Use the Remainder Theorem: what is the remainder when x⁴ − 5x² + 4 is divided by (x + 2)? — 0.p(−2) = 16 − 20 + 4 = 0. The remainder is 0, which means (x + 2) is a factor.
According to the Rational Root Theorem, which are candidate rational roots of 2x³ − 3x + 1? — ±1, ±2, ±1/2.Rational Root Theorem: candidates are ±(factors of constant)/(factors of leading coeff) = ±(1)/(1, 2) = ±1, ±1/2.
After synthetic division of p(x) by (x − c), the last number in the row is: — The value of p(c).In synthetic division, the last number in the bottom row is the remainder, which equals p(c) by the Remainder Theorem.
What is the remainder when p(x) = x³ − 3x² + 2x − 5 is divided by (x − 3)? — −1.p(3) = 27 − 27 + 6 − 5 = 1. Wait: 27 − 27 = 0, 0 + 6 = 6, 6 − 5 = 1. The remainder is 1.
How many steps does synthetic division of a degree-3 polynomial by a linear factor require? — 3.Synthetic division processes the coefficients one by one. For a degree-3 polynomial (4 coefficients), there are 3 multiply-and-add steps after bringing down the first.
If p(x) = x³ + ax² + bx + c has (x − 2) and (x + 1) as factors, what is p(2) + p(−1)? — 0.If (x − 2) and (x + 1) are factors, then p(2) = 0 and p(−1) = 0. Therefore p(2) + p(−1) = 0 + 0 = 0.
What is the result of dividing x³ − 1 by (x − 1)? — x² + x + 1.x³ − 1 = (x − 1)(x² + x + 1). After division, the quotient is x² + x + 1 with remainder 0.
What is a "depressed polynomial" after synthetic division? — A polynomial with degree reduced by 1.After synthetic division of a degree-n polynomial by a linear factor, the quotient has degree n−1. This quotient is called the "depressed polynomial."
The quotient when dividing a degree-5 polynomial by (x² + 1) has degree: — 3.When dividing a degree-5 polynomial by a degree-2 polynomial, the quotient has degree 5 − 2 = 3.
Which scenario confirms (x + 4) is NOT a factor of p(x)? — p(−4) ≠ 0.(x + 4) = (x − (−4)), so c = −4. If p(−4) ≠ 0, then by the Factor Theorem, (x + 4) is NOT a factor.
What is p(1) for p(x) = xⁿ − 1 for any positive integer n? — 0.p(1) = 1ⁿ − 1 = 1 − 1 = 0 for any n. This confirms (x − 1) is always a factor of xⁿ − 1.
What is the sum of cubes factorization formula for a³ + b³? — (a + b)(a² − ab + b²).a³ + b³ = (a + b)(a² − ab + b²). Note the middle term of the trinomial is NEGATIVE for the sum of cubes.
Which of the following is the correct factored form of x³ − 8x² + 16x? — x(x − 4)².Factor GCF x: x(x² − 8x + 16) = x(x − 4)². Check: (x−4)² = x²−8x+16. ✓
How many real roots does x³ + 1 = 0 have? — 1.x³ + 1 = (x+1)(x²−x+1) = 0. x = −1 is a real root. The quadratic x²−x+1 has discriminant 1−4=−3<0, giving 2 complex roots. Total real roots: 1.
Factor: 5x⁴ − 20x². — 5x²(x − 2)(x + 2).GCF = 5x². Then 5x²(x² − 4) = 5x²(x − 2)(x + 2). Factor completely using difference of squares.
If p(x) = (x − 1)²(x + 3), how many distinct roots does p(x) have? — 2.Distinct roots: x = 1 (from (x−1)², a double root) and x = −3. There are 2 distinct real roots.
The polynomial x² + 4 (over the reals): — Is irreducible over the reals.x² + 4 has no real factors because x² + 4 > 0 for all real x (no real roots). It is irreducible over ℝ. Over ℂ it factors as (x+2i)(x−2i).
After factoring by grouping 2x³ − x² + 6x − 3, what is the complete factorization? — (x² + 3)(2x − 1).Group: x²(2x−1) + 3(2x−1) = (x²+3)(2x−1).
For f(x) = (x² + 2x)/(x + 2), is there a hole? — Yes, there is a hole at x = −2.Factor numerator: x(x+2). The (x+2) cancels with the denominator → hole at x = −2.
What is the domain of f(x) = (x − 1)/(x² + x − 6)? — All reals except x = 2 and x = −3.x² + x − 6 = (x+3)(x−2). Domain excludes where denominator = 0: x = 2 and x = −3.
Which rational function has a hole at x = 1 and a vertical asymptote at x = −2? — (x−1)/[(x+2)(x−1)].(x−1)/[(x+2)(x−1)]: the (x−1) cancels → hole at x=1. Remaining denominator (x+2) → vertical asymptote at x=−2.
If deg(numerator) = deg(denominator) + 1, the graph of the rational function has: — An oblique (slant) asymptote.When the degree of the numerator is exactly 1 more than the denominator, polynomial long division yields a linear quotient, giving an oblique (slant) asymptote.
What is the y-intercept of f(x) = (2x + 6)/(x² − 9)? — y = −2/3.y-intercept: f(0) = (0+6)/(0−9) = 6/(−9) = −2/3.
f(x) = (x² − 1)/(x − 1) is equivalent to which function (with a restriction)? — f(x) = x + 1, x ≠ 1.(x²−1)/(x−1) = (x+1)(x−1)/(x−1) = x+1, but only when x ≠ 1 (since the original is undefined there).
How many vertical asymptotes does f(x) = (x² + 1)/(x² − 5x + 6) have? — 2.Factor denominator: (x−2)(x−3). Numerator x²+1 is never zero. Neither factor cancels. So there are 2 vertical asymptotes: x = 2 and x = 3.
A rational function f(x) has a horizontal asymptote at y = 0. What does this tell us about end behavior? — f(x) → 0 as x → ±∞.A horizontal asymptote y = 0 means that as x → +∞ or x → −∞, f(x) approaches 0 (the x-axis is an asymptote for the end behavior).
What is the behavior of f(x) = 1/x as x approaches 0 from the right (x → 0⁺)? — f(x) → +∞.As x → 0⁺ (small positive values), 1/x becomes very large positive. So f(x) → +∞.
For f(x) = (2x − 6)/(x² − 9), where is the hole? — x = 3.Factor: numerator = 2(x−3), denominator = (x+3)(x−3). The (x−3) cancels → hole at x = 3. Remaining asymptote: x = −3.
A rational function has no x-intercepts. What does this mean? — The numerator is never zero.x-intercepts occur where the numerator equals zero (and denominator is nonzero). If the numerator is never zero (e.g., x²+1), there are no x-intercepts.
Does f(x) = (x + 1)/(x² + 1) have any vertical asymptotes? — No, x² + 1 > 0 for all real x.x² + 1 > 0 for all real numbers (minimum value is 1 at x = 0). So the denominator is never zero, and there are no vertical asymptotes.
What is the domain restriction when simplifying (x² − 4)/(x − 2)? — x ≠ 2 only.The original expression is undefined when x − 2 = 0, so x ≠ 2. Even after simplifying to x + 2, the restriction x ≠ 2 remains.
Simplify: [(x²−1)/(x+2)] · [(x+2)/(x+1)] — x − 1.[(x+1)(x−1)/(x+2)] · [(x+2)/(x+1)]. Cancel (x+2) and (x+1): result = x − 1.
Add: (2/x) + (3/(x+1)). What is the result? — (2x+5)/(x(x+1)).LCD = x(x+1). 2(x+1)/(x(x+1)) + 3x/(x(x+1)) = (2x+2+3x)/(x(x+1)) = (5x+2)/(x(x+1)).
What is the excluded value in the expression (x+4)/[(x−2)(x+4)]? — x = 2 and x = −4.The expression is undefined when either (x−2) = 0 or (x+4) = 0, giving x = 2 or x = −4. Even though (x+4) cancels when simplifying, x = −4 is still excluded.
Subtract: (3/(x+1)) − (2/(x−1)). What is the numerator? — x − 5.LCD = (x+1)(x−1). [3(x−1) − 2(x+1)] / [(x+1)(x−1)] = [3x−3−2x−2]/... = (x−5)/[(x+1)(x−1)].
The form A/(x−a) + B/(x−b) is the partial fraction setup for which type of denominator? — Distinct linear factors.When the denominator has two distinct linear factors (x−a)(x−b), the partial fraction decomposition uses A/(x−a) + B/(x−b).
Solve: (x+1)/(x−3) = 2. — x = 7.Cross multiply (or multiply both sides by (x−3)): x+1 = 2(x−3) = 2x−6. Then 1+6 = 2x−x → x = 7. Check: 8/4 = 2 ✓.
What is an "extraneous solution" in rational equations? — A solution that makes the denominator zero.An extraneous solution is a value that satisfies the transformed equation but makes the original expression undefined (denominator = 0). Always check solutions in the original equation.