In general, a binomial of the form x + f, where f is a constant, is a factor of a polynomial when the remainder of dividing the polynomial by x + f is 0. Let R be the remainder resulting from the division of the polynomial P(x) = ax³ + bx² + cx + d by x + 1. So the polynomial P(x) can be rewritten as P(x) = (x + 1)q(x) + R, where q(x) is a polynomial of second degree and R is a constant. Since –1 is a root of the equation P(x) = 0, it follows that P(–1) = 0.

Since P(–1) = 0 and P(–1) = R, it follows that R = 0. This means that x + 1 is a factor of P(x).Choices A, C, and D are incorrect because none of these choices can be a factor of the polynomial P(x) = ax³ + bx² + cx + d. For example, if x – 1 were a factor (choice A), then P(x) = (x –1)h(x), for some polynomial function h. It follows that P(1) = (1 – 1)h(1) = 0, so 1 would be another root of the given equation, and thus the given equation would have at least 4 roots. However, a third-degree equation cannot have more than three roots. Therefore, x – 1 cannot be a factor of P(x).

Marisa will hire x junior directors and y senior directors. Since she needs to hire at least 10 staff members, x + y ≥ 10. Each junior director will be paid $640 per week, and each senior director will be paid $880 per week. Marisa’s budget for paying the new staff is no more than $9,700 per week; in terms of x and y, this condition is 640x + 880y ≤ 9,700. Since Marisa must hire at least 3 junior directors and at least 1 senior director, it follows that x ≥ 3 and y ≥ 1. All four of these conditions are represented correctly in choice B.

Choices A and C are incorrect. For example, the first condition, 640x + 880y ≥ 9,700, in each of these options implies that Marisa can pay the new staff members more than her budget of $9,700. Choice D is incorrect because Marisa needs to hire at least 10 staff members, not at most 10 staff members, as the inequality x + y ≤ 10 implies.

Ken earned $8 per hour for the first 10 hours he worked, so he earned a total of $80 for the first 10 hours he worked. For the rest of the week, Ken was paid at the rate of $10 per hour. Let x be the number of hours he will work for the rest of the week. The total of Ken’s earnings, in dollars, for the week will be 10x + 80. He saves 90% of his earnings each week, so this week he will save 0.9(10x + 80) dollars. The inequality 0.9(10x + 80) ≥ 270 represents the condition that he will save at least $270 for the week. Factoring 10 out of the expression 10x + 80 gives 10(x + 8). The product of 10 and 0.9 is 9, so the inequality can be rewritten as 9(x + 8) ≥ 270. Dividing both sides of this inequality by 9 yields x + 8 ≥ 30, so x ≥ 22. Therefore, the least number of hours Ken must work the rest of the week to save at least $270 for the week is 22.

Choices A and B are incorrect because Ken can save $270 by working fewer hours than 38 or 33 for the rest of the week. Choice D is incorrect. If Ken worked 16 hours for the rest of the week, his total earnings for the week will be $80 + $160 = $240, which is less than $270. Since he saves only 90% of his earnings each week, he would save even less than $240 for the week.

Each of the options is a quadratic expression in vertex form. To rewrite the given expression in this form, the number 9 needs to be added to the first two terms, because x² + 6x + 9 is equivalent to (x + 3)². Rewriting the number 4 as 9 – 5 in the given expression yields x² + 6x + 9 – 5, which is equivalent to (x + 3)² – 5.

Choice A is incorrect. Squaring the binomial and simplifying the expression in option A gives x² + 6x + 9 + 5. Combining like terms gives x² + 6x + 14, not x² + 6x + 4. Choice C is incorrect. Squaring the binomial and simplifying the expression in choice C gives x² – 6x + 9 + 5. Combining like terms gives x² – 6x + 14, not x² + 6x + 4. Choice D is incorrect. Squaring the binomial and simplifying, the expression in choice D gives x² – 6x + 9 – 5. Combining like terms gives x² – 6x + 4, not x² + 6x + 4.

The value of the camera equipment depreciates from its original purchase value at a constant rate for 12 years. So if x is the amount, in dollars, by which the value of the equipment depreciates each year, the value of the camera equipment, in dollars, t years after it is purchased would be 32,400 – xt. Since the value of the camera equipment after 12 years is $0, it follows that 32,400 – 12x = 0.To solve for x, rewrite the equation as 32,400 = 12x. Dividing both sides of the equation by 12 gives x = 2,700. It follows that the value of the camera equipment depreciates by $2,700 each year. Therefore, the value of the equipment after 4 years, represented by the expression 32,400 – 2,700(4), is $21,600.

Choice A is incorrect. The value given in choice A is equivalent to $2,700 × 4. This is the amount, in dollars, by which the value of the camera equipment depreciates 4 years after it is purchased, not the dollar value of the camera equipment 4 years after it is purchased. Choice B is incorrect. The value given in choice B is equal to $2,700 × 6, which is the amount, in dollars, by which the value of the camera equipment depreciates 6 years after it is purchased, not the dollar value of the camera equipment 4 years after it is purchased.Choice D is incorrect. The value given in choice D is equal to $32,400 – $2,700. This is the dollar value of the camera equipment 1 year after it is purchased.

Substituting –1 for x in the equation that defines f gives f(−1) = \(\frac{(−1)^{2} −6(−1) + 3}{(−1)−1}\). Simplifying the expressions in the numerator and denominator yields \(\frac{1 + 6 + 3}{−2}\), which is equal to \(\frac{10}{-2}\)​ or –5.

Choices B, C, and D are incorrect and may result from misapplying the order of operations when substituting –1 for x.

Applying the distributive property, the original expression is equivalent to 5 + 12i – 9i² + 6i. Since i = √−1 ​, it follows that i² = −1. Substituting –1 for i² into the expression and simplifying yields 5 + 12i + 9 + 6i, which is equal to 14 + 18i.

Choices A, B, and C are incorrect and may result from substituting 1 for i² or errors made when rewriting the given expression.

Choice B is correct. The first equation can be rewritten as y – x = 3 and the second as ^x⁄₄ + y = 3, which implies that −x =^x⁄₄​, and so x = 0. The ordered pair (0, 3) satisfies the first equation and also the second, since 0 + 2(3) = 6 is a true equality.

Alternatively, the first equation can be rewritten as y = x + 3. Substituting x + 3 for y in the second equation gives ^x⁄₂ + 2(x + 3) = 6. This can be rewritten using the distributive property as ^x⁄₂ ​ + 2x + 6 = 6.It follows that 2x + ^x⁄₂ ​must be 0. Thus, x = 0. Substituting 0 for x in the equation y = x + 3 gives y = 3. Therefore, the ordered pair (0, 3) is the solution to the system of equations shown.

Choice A is incorrect; it satisfies the first equation but not the second. Choices C and D are incorrect because neither satisfies the first equation, x = y − 3.

Using the distributive property to multiply 3 and (x + 5) gives 3x + 15 − 6, which can be rewritten as 3x + 9.

Choice A is incorrect and may result from rewriting the given expression as 3(x + 5 − 6). Choice B is incorrect and may result from incorrectly rewriting the expression as (3x + 5) − 6. Choice D is incorrect and may result from incorrectly rewriting the expression as 3(5x) − 6.

Alternatively, evaluating the given expression and each answer choice for the same value of x, for example x = 0, will reveal which of the expressions is equivalent to the given expression

Maria spends x minutes running each day and y minutes biking each day. Therefore, x + y represents the total number of minutes Maria spent running and biking each day. Because x + y = 75, it follows that 75 is the total number of minutes that Maria spent running and biking each day.

Choices A and B are incorrect. The problem states that Maria spends time in both activities each day, therefore x and y must be positive. If 75 represents the number of minutes Maria spent running each day, then Maria spent no minutes biking each day. Similarly, if 75 represents the number of minutes Maria spent biking each day, then Maria spent no minutes running each day. The number of minutes Maria spends running each day and biking each day may vary; however, the total number of minutes she spends each day on these activities is constant and equal to 75. Choice D is incorrect. The number of minutes Maria spent biking for each minute spent running cannot be determined from the information provided.