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By the remainder theorem, dividing p(x) by (x - 2) leaves remainder p(2). Substituting gives 2(8) - 5(4) + 4(2) - 7 = 16 - 20 + 8 - 7 = -3.
By the factor theorem, (x - c) is a factor exactly when f(c) = 0. Here f(2) = 8 + 8 - 10 - 6 = 0, so x - 2 is a factor.
The degree of a product of polynomials is the sum of their degrees. The first factor has degree 2 and the second has degree 3, so the product has degree 2 + 3 = 5.
Synthetic division using x = 1 gives coefficients 2, -3, 4, -5, producing quotient coefficients 2, -1, 3 and remainder -2. Therefore the quotient is 2x^2 - x + 3.
The general term is C(5,r) 2^(5-r) (-3x)^r. For x^3, r = 3, giving C(5,3) × 2^2 × (-3)^3 = 10 × 4 × (-27) = -1080.
Write (5x + 1)/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2), so 5x + 1 = A(x + 2) + B(x - 1). Setting x = 1 gives A = 2, and setting x = -2 gives B = 3.
Since (x - 1) is a factor, the factor theorem gives f(1) = 0. Substituting x = 1 gives 1 + a - 3 + 2 = a, so a = 0.
Dividing the cubic by the known factor (x - 3) gives x^2 - x - 2. This quadratic factorises as (x - 2)(x + 1), so the other roots are 2 and -1.
The general term is C(6,r) x^(6-r) (2/x)^r = C(6,r) 2^r x^(6-2r). For the term independent of x, set 6 - 2r = 0, so r = 3. The term is C(6,3) × 2^3 = 20 × 8 = 160.
Using p(1) = 0 gives 1 - 2 + a + b - 6 = 0, so a + b = 7. Using p(-1) = 12 gives 1 + 2 + a - b - 6 = 12, so a - b = 15. Solving these simultaneously gives a = 11 and b = -4.
For a quadratic ax^2 + bx + c = 0, the discriminant b^2 - 4ac determines the nature of the roots. Here b^2 - 4ac = (-4)^2 - 4(2)(5) = 16 - 40 = -24, which is negative, so there are no real roots. Since the coefficients are real, the roots form a complex conjugate pair.
A linear factor x - k corresponds to a root x = k. Substituting x = 2 into x^3 - 8 gives 8 - 8 = 0, so x - 2 is a factor. Equivalently, the difference of cubes gives x^3 - 8 = (x - 2)(x^2 + 2x + 4).
Complete the square by taking half of the coefficient of x, which is 3, so a = 3. Then x^2 + 6x + 5 = (x + 3)^2 - 9 + 5 = (x + 3)^2 - 4, giving b = -4. Hence (a, b) = (3, -4).
Write p(x) = (x - 2)(x + 1)q(x) + ax + b. Substituting x = 2 gives 2a + b = p(2) = -5. Substituting x = -1 gives -a + b = p(-1) = -11. Solving these equations gives a = 2 and b = -9, so a + b = -7.
For 2x^2 - 7x + 3 = 0, the sum of roots is α + β = 7/2 and the product is αβ = 3/2. Then 1/α + 1/β = (α + β)/(αβ) = (7/2)/(3/2) = 7/3.
The factor (x - 2) is repeated twice, so it requires both A/(x - 2) and B/(x - 2)^2. The factor (x + 1) is not repeated, so it needs only the single term C/(x + 1). Therefore the correct form includes three partial fractions.
By the binomial theorem, the coefficient of x^2 in (1 + ax)^6 is C(6, 2)a^2 = 15a^2. Setting 15a^2 = 60 gives a^2 = 4. Since a > 0, we have a = 2.
Factorise x^3 - 4x = x(x - 2)(x + 2). The critical values are x = -2, 0 and 2. A sign chart shows the product is positive on -2 < x < 0 and on x > 2, so these are the solution intervals.
A repeated root at x = 2 means (x - 2)^2 is a factor, so p(x) = (x - 2)^2(x - r). Since p(1) = 0, we get (1 - 2)^2(1 - r) = 1 - r = 0, so r = 1. Thus p(x) = (x - 2)^2(x - 1), whose constant term is 4 × (-1) = -4.
The coefficient of x^5 in (1 - 2x)^7 is C(7, 5)(-2)^5 = -672, and the coefficient of x^4 is C(7, 4)(-2)^4 = 560. Multiplying by (1 + x) gives coefficient of x^5 equal to -672 + 560 = -112.
A quadratic has a repeated root exactly when its discriminant b^2 - 4ac is zero. Here a = 1, b = k and c = 9, so k^2 - 36 = 0, giving k^2 = 36 and hence k = ±6. Both values give a perfect square and therefore a double root.
The remainder theorem gives the remainder on division by 2x - 1 as p(1/2), because 2x - 1 = 0 when x = 1/2. Substituting gives 4(1/8) - 2(1/4) + 5(1/2) - 3 = 1/2 - 1/2 + 5/2 - 3 = -1/2. Thus the remainder is -1/2.
By Vieta’s formulas, α + β = 6 and αβ = 4. Use the identity α^2 + β^2 = (α + β)^2 - 2αβ, so the value is 6^2 - 2(4) = 36 - 8 = 28. This avoids solving for the individual roots.
Factor the quartic as (x^2 - 1)(x^2 - 4) ≤ 0. Let y = x^2; then (y - 1)(y - 4) ≤ 0, so 1 ≤ y ≤ 4. Hence 1 ≤ x^2 ≤ 4, which gives x ∈ [-2, -1] ∪ [1, 2].
The square root requires its argument to be non-negative, so 4 − x ≥ 0, giving x ≤ 4. Because the square root is in a denominator it cannot be zero, so 4 − x ≠ 0, that is x ≠ 4. Combining these two conditions leaves x < 4.
The inverse function undoes the operations of f, so f⁻¹(x) = (x + 7)/3. Substituting x = 5 gives (5 + 7)/3 = 4. Equivalently, solving 3x − 7 = 5 also yields x = 4.
Replacing x by x − 3 acts inside the function and shifts the graph 3 units in the positive x-direction, that is to the right. Adding 2 outside the function raises every y-coordinate by 2, moving the graph 2 units up.
Completing the square gives x² − 6x + 5 = (x − 3)² − 4. Since a square is never negative, (x − 3)² − 4 ≥ −4, and the value −4 is attained at x = 3. Hence the range is [−4, ∞).
The modulus equation splits into two linear cases: 2x − 5 = 7 gives x = 6, and 2x − 5 = −7 gives x = −1. Both values satisfy the original equation, so the solution set is {−1, 6}.
Write y = x² + 1 and solve for x: x² = y − 1, so x = √(y − 1) because the restriction x ≥ 0 selects the positive square root. Interchanging x and y gives f⁻¹(x) = √(x − 1), defined for x ≥ 1.
Factorising gives (x − 2)(x − 3) < 0, so the roots are 2 and 3. The parabola y = x² − 5x + 6 opens upwards, so it lies below the x-axis strictly between the roots. Hence the solution is 2 < x < 3.
The composite is h(x) = √(1/(x − 2)). For the square root to be real we need 1/(x − 2) ≥ 0, which requires x − 2 > 0 since the numerator is positive. Also x = 2 must be excluded, so the domain is x > 2.
A quadratic has no real roots exactly when its discriminant is negative. Here the discriminant is (2k)² − 4(k + 2) = 4k² − 4k − 8, so we need k² − k − 2 < 0. Factorising gives (k − 2)(k + 1) < 0, and the upward parabola is negative between its roots, so −1 < k < 2.
Write y = (2x + 1)/(x − 3) and rearrange: y(x − 3) = 2x + 1, so xy − 3y = 2x + 1. Collecting x terms gives x(y − 2) = 3y + 1, hence x = (3y + 1)/(y − 2). Interchanging x and y gives f⁻¹(x) = (3x + 1)/(x − 2), with x ≠ 2.
First evaluate the inner function: g(2) = 2² = 4. Then apply f to this result: f(4) = 2(4) − 3 = 5. Composition is worked from the inside outwards.
The expression under the square root must be non-negative, so 2x − 6 ≥ 0. Solving gives x ≥ 3, and the endpoint 3 is included because √0 is defined.
A function is even when f(−x) = f(x) for every x in its domain. For f(x) = x² + 1, substituting −x gives (−x)² + 1 = x² + 1, so the condition holds. The other listed functions do not satisfy this symmetry condition.
The inequality |2x + 1| ≤ 7 is equivalent to −7 ≤ 2x + 1 ≤ 7. Subtract 1 to get −8 ≤ 2x ≤ 6, then divide by 2 to obtain −4 ≤ x ≤ 3. Both endpoints are included because the modulus inequality is non-strict.
Replacing x by 2x affects the horizontal direction and compresses the graph towards the y-axis. A point at x = a on y = f(x) moves to x = a/2 on y = f(2x), which is a horizontal stretch of scale factor 1/2. Vertical transformations would instead modify the whole function value.
The radicand must satisfy x² − 9 ≥ 0, which factors as (x − 3)(x + 3) ≥ 0. This inequality holds outside the interval between the roots, including the roots themselves, so the domain is x ≤ −3 or x ≥ 3.
For x > 2, the denominator x − 2 is positive, so 1/(x − 2) is positive. As x approaches 2 from above the value becomes arbitrarily large, and as x increases without bound the value approaches 0 but never reaches it. Hence the range is (0, ∞).
Substitute g(x) into f: f(g(x)) = 2[x/(x − 1)] + 1. Writing 1 as (x − 1)/(x − 1) gives [2x + (x − 1)]/(x − 1) = (3x − 1)/(x − 1), with x ≠ 1.
The equation splits into x² − 4x = 2 or x² − 4x = −2. The first quadratic has discriminant 16 + 8 = 24 > 0, giving two distinct real roots; the second has discriminant 16 − 8 = 8 > 0, also giving two distinct real roots. No root is shared, so there are four real solutions in total.
Divide x² + 1 by x − 2: x² + 1 = (x − 2)(x + 2) + 5, so y = x + 2 + 5/(x − 2). As x tends to ±∞ the fractional term tends to 0, leaving the oblique asymptote y = x + 2.
By definition, log_2 32 is the exponent to which 2 must be raised to obtain 32. Since 2^5 = 32, the logarithm equals 5.
Use the quotient law log_a m - log_a n = log_a(m/n). Thus log_3 18 - log_3 2 = log_3(18/2) = log_3 9, and since 3^2 = 9, the value is 2.
Write both sides as powers of 2: 4^x = (2^2)^x = 2^(2x), and 8 = 2^3. Equating exponents gives 2x = 3, so x = 3/2.
Convert the logarithmic equation to exponential form: x + 3 = 2^4 = 16. Hence x = 13, which is valid because x + 3 = 16 is positive.
Take natural logarithms of both sides: (x + 1)ln3 = ln7. Therefore x + 1 = ln7/ln3, so x = ln7/ln3 - 1.
Combine the logarithms: log_10[x(x - 3)] = 1, so x(x - 3) = 10. Solving x^2 - 3x - 10 = 0 gives (x - 5)(x + 2) = 0, so x = 5 or x = -2; reject -2 because logarithms require positive arguments.