SAT Math Full Practice Test, Beginner Level

The answer is (A), $10$ card kits.

1. Let $k$ be the number of completed kits, a nonnegative integer. The whole roll is accounted for by $12$ storage labels, $6k$ kit labels and $24$ unused labels, so $96 = 12 + 6k + 24$.
2. The labels available for kits total $96 - 12 - 24 = 60$. The $24$ unused labels are part of the original roll but are not used on kits.
3. Each kit uses $6$ labels, so $k = \frac{60}{6} = 10$.
4. Check the original inventory with $12 + 6(10) + 24 = 96$. The count is a nonnegative whole number, so choice (A) satisfies every condition.

Choice (B), $12$, comes from $\frac{96 - 24}{6}$, which gives the $12$ storage box labels to kits as well. Choice (C), $14$, comes from $\frac{96 - 12}{6}$, which uses the $24$ labels that are stated to remain on the roll. Choice (D), $18$, comes from $\frac{96 - 12 + 24}{6}$. Adding the remaining labels counts them as extra inventory.

The answer is $240$.

1. The ratio divides all the tiles into $3$ equal-sized green groups and $5$ equal-sized cream groups, for $3 + 5 = 8$ equal groups in the whole kit.
2. Each group contains $\frac{640}{8} = 80$ tiles.
3. There are $3$ green groups, so the green count is $3(80) = 240$.
4. The cream count is $5(80) = 400$. The counts total $640$, and $240 : 400$ reduces to $3 : 5$, which checks both given conditions.

The answer is (C), $4x + 4$ square centimeters.

1. The remaining area equals the area of the outer square minus the area of the removed square, $(x + 2)^2 - x^2$.
2. The outer area expands to $x^2 + 4x + 4$. Subtracting $x^2$ leaves $4x + 4$.
3. A geometric check splits the remaining L shape into two rectangles that are $x$ by $2$ and a square that is $2$ by $2$. Their areas total $2x + 2x + 4 = 4x + 4$ with no overlap.
4. The expression holds for every permitted $x$, so choice (C) is correct. At $x = 3$, the outer and removed areas are $25$ and $9$, and the remaining area $16$ agrees with $4(3) + 4$.

Choice (A), $4x + 2$, counts the two strips correctly but treats the $2$ by $2$ corner square as having area $2$ instead of $4$. Choice (B), $2x + 4$, leaves out one of the two $x$ by $2$ strips. Choice (D), $x^2 + 4$, keeps the removed $x$ by $x$ square and leaves out the two strips. It happens to agree at $x = 4$, which does not make it correct for every positive $x$.

The answer is (D), $48$ degrees.

1. Angles $AOC$ and $BOD$ are vertical angles, because each pair of their sides consists of opposite rays. Vertical angles have equal measures, so $3x + 6 = 5x - 22$.
2. Subtract $3x$ from both sides to get $6 = 2x - 22$. Add $22$ to both sides to get $28 = 2x$. Thus $x = 14$.
3. The requested quantity is angle $AOC$, whose measure is $3x + 6$ degrees. Substituting $x = 14$ gives $3(14) + 6 = 48$ degrees.
4. The other given angle is $5(14) - 22 = 48$ degrees, which confirms the equal angle condition. An adjacent angle would measure $180 - 48 = 132$ degrees, but it is not the requested angle.

Choice (A), $14$, is the value of $x$, not the measure $3x + 6$ of the requested angle. Choice (B), $42$, is $3(14)$ and leaves out the $+6$ in the given angle expression. Choice (C), $132$, is the supplement of the requested $48$ degree angle and belongs to an adjacent angle.

The answer is $9$.

1. The $4$ multiplies the whole quantity inside the parentheses. Undo the outside addition first by subtracting $5$, which leaves four equal copies of that quantity, $4(x - 3) = 24$.
2. Divide both sides by $4$ to find one copy, $x - 3 = 6$. Adding $3$ to both sides gives $x = 9$.
3. Distributing the $4$ also works, but it must multiply both terms, so $4(x - 3) + 5 = 4x - 12 + 5 = 4x - 7$. Multiplying only the $x$ would change the equation.
4. Check $9$ in the original grouped expression. Since $9 - 3 = 6$, the left side is $4(6) + 5 = 29$, which matches the right side.

The answer is (B), $t^7$.

1. An exponent counts repeated factors of the base. The first expression has three factors of $t$ and the second has four, and multiplying the expressions joins the two groups, $(t \cdot t \cdot t)(t \cdot t \cdot t \cdot t)$.
2. There are seven factors altogether, so $t^3 \cdot t^4 = t^{3+4} = t^7$. This is why exponents add when powers of the same base are multiplied.
3. Multiplying $3$ by $4$ describes a different operation, raising $t^3$ to the fourth power, since $(t^3)^4 = t^{12}$. Joining the groups also does not create a coefficient of $2$.
4. As a numerical check, at $t = 2$ the product is $8 \cdot 16 = 128$, and $2^7 = 128$ as well. The factor count shows that the identity holds for every real $t$, including zero.

Choice (A) multiplies the exponents as if the expression were $(t^3)^4$. Choice (C) adds a coefficient of $2$ that nothing supports. Choice (D) subtracts the exponents as if dividing powers.

The answer is (D), the increase in volume for each additional minute.

1. The variable $t$ counts minutes. Multiplying it by $3$ gives the water added during those minutes, so the coefficient has units of liters per minute.
2. At $0$ minutes there are $14$ liters, after $1$ minute there are $17$ and after $2$ minutes there are $20$, since $V(0) = 14$, $V(1) = 17$ and $V(2) = 20$. Each one-minute step adds $3$ liters.
3. The graph starts at $14$ because that is the initial amount. Its slope is $\frac{3}{1} = 3$ liters per minute, because it rises $3$ liters for each one-minute increase in time.
4. Choice (D) states that rate. After $3$ minutes the volume is $V(3) = 23$ liters, and adding $1$ liter takes $\frac{1}{3}$ minute. Those are different quantities from the increase of $3$ liters each minute.

Choice (A) confuses the coefficient $3$ with the initial value $14$. Choice (B) confuses a rate with a total volume, which would be $V(3) = 23$. Choice (C) reverses the rate, since the time per liter is $\frac{1}{3}$ minute.

The answer is (C), $92$ members.

1. The percentage applies to the original $80$ members. Fifteen percent means $15$ out of each $100$, so the added membership is $\frac{15}{100} \cdot 80$.
2. Ten percent of $80$ is $8$, and five percent is $4$. Together they make $0.15 \cdot 80 = 12$ additional members. The original $80$ still has to be included.
3. Adding gives $80 + 12 = 92$ members, so choice (C) is correct. Adding $15$ directly would treat a percentage as a fixed count of members.
4. Check the change in the other direction, $\frac{92 - 80}{80} = \frac{12}{80} = 0.15$, which is $15$ percent.

Choice (A) reports the increase rather than the final membership. Choice (B) adds $15$ members rather than $15$ percent of $80$. Choice (D) subtracts the change of $12$ members.

The answer is $11$.

1. The area of a square is its side length multiplied by itself, which gives the equation $s^2 = 121$ with $s > 0$.
2. Since $11^2 = 121$ and also $(-11)^2 = 121$, the equation by itself has two real solutions.
3. The side length condition selects the positive value, $s = 11$. The negative value solves the equation but does not fit the stated meaning of $s$.
4. Checking the dimensions gives $11 \text{ cm} \times 11 \text{ cm} = 121 \text{ cm}^2$, so the response is $11$.

The answer is (B), $67$ degrees.

1. The three interior angles of a triangle add to $180$ degrees. Writing $R$ for the measure of angle $R$ in degrees gives $46 + 67 + R = 180$.
2. Since $46 + 67 = 113$, the remaining angle is $R = 180 - 113 = 67$ degrees.
3. Choice (B) is correct. The $113$ in choice (A) is the sum of the two given angles, so it does not yet answer the question about the remaining angle.
4. As a final check, $46 + 67 + 67 = 180$, so the completed triangle satisfies the angle sum.

Choice (A) reports the sum of the given angles. Choice (C) copies angle $P$ without using the sum. Choice (D) reports the total of all three angles instead of one angle.

The answer is (A), $k \le 8$.

1. At most $41$ means that spending exactly $41$ dollars is allowed. The entry fee and the kit costs are both part of that total, so the inequality uses a less than or equal to sign, $9 + 4k \le 41$.
2. Subtracting $9$ leaves $4k \le 32$ for the kits. Dividing by the positive cost of $4$ dollars per kit gives $k \le 8$.
3. The boundary is included, so choice (A) describes the allowed counts. The permitted values are the integers from $0$ through $8$, including $8$ itself.
4. Eight kits cost $9 + 4(8) = 41$ dollars, exactly the limit. Nine kits would cost $9 + 4(9) = 45$ dollars, which is more than $41$. A strict inequality would wrongly exclude a permitted purchase.

Choice (B) excludes the allowed boundary of $8$ kits. Choice (C) reverses the permitted direction. Choice (D) leaves out most of the fixed fee and so allows counts that cost more than $41$ dollars.

The answer is $9$.

1. An $x$-intercept lies on the $x$-axis, so its $y$-coordinate is $0$. Set $(x + 4)(x - 5) = 0$.
2. A product of two real numbers is zero only if at least one factor is zero. Thus $x + 4 = 0$ or $x - 5 = 0$, which gives $x = -4$ or $x = 5$.
3. The intercepts are $(-4, 0)$ and $(5, 0)$. Their distance along the axis is $5 - (-4) = 9$ coordinate units.
4. Equivalently, the left intercept is $4$ units from zero and the right intercept is $5$ units from zero, which gives $4 + 5 = 9$. Both roots give $y = 0$ in the original function.

The answer is (D), $5$.

1. Both equations contain the same $2x$. Subtracting the entire second equation from the first removes that term and leaves an equation in $y$ alone, $(2x + 3y) - (2x - y) = 31 - 11$.
2. The $y$ terms become $3y - (-y) = 4y$, and on the right $31 - 11 = 20$. So $4y = 20$, and dividing by $4$ gives $y = 5$.
3. Subtracting only $y$ would miss the negative sign in the second equation. A whole equation is subtracted, so every term takes part with its sign, $-(2x - y) = -2x + y$.
4. To check, put $y = 5$ into the second equation. Then $2x - 5 = 11$, so $x = 8$, and the first equation gives $2(8) + 3(5) = 16 + 15 = 31$. Both conditions hold, and choice (D) reports the requested $y$.

Choice (A) reports $x$ instead of $y$. Choice (B) reports $4y$ before dividing. Choice (C) uses $3y - y$ rather than $3y - (-y)$.

The answer is (C), $\frac{9}{20}$.

1. The condition says the selected volunteer is already in the evening group. That group is the reference group for the probability, so first find its total, $18 + 22 = 40$.
2. Of the $40$ evening volunteers, $18$ are trained. The probability is therefore $P(\text{trained} \mid \text{evening}) = \frac{18}{40}$.
3. Dividing numerator and denominator by $2$ gives $\frac{9}{20}$, which is choice (C). Using all $80$ volunteers in the denominator would answer a different question about the whole group.
4. Check with the untrained part of the same evening group. Since $\frac{22}{40} = \frac{11}{20}$ and $\frac{18}{40} + \frac{22}{40} = 1$, the two categories account for every evening volunteer.

Choice (A) uses the whole group of $80$ as the denominator. Choice (B) divides the trained evening volunteers by all $44$ trained volunteers, which reverses the condition. Choice (D) gives the untrained fraction within the evening group.

The answer is (C), $12$.

1. The input $-2$ replaces every $t$ in the rule. Keeping parentheses around it preserves its sign in both the square and the product, $f(-2) = (-2)^2 - 3(-2) + 2$.
2. The square $(-2)^2$ means $-2$ times $-2$, which is $4$. The next term is $-3(-2) = 6$.
3. Adding gives $f(-2) = 4 + 6 + 2 = 12$, so choice (C) is correct. A minus sign outside the square would give a different expression, but here the sign belongs to the input being squared.
4. An independent check comes from factoring the rule as $(t - 1)(t - 2)$. At $t = -2$ the factors are $-3$ and $-4$, and $(-3)(-4) = 12$.

Choice (A) evaluates $4 - 6 + 2$ after losing the sign of the input in $-3t$. Choice (B) reports only the squared term. Choice (D) treats both the square and the product as negative, $-4 - 6 + 2$.

The answer is $5$.

1. Because the function is linear, equal steps in the input give equal changes in the output. From input $2$ to input $5$ the input rises by $3$ and the output rises by $9$, so the rate is $\frac{20 - 11}{5 - 2} = \frac{9}{3} = 3$.
2. The next pair confirms the rate, since $\frac{29 - 20}{8 - 5} = \frac{9}{3} = 3$.
3. Going back from input $2$ to input $0$ is a move of two input units, so the output decreases by $6$, which gives $f(0) = 11 - 3(2) = 5$.
4. The first listed output is not the starting value, because its input is $2$. The rule $f(x) = 3x + 5$ gives $11$, $20$ and $29$ at the three inputs, so the response is $5$.

The answer is (B), $15$ cm.

1. The right angle is at $N$, and the side opposite it is $MO$. So $MO$ is the hypotenuse and stands alone on one side of the Pythagorean equation, $MO^2 = MN^2 + NO^2$.
2. Substituting the two legs gives $MO^2 = 9^2 + 12^2 = 81 + 144 = 225$.
3. A length is positive, so take the positive square root, $MO = \sqrt{225} = 15$ centimeters. That is choice (B).
4. Adding the two legs would give a path along two sides, not the straight hypotenuse. The result is longer than either leg, and $15^2 = 225 = 9^2 + 12^2$ checks the required relationship.

Choice (A) adds the two legs. Choice (C) subtracts the leg lengths. Choice (D) reports the squared length as a length.

The answer is (A), the line through $(0, 6)$ and $(3, 0)$.

1. Two points determine a straight line, so generate two points from the equation and compare them with the choices. Setting $x = 0$ gives $y = 6$, the point $(0, 6)$.
2. For a second point, set $y = 0$. Then $0 = -2x + 6$, so $2x = 6$ and $x = 3$, the point $(3, 0)$.
3. Choice (A) contains both points, and its slope is $\frac{0 - 6}{3 - 0} = -2$. Choice (C) starts at the same $y$-intercept but rises as $x$ increases, while the negative coefficient in the equation requires a falling line.
4. Substituting each point checks the line directly, since $6 = -2(0) + 6$ and $0 = -2(3) + 6$. The matching intercept and slope confirm choice (A).

Choice (B) has $y$-intercept $-6$ and a positive slope. Choice (C) has the correct intercept but slope $2$. Choice (D) has $y$-intercept $3$ and slope $-\frac{1}{2}$.

The answer is (D), $8\%$.

1. Increasing $n$ by $1$ adds one more factor of $1.08$. Each week’s quantity is therefore $1.08$ times the quantity of the week before, $Q(n + 1) = 1.08\,Q(n)$.
2. The multiplier splits as $1.08 = 1 + 0.08$. The $1$ keeps the previous amount, and $0.08 = \frac{8}{100}$ is the added fraction.
3. The weekly increase is $(1.08 - 1) \times 100\% = 8\%$, so choice (D) is correct. The value $108\%$ describes the whole new amount as a percentage of the old amount, which includes the original $100\%$.
4. At week $0$ the value is $50$, and at week $1$ it is $54$. The increase of $4$ divided by the original $50$ is $\frac{4}{50} = 0.08$, which checks the weekly rate.

Choice (A) uses the initial quantity as a percentage. Choice (B) reads the multiplier itself as a percentage increase. Choice (C) reports the whole new amount relative to the previous amount.

The answer is (B), $12$.

1. The question asks for a prediction from the model. The plotted point $(4, 11)$ is an observation, while the line gives the predicted value of the model at that input.
2. Substituting $4$ into the rule gives $2(4) + 4 = 8 + 4 = 12$.
3. Choice (B) is correct. The value $11$ belongs to the actual data point at that input. A fitted model can differ from an individual observation, so the two values play different roles.
4. The prediction is consistent with the rate of the line. At $x = 3$ the model gives $10$, and one more unit of input adds $2$, which gives $12$ at $x = 4$.

Choice (A) reports the observation at $x = 4$ instead of the model prediction. Choice (C) uses only $2x$ and drops the intercept. Choice (D) copies the last observed $y$-value, the one at $x = 5$.

The answer is $7$.

1. The formula combines a fixed $8$ dollars with $5$ dollars for each hour. To find the hours from a total cost, first remove the fixed charge, $C - 8 = 5h$.
2. Dividing the remaining charge by $5$ gives the hours, $h = \frac{C - 8}{5}$. The whole quantity $C - 8$ belongs in the numerator, because that is the amount charged at the hourly rate.
3. With $C = 43$ this gives $h = \frac{43 - 8}{5} = \frac{35}{5} = 7$ hours. Dividing the total first and then subtracting $8$ would mix dollar and hour quantities.
4. Substituting $7$ hours into the original rule gives $8 + 5(7) = 43$ dollars. The duration is nonnegative and the original cost is recovered, so the response is $7$.

The answer is (D), $x$, with $x \ne -7$.

1. The original denominator cannot equal zero. Since $x + 7 = 0$ at $x = -7$, that input is excluded before anything is simplified.
2. Both terms of the numerator contain $x$, so it factors as $x(x + 7)$. This creates a common factor with the denominator, $\frac{x^2 + 7x}{x + 7} = \frac{x(x + 7)}{x + 7}$.
3. For the allowed inputs, $x + 7$ is not zero, so the common factor cancels and leaves $x$. The restriction stays attached to the result, which gives choice (D).
4. At $x = 2$ the original fraction is $\frac{4 + 14}{9} = 2$. At $x = -7$ the original expression is undefined, even though the simplified expression $x$ alone could be evaluated there. That is why choice (A) wrongly enlarges the domain.

Choice (A) drops the excluded value of the original expression. Choice (B) does not keep the numerator after factoring and canceling. Choice (C) cancels terms that do not match and gives the wrong excluded value.

The answer is (B), $6$.

1. Equal charges mean the two complete cost expressions have the same value at the same duration, $18 + 3h = 6 + 5h$. The fixed fees matter as well as the hourly rates.
2. Subtract $6$ from both sides, then subtract $3h$. This leaves $12 = 2h$, so the matching duration is $h = 6$ hours.
3. Another view gives the same result. Service A starts $18 - 6 = 12$ dollars higher but rises $5 - 3 = 2$ dollars less per hour, so it takes $\frac{12}{2} = 6$ hours to make up the difference.
4. At $6$ hours, service A charges $18 + 3(6) = 36$ dollars and service B charges $6 + 5(6) = 36$ dollars. Choice (B) gives the duration. The $12$ was the difference in fixed fees, not a number of hours.

Choice (A) divides the gap in fixed fees by $4$ rather than by the actual difference in rates, $2$. Choice (C) reports the initial dollar difference as a duration. Choice (D) reports the fixed fee of service A rather than the matching duration.

The answer is $24$.

1. For the real square root to exist, the quantity inside must be nonnegative, $2x + 1 \ge 0$, so $x \ge -\frac{1}{2}$. The right side is the nonnegative number $7$, so squaring both sides is a useful step.
2. Squaring gives $2x + 1 = 49$. Subtracting $1$ leaves $2x = 48$, and dividing by $2$ gives $x = 24$.
3. A value found after squaring should be checked in the original equation. At $x = 24$ the inside quantity is $49$, and $\sqrt{2(24) + 1} = \sqrt{49} = 7$.
4. The candidate meets the domain condition and the original equation. The squared equation has only this one candidate, so the response is $24$. The square root symbol always means the nonnegative root.

The answer is (C), $18$.

1. The stated correspondence tells which sides match, even if the triangles are drawn in different orientations. Side $AB$ matches $DE$, side $BC$ matches $EF$ and side $AC$ matches $DF$.
2. The side that measured $4$ has become $12$, so the scale factor is $k = \frac{DE}{AB} = \frac{12}{4} = 3$. Similarity applies that same positive factor to every pair of corresponding sides.
3. Side $BC$ measures $6$, so its corresponding side is $EF = 3(6) = 18$, which is choice (C). The side of length $7$ corresponds to $DF$ instead.
4. Check that the corresponding ratios agree, $\frac{12}{4} = \frac{18}{6} = \frac{21}{7} = 3$. The remaining side would be $21$, which shows why that number belongs to a different side.

Choice (A) adds the change $12 - 4 = 8$ to the side $6$ instead of multiplying by the scale factor $3$. Choice (B) adds the corresponding lengths $4$ and $12$ instead of scaling the requested side. Choice (D) finds $DF$ rather than $EF$.

The answer is (B), $9$ tasks.

1. At least $90$ includes $90$ and every greater total. Let $n$ count whole tasks. The starting points and the task points together must meet that lower bound, $14 + 9n \ge 90$.
2. Subtracting $14$ gives $9n \ge 76$. Dividing by $9$ gives $n \ge \frac{76}{9} = 8\tfrac{4}{9}$.
3. A whole number must reach or pass that boundary. Since $8$ is below it, the least allowed count is $9$. Rounding to the nearest whole number would not respect this lower bound.
4. Eight tasks give $14 + 9(8) = 86$ points, which is too few. Nine tasks give $14 + 9(9) = 95$ points, which meets the requirement. These neighboring checks show that choice (B) is the least valid count.

Choice (A) rounds the cutoff down even though it is a minimum. Choice (C) is enough tasks but not the least number that is enough. Choice (D) reports the additional points needed rather than a count of tasks.

The answer is (D), $(3, 7)$.

1. A square is never negative, and its coefficient here is positive, so $2(x - 3)^2 \ge 0$ and the expression is smallest where the squared quantity is zero. That happens when $x - 3 = 0$, at $x = 3$.
2. At $x = 3$ the squared part contributes zero, so $g(3) = 2(0)^2 + 7 = 7$. The vertex has input $3$ and output $7$, which is the point $(3, 7)$.
3. The $-3$ inside the square does not make the input of the vertex $-3$. The input comes from setting the inside quantity equal to zero, and an ordered pair lists that input first and its output second.
4. As a symmetry check, the inputs $2$ and $4$ are each one unit from $3$, and $g(2) = 9 = g(4)$. Their points lie at the same height on opposite sides of the vertex.

Choice (A) copies the sign inside the square without solving $x - 3 = 0$. Choice (B) changes the sign of the vertical shift. Choice (C) reverses the input and output coordinates.

The answer is (B). The mean increases by $1$, and the median stays $4$.

1. There are seven values, and their original sum is $28$. The original mean is $\frac{28}{7} = 4$. The middle value of the sorted list is the fourth value, $4$, so the original median is also $4$.
2. Replacing $6$ by $13$ adds $7$ to the sum without changing the number of values. The new sum is $35$ and the new mean is $\frac{35}{7} = 5$, an increase of $1$.
3. The new ordered list is $2$, $3$, $3$, $4$, $5$, $5$, $13$. Its fourth value is still $4$, so the median does not change.
4. Choice (B) describes both changes. In the mean, the increase in the sum is shared over seven values, while the middle position of the ordered list keeps the same value.

Choice (A) assumes that changing an extreme value also shifts the middle value, but the fourth value remains $4$. Choice (C) uses $7$, the increase in the sum, while the increase in the mean is $\frac{7}{7} = 1$ because there are seven values. Choice (D) has the two measures backward. The mean uses the total and therefore changes, while the median uses the middle position, which stays the same here.

The answer is $5$.

1. Let $L$ be the number of large frames and $S$ the number of small frames. The counts satisfy $S + L = 12$ and $3S + 5L = 46$.
2. If all $12$ frames used the small frame requirement, the clip count would be $12(3) = 36$. Each large frame uses $5 - 3 = 2$ clips beyond that baseline.
3. The actual display uses $46 - 36 = 10$ clips above the baseline, so $2L = 10$ and $L = 5$.
4. Algebraically, $S = 12 - L$ gives $3(12 - L) + 5L = 46$, which simplifies to $36 + 2L = 46$ and the same result, $L = 5$.
5. Then $S = 7$. Check both original totals, $7 + 5 = 12$ frames and $7(3) + 5(5) = 21 + 25 = 46$ clips.

The answer is (A), $7$ cm.

1. The base area is the length times the width, $5 \times 8 = 40$ square centimeters. Each centimeter of height adds one layer with that base area.
2. Volume is base area times height, so $40h = 280$. Dividing by $40$ gives a height of $h = 7$ centimeters.
3. Choice (A) is correct. Dividing only by $8$ or only by $5$ would still leave an area, because one base dimension has not been removed. Dividing by the full base area gives a length, $\frac{280 \text{ cm}^3}{40 \text{ cm}^2} = 7 \text{ cm}$.
4. Multiplying the three internal dimensions checks the result, $5 \times 8 \times 7 = 280$ cubic centimeters, which matches the given volume.

Choice (B) divides the volume only by the width $8$. Choice (C) reports the base area as a height. Choice (D) divides the volume only by the length $5$.

The answer is (C), $(x + 4)(x + 5)$.

1. Multiplying two factors of the form $x + a$ and $x + b$ gives $(x + a)(x + b) = x^2 + (a + b)x + ab$. Both the sum and the product have to match.
2. The product must be $20$ and the sum $9$. The positive pairs $1$ and $20$, $2$ and $10$, and $4$ and $5$ all multiply to $20$, but only $4 + 5 = 9$.
3. That gives the factors $x + 4$ and $x + 5$, which is choice (C). The two negative constants in choice (D) multiply to $20$ but add to $-9$, which gives the wrong middle term.
4. Expanding the answer checks the identity for every real input, $(x + 4)(x + 5) = x^2 + 5x + 4x + 20 = x^2 + 9x + 20$.

Choice (A) matches the product but has middle coefficient $12$. Choice (B) matches the product but has middle coefficient $21$. Choice (D) has middle coefficient $-9$.

The answer is (D), none.

1. The same $3x$ appears on both sides, so subtracting it keeps the solution set and removes the variable. What remains is $2 = 8$.
2. Since $2 \ne 8$ and this false statement does not depend on $x$, no value of $x$ can make the original equation true.
3. Canceling the variable does not by itself decide the answer. If the remaining constants had been equal, every $x$ would work. Here the difference is $(3x + 8) - (3x + 2) = 6$ for every input.
4. The two sides graph as parallel lines with slope $3$ and intercepts $2$ and $8$, so they stay $6$ units apart and never meet. That agrees with the algebra and gives choice (D).

Choice (A) assumes that canceling always produces an identity. Choice (B) assumes that every linear-looking equation has exactly one solution. Choice (C) mistakes the two expressions for two solutions.

The answer is (B), $0$ and $6$.

1. Both terms contain $x$, so factoring out $x$ shows the structure, $x(x - 6) = 0$.
2. A product is zero when at least one factor is zero. The first factor gives $x = 0$, and the second gives $x - 6 = 0$, so $x = 6$.
3. Dividing the original equation by $x$ without a check could lose the solution $0$, since division by $x$ requires $x \ne 0$. Factoring keeps that possibility and tests it directly.
4. Zero gives $0^2 - 6(0) = 0$, and six gives $6^2 - 6(6) = 0$. Negative six would give $(-6)^2 - 6(-6) = 72$ instead, so choice (B) is the complete solution set.