = 3. The hole is at (1, 3).
What does the graph of f(x) = 1/(x−2)² look like near x = 2? — Goes to +∞ on both sides of x = 2.1/(x−2)² is always positive (square in denominator). As x → 2 from either side, the denominator → 0⁺, so f(x) → +∞ on both sides.
The graph of f(x) = 1/x has which type of symmetry? — Origin symmetry (odd function).f(−x) = 1/(−x) = −1/x = −f(x), so f is an odd function. This means it has rotational symmetry about the origin (180° rotation maps the graph to itself).
For the partial fraction decomposition 5/(x(x−5)) = A/x + B/(x−5), what is A? — −1.Multiply both sides by x: A = 5/(x−5) evaluated at x=0 = 5/(−5) = −1. So A = −1.
For 5/(x(x−5)) = A/x + B/(x−5), what is B? — 1.Multiply both sides by (x−5): B = 5/x at x=5 = 5/5 = 1. So B = 1.
If the denominator has a repeated linear factor (x−a)², what partial fractions do you use? — A/(x−a) + B/(x−a)².For a repeated linear factor (x−a)², you need two terms: A/(x−a) + B/(x−a)². This accounts for the higher multiplicity.
Partial fraction decomposition for (3x + 1)/[(x+1)(x²+1)] would require: — A/(x+1) + (Bx+C)/(x²+1).For distinct denominators with a linear factor (x+1) and an irreducible quadratic (x²+1), use A/(x+1) + (Bx+C)/(x²+1).
Expand partial fractions: 1/(x²−1) = A/(x−1) + B/(x+1). What are A and B? — A = 1/2, B = −1/2.1 = A(x+1) + B(x−1). Set x=1: 1=2A → A=1/2. Set x=−1: 1=−2B → B=−1/2.