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When adding polynomials, combine like terms by adding coefficients of matching powers. The x^2 terms give 3 + 4 = 7x^2, the x terms give -5 + 7 = 2x, and the constants give 2 - 6 = -4.
Apply the distributive property (FOIL): multiply 2x by x and 5, then -3 by x and 5. This yields 2x^2 + 10x - 3x - 15, and combining the middle terms gives 2x^2 + 7x - 15.
The expression is a difference of squares, since x^2 - 9 = x^2 - 3^2. The difference-of-squares pattern a^2 - b^2 = (a - b)(a + b) gives (x - 3)(x + 3).
When multiplying powers with the same base, add the exponents; when dividing, subtract them. So 2^3 · 2^4 = 2^7, and 2^7 ÷ 2^5 = 2^2 = 4.
The absolute value of a number is its distance from zero, so |3 - 8| = |-5| = 5 and |2 - 9| = |-7| = 7. Subtracting gives 5 - 7 = -2.
Factor 50 as 25 · 2 so that a perfect square can be extracted. Since √(25 · 2) = √25 · √2 = 5√2, the simplified form is 5√2.
Substitute x = -2 into the function: 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1. This simplifies to 8 + 6 + 1 = 15.
Subtract 3x from both sides to get 2x - 7 = 9, then add 7 to both sides to get 2x = 16. Dividing by 2 gives x = 8.
Factor both numerator and denominator: x^2 - 4 = (x - 2)(x + 2) and x^2 + x - 6 = (x + 3)(x - 2). Cancel the common factor (x - 2) to obtain (x + 2)/(x + 3).
The quadratic factors as (x - 1)(x - 5) = 0 because the product of the roots is 5 and their sum is 6. Setting each factor to zero gives x = 1 and x = 5.
For a quadratic ax^2 + bx + c = 0, the sum of the roots equals -b/a. Here a = 2 and b = -8, so the sum is -(-8)/2 = 4.
Multiply the coefficients: 2 · (-3) = -6. Then add exponents of like bases: x^3 · x = x^4 and y^2 · y^4 = y^6, giving -6x^4 y^6.
By the Remainder Theorem, the remainder of dividing p(x) by x - 2 equals p(2). Evaluating gives 8 - 16 + 4 - 5 = -9.
Multiply numerator and denominator by xy to clear the fractions: the numerator becomes y + x and the denominator becomes y - x. Thus the simplified expression is (x + y)/(y - x).
Subtract 5 from both sides to get -2x > 6. Dividing both sides by -2 reverses the inequality symbol, giving x < -3.
The composition (f ∘ g)(2) means f(g(2)). First g(2) = 4 - 1 = 3, then f(3) = 9 + 2 = 11.
The numerator factors as (x - 2)(x - 3). Canceling the common factor (x - 2) leaves x - 3, valid wherever x ≠ 2.
Multiply numerator and denominator by the conjugate √5 + 1. The denominator becomes 5 - 1 = 4, so the expression is 6(√5 + 1)/4 = (3/2)(√5 + 1).
Rewrite both sides with base 2: 4^(x-1) = 2^(2x-2) and 8^(x+1) = 2^(3x+3). Equating exponents gives 2x - 2 = 3x + 3, so x = -5.
The discriminant is b^2 - 4ac = 16 - 24 = -8. Because the discriminant is negative, the quadratic has no real roots and instead has two nonreal complex conjugate roots.
By the Factor Theorem, dividing p(x) by (x + 3) gives x^2 - x - 2, which factors as (x - 2)(x + 1). Thus the remaining factors are (x - 2) and (x + 1).
Factor x^2 - 4 as (x - 2)(x + 2) and use the common denominator (x - 2)(x + 2). Rewriting the first term gives x(x + 2) = x^2 + 2x, and subtracting 4 yields (x^2 + 2x - 4)/((x - 2)(x + 2)).
Since x = 1 is a root, divide by (x - 1) to get x^2 - x - 6, which factors as (x - 3)(x + 2). The complete factorization is therefore (x - 1)(x - 3)(x + 2).
A quadratic has exactly one real solution when its discriminant b^2 - 4ac equals zero. Here k^2 - 36 = 0, so k = 6 or k = -6.
Factor to get (x - 3)(x + 3)/((x + 5)(x - 3)) and multiply by the reciprocal (x + 5)/(x - 3). Canceling (x - 3) and (x + 5) leaves (x + 3)/(x - 3).
Factor by grouping after splitting the middle term: the product ac = (6)(-10) = -60 must split into 15 and -4, giving 6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (2x + 5)(3x - 2). Checking the expansion returns 6x^2 + 11x - 10, confirming the factorization.
Apply the power rule to each factor: 8^(2/3) = (8^(1/3))^2 = 2^2 = 4, and (x^6)^(2/3) = x^(6 · 2/3) = x^4. Multiplying the simplified factors gives 4x^4, and this holds for negative x as well because the exponent 4 is even.
Simplify each radical by extracting perfect squares: 5√12 = 5 · 2√3 = 10√3, 2√27 = 2 · 3√3 = 6√3, and √75 = 5√3. Combining like radicals gives 10√3 + 6√3 - 5√3 = 11√3, so k = 11.
The Remainder Theorem states that dividing p(x) by (x - c) leaves remainder p(c). Here the divisor is x + 4 = x - (-4), so c = -4 and the remainder equals p(-4), which is given as 7.
Factor out the common factor x to get x(x^2 - 4) = 0, then factor the difference of squares as x(x - 2)(x + 2) = 0. The Zero Product Property gives x = 0, x = 2, and x = -2, and a cubic has at most three real roots, so this list is complete.
Synthetic division with c = 1 on coefficients 2, -3, 4, -5 gives 2, -1, 3, and remainder -2. The quotient is therefore 2x^2 - x + 3 with remainder -2, written as 2x^2 - x + 3 - 2/(x - 1). Checking with the Remainder Theorem, p(1) = 2 - 3 + 4 - 5 = -2, which matches.
Complete the square: x^2 - 6x + 5 = (x - 3)^2 - 4. Since (x - 3)^2 ≥ 0, the minimum value is -4, attained at x = 3, and the parabola opens upward, so the range is y ≥ -4.
Distribute the cube root across the factors: 27^(1/3) = 3, (x^9)^(1/3) = x^3, and (y^-3)^(1/3) = y^-1 = 1/y. Combining gives 3x^3/y, and the negative exponent moves the y-factor to the denominator.
Composition requires substituting g(x) into f: f(g(x)) = (x - 3)^2 + 2(x - 3). Expanding gives x^2 - 6x + 9 + 2x - 6 = x^2 - 4x + 3, which is the composed polynomial.
Write y = (2x - 5)/3, then solve for x: multiply both sides by 3 to get 3y = 2x - 5, add 5 to get 3y + 5 = 2x, and divide by 2 to get x = (3y + 5)/2. Replacing y with x gives f^-1(x) = (3x + 5)/2.
A system of two linear equations has no solution when the lines are parallel, meaning the coefficients are proportional but the constants are not. Setting the ratio of x-coefficients 4/2 = 2 equal to the ratio of y-coefficients k/3 gives k = 6; then 20/12 ≠ 2, so the lines are distinct and parallel.
Factor each piece: x^2 + 7x + 12 = (x + 3)(x + 4) and x^2 - 9 = (x - 3)(x + 3). After multiplying, cancel the common factors (x + 4) and (x + 3), leaving (x + 3)/(x - 3).
Multiply numerator and denominator by the conjugate 3 + √2. The denominator becomes 9 - 2 = 7, and the numerator becomes (3 + √2)^2 = 9 + 6√2 + 2 = 11 + 6√2. Thus the expression equals 11/7 + (6/7)√2, so a + b = 11/7 + 6/7 = 17/7, which simplifies the check; recomputing a + b as 11/7 + 6/7 gives 17/7, so the listed value 13/7 matches only if the numerator is 7 + 6√2, indicating a transcription issue.
Square both sides to get 2x + 7 = x^2 + 4x + 4, which rearranges to x^2 + 2x - 3 = 0 and factors as (x + 3)(x - 1) = 0, giving x = -3 or x = 1. Checking in the original equation, x = -3 makes the right side -1 while the left side is 1, so it is extraneous; only x = 1 works, since both sides equal 3.
Rewrite 32 as 2^5 so that 2^(x + 1) = 2^5, giving x + 1 = 5 and x = 4. Then 4^x = 4^4 = 256, which follows from (2^2)^4 = 2^8.
Factor as (x - 3)(x + 2) < 0; the critical values are 3 and -2. The product of the two factors is negative between the roots because one factor is positive and the other negative there, so the solution is -2 < x < 3.
Absolute value equals the expression itself when it is nonnegative and its negation when it is negative. If x < 4, then x - 4 < 0, so |x - 4| = -(x - 4) = 4 - x.
Factor the difference of cubes: x^3 - 8 = (x - 2)(x^2 + 2x + 4). Canceling (x - 2) leaves x^2 + 2x + 4 for x ≠ 2, and evaluating at x = 2 gives 4 + 4 + 4 = 12, the value that removes the discontinuity.
Treat x^4 - 16 as a difference of squares: (x^2 - 4)(x^2 + 4). The first factor further splits into (x - 2)(x + 2), but x^2 + 4 is irreducible over the integers and is therefore a factor.
Use the identity x^2 + y^2 = (x + y)^2 - 2xy. Substituting gives 7^2 - 2(10) = 49 - 20 = 29, which matches the values x = 5 and y = 2 that satisfy both given equations.
A rational expression is undefined wherever its denominator equals zero. Factoring x^2 - 5x + 6 gives (x - 2)(x - 3), which is zero at x = 2 and x = 3; the numerator's zero at x = 1 does not make the expression undefined.
By the binomial theorem, the term containing x^2 is C(4,2)(2x)^2(-3)^2 = 6 · 4x^2 · 9 = 216x^2, so the coefficient is 216. Both the x-factor and the constant are squared, so the sign is positive.
Raise each factor to the sixth power by multiplying exponents: (a^(1/2))^6 = a^3 and (b^(2/3))^6 = b^4. The product is a^3 b^4.
Complete the square: f(x) = (x - 2)^2 - 3, so the vertex is at x = 2, which lies inside the interval. Since the parabola opens upward, the minimum is f(2) = -3; the endpoint values f(0) = 1 and f(5) = 6 are larger.
End behavior is governed by the leading term -2x^4. Because the degree 4 is even, both ends go the same direction, and because the leading coefficient -2 is negative, both ends fall toward -∞.