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Grade 11 Review Quiz — Easy Version 2
Grade 11 · MathSheets Name: ____________________ Date: __________
Answer every question. The lesson name before each question tells you which skill it checks — open that lesson if you need a reminder.
How to solve
This quiz mixes questions from different lessons, like a real test. Read each question carefully: the lesson name in front tells you what kind of problem it is. Check your answers, then practise the lessons where you made mistakes.
Example
Exponential Growth, Decay & Compound Interest: Does y = 800(0.92)t show growth or decay, and by what percent per period? Answer like: growth, 4%
Answer: decay, 8%
Not quite. The answer is
💡 Only the leading term matters at the ends. The term with the highest power of x is −x³, so the degree is 3 and the leading coefficient is −1. The degree 3 is odd, so the ends point opposite ways. The leading coefficient is negative, so the right end goes down and the left end goes up. The answer is up, down.
✓ Correct!
Not quite. The answer is
💡 Only the leading term matters at the ends. The term with the highest power of x is −x³, so the degree is 3 and the leading coefficient is −1. The degree 3 is odd, so the ends point opposite ways. The leading coefficient is negative, so the right end goes down and the left end goes up. The answer is up, down.
💡 How to solve: Polynomial Degree & End Behavior
Only the leading term matters at the ends. Even degree: both ends point the same way (up if the leading coefficient is positive, down if negative). Odd degree: the ends point opposite ways (the right end goes up if the leading coefficient is positive). A polynomial of degree n has at most n - 1 turning points.
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Find the degree and the leading coefficient of f(x) = 2x² - 3x4 - x + x³. Answer as: degree, leading coefficient.
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💡 The ball lands when its height is 0: −16t2 + 112t = 0. Factor out t: t(−16t + 112) = 0, so t = 0 or t = 112 ÷ 16 = 7. t = 0 is the moment it is kicked, so it lands after 7 seconds.
✓ Correct!
Not quite. The answer is
💡 The ball lands when its height is 0: −16t2 + 112t = 0. Factor out t: t(−16t + 112) = 0, so t = 0 or t = 112 ÷ 16 = 7. t = 0 is the moment it is kicked, so it lands after 7 seconds.
💡 How to solve: Word Problems
For doubling, multiply the starting amount by 2 for each hour: amount = start × 2n. For two numbers that differ by d and multiply to p, use x(x + d) = p and look for two factors of p that are d apart. Write the two numbers as a and b, smaller first.
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A video has 5 views. The number of views triples every day. How many views will it have after 2 days?
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💡 By the Remainder Theorem the remainder is f(3). Synthetic division by 3 with the coefficients 3, -2, 2, -2: Bring down 3. 3 × 3 = 9, and -2 + 9 = 7. 7 × 3 = 21, and 2 + 21 = 23. 23 × 3 = 69, and -2 + 69 = 67. The last number is the remainder: 67.
✓ Correct!
Not quite. The answer is
💡 By the Remainder Theorem the remainder is f(3). Synthetic division by 3 with the coefficients 3, -2, 2, -2: Bring down 3. 3 × 3 = 9, and -2 + 9 = 7. 7 × 3 = 21, and 2 + 21 = 23. 23 × 3 = 69, and -2 + 69 = 67. The last number is the remainder: 67.
💡 How to solve: Remainder & Factor Theorems
Remainder Theorem: the remainder of f(x) ÷ (x - c) equals f(c), so synthetic division by c (or substituting c) gives it. Factor Theorem: (x - c) is a factor exactly when f(c) = 0. When it is, the quotient from synthetic division holds the other factors.
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Find the remainder when f(x) = 2x³ + 2x² - x + 3 is divided by (x - 3).
| 3 | 2 | 2 | -1 | 3 |
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💡 Product rule: log4(8) + log4(8) = log4(8 × 8) = log4(64). Rewrite: log4(64) = x means 4x = 64, and 4 × 4 × 4 = 64, so x = 3. The answer is 3.
✓ Correct!
Not quite. The answer is
💡 Product rule: log4(8) + log4(8) = log4(8 × 8) = log4(64). Rewrite: log4(64) = x means 4x = 64, and 4 × 4 × 4 = 64, so x = 3. The answer is 3.
💡 How to solve: Properties of Logarithms
Product rule: logb(MN) = logb(M) + logb(N). Quotient rule: logb(M/N) = logb(M) - logb(N). Power rule: logb(Mk) = k·logb(M). Combine or split the logs, then use logb(bk) = k.
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Fill in the blank: log5(25x) = ___ + log5(x)
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💡 Set each factor equal to 0. x + 3 = 0 gives x = -3, x - 3 = 0 gives x = 3 and x - 4 = 0 gives x = 4. In increasing order: -3, 3, 4.
✓ Correct!
Not quite. The answer is
💡 Set each factor equal to 0. x + 3 = 0 gives x = -3, x - 3 = 0 gives x = 3 and x - 4 = 0 gives x = 4. In increasing order: -3, 3, 4.
💡 How to solve: Zeros & Multiplicity
Each factor (x - r)m gives the zero x = r with multiplicity m. If m is odd the graph crosses the x-axis at r; if m is even it touches the axis and turns back. The degree is the sum of all the multiplicities.
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List the zeros of f(x) = (x - 3)³(x - 5)³ in increasing order, separated by commas.
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💡 The first term is a_1 = 9 and the common difference is d = 11 − 9 = 2. The last term is a_12 = a_1 + (n − 1)d = 9 + 11(2) = 31. Use S_n = n(a_1 + a_n) ÷ 2: S_12 = 12(9 + 31) ÷ 2 = 12 × 40 ÷ 2 = 240. The sum is 240.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 9 and the common difference is d = 11 − 9 = 2. The last term is a_12 = a_1 + (n − 1)d = 9 + 11(2) = 31. Use S_n = n(a_1 + a_n) ÷ 2: S_12 = 12(9 + 31) ÷ 2 = 12 × 40 ÷ 2 = 240. The sum is 240.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
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Find the sum of the first 16 terms of the arithmetic series 2 + 4 + 6 + ….
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💡 Find two numbers that multiply to -8 and add to 2: 4 and -2. So x² + 2x - 8 = (x + 4)(x - 2). The roots are -4 and 2. The parabola opens up, so it is below the x-axis between the roots and above it outside them. ≥ 0 means above or on the axis: x ≤ -4 or x ≥ 2.
✓ Correct!
Not quite. The answer is
💡 Find two numbers that multiply to -8 and add to 2: 4 and -2. So x² + 2x - 8 = (x + 4)(x - 2). The roots are -4 and 2. The parabola opens up, so it is below the x-axis between the roots and above it outside them. ≥ 0 means above or on the axis: x ≤ -4 or x ≥ 2.
💡 How to solve: Quadratic Inequalities
Get 0 on one side, then factor to find the roots, e.g. x2 - 5x + 6 = (x - 2)(x - 3) has roots 2 and 3. A parabola that opens up is below the x-axis between its roots and above it outside them, so x2 - 5x + 6 < 0 gives 2 < x < 3 and x2 - 5x + 6 > 0 gives x < 2 or x > 3. If the x2 coefficient is negative the parabola opens down and the two cases swap. Use ≤ or ≥ in the answer when the inequality includes "or equal to".
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Solve: (x - 5)(x - 6) ≤ 0
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💡 Multiply both sides by (x + 1): x² + 3x = -2. Move everything to one side: x² + 3x + 2 = 0. Find two numbers that multiply to 2 and add to 3: 1 and 2. So x² + 3x + 2 = (x + 1)(x + 2). So x = -1 or x = -2. Check: x = -1 makes the denominator (x + 1) equal to 0, so it is extraneous. x = -2 works. So x = -2.
✓ Correct!
Not quite. The answer is
💡 Multiply both sides by (x + 1): x² + 3x = -2. Move everything to one side: x² + 3x + 2 = 0. Find two numbers that multiply to 2 and add to 3: 1 and 2. So x² + 3x + 2 = (x + 1)(x + 2). So x = -1 or x = -2. Check: x = -1 makes the denominator (x + 1) equal to 0, so it is extraneous. x = -2 works. So x = -2.
💡 How to solve: Solving Rational Equations
Multiply both sides by the least common denominator to clear the fractions, then solve the equation that is left. Check each answer in the original equation: a value that makes any denominator 0 is extraneous and must be rejected.
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Solve: -2x = -3x - 1
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💡 Both equations equal y, so set them equal: x² - 4 = 0. Move everything to the left: x² - 4 = 0. It is a difference of squares: x² - 4 = (x + 2)(x - 2). So x = -2 or x = 2. Put each x into the line: When x = -2, y = 0. When x = 2, y = 0. The solutions are (-2, 0) and (2, 0).
✓ Correct!
Not quite. The answer is
💡 Both equations equal y, so set them equal: x² - 4 = 0. Move everything to the left: x² - 4 = 0. It is a difference of squares: x² - 4 = (x + 2)(x - 2). So x = -2 or x = 2. Put each x into the line: When x = -2, y = 0. When x = 2, y = 0. The solutions are (-2, 0) and (2, 0).
💡 How to solve: Linear-Quadratic Systems of Equations
Both equations equal y, so set them equal to each other: x2 - 2x - 3 = x + 1. Move everything to one side (x2 - 3x - 4 = 0), factor ((x - 4)(x + 1) = 0) and solve for x. Put each x back into the line to get its y, giving the points (-1, 0) and (4, 5). The discriminant b2 - 4ac of the combined quadratic tells how many solutions there are: positive means 2, zero means 1 (the line is tangent to the parabola), negative means 0.
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Solve the system: y = x² - 14 and y = -5
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💡 Work from the inside out. First find the inside function: f(−1) = −1 − 3 = −4. Then put −4 into g: g(−4) = −4 + 3 = −1. So g(f(−1)) = −1.
✓ Correct!
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💡 Work from the inside out. First find the inside function: f(−1) = −1 − 3 = −4. Then put −4 into g: g(−4) = −4 + 3 = −1. So g(f(−1)) = −1.
💡 How to solve: Function Composition
Work from the inside out. For f(g(k)), first find g(k), then put that result into f. For g(f(k)), find f(k) first, then put that result into g.
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If f(x) = 3x - 2 and g(x) = 3x, find f(g(4)).
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💡 Write each term as a cube: x3 is already a cube and 1 = 13. Use the difference of cubes A3 − B3 = (A − B)(A2 + AB + B2) with A = x and B = 1: A2 = x2, AB = x and B2 = 1. So the answer is (x - 1)(x2 + x + 1).
✓ Correct!
Not quite. The answer is
💡 Write each term as a cube: x3 is already a cube and 1 = 13. Use the difference of cubes A3 − B3 = (A − B)(A2 + AB + B2) with A = x and B = 1: A2 = x2, AB = x and B2 = 1. So the answer is (x - 1)(x2 + x + 1).
💡 How to solve: Factoring Sums & Differences of Cubes
Use a3 + b3 = (a + b)(a2 - ab + b2) and a3 - b3 = (a - b)(a2 + ab + b2). First take out any common factor, then write each term as a cube, for example 8x3 = (2x)3 and 27 = 33.
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Factor: x³ + 125
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💡 Use synthetic division by -1 (the zero of (x + 1)) with the coefficients 1, 2, 1, 0 (0 for a missing term). Bring down 1. 1 × (-1) = -1, and 2 + (-1) = 1. 1 × (-1) = -1, and 1 + (-1) = 0. 0 × (-1) = 0, and 0 + 0 = 0. The remainder is 0. The bottom row 1, 1, 0 gives the quotient, one degree lower: x² + x.
✓ Correct!
Not quite. The answer is
💡 Use synthetic division by -1 (the zero of (x + 1)) with the coefficients 1, 2, 1, 0 (0 for a missing term). Bring down 1. 1 × (-1) = -1, and 2 + (-1) = 1. 1 × (-1) = -1, and 1 + (-1) = 0. 0 × (-1) = 0, and 0 + 0 = 0. The remainder is 0. The bottom row 1, 1, 0 gives the quotient, one degree lower: x² + x.
💡 How to solve: Dividing Polynomials
Use synthetic division: for dividing by (x - r), use r (for x + r, use -r). Bring down the first coefficient, multiply by r, add to the next coefficient, and repeat. The results are the coefficients of the quotient x² + bx + c.
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Divide: (x³ + 3x² - 7x - 12) ÷ (x + 4)
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💡 Factor out x: x(x² - 16) = 0. The difference of squares x² - 16 = (x + 4)(x - 4). So x(x + 4)(x - 4) = 0, and setting each factor equal to 0 gives x = 0, x = -4 or x = 4. In increasing order: -4, 0, 4.
✓ Correct!
Not quite. The answer is
💡 Factor out x: x(x² - 16) = 0. The difference of squares x² - 16 = (x + 4)(x - 4). So x(x + 4)(x - 4) = 0, and setting each factor equal to 0 gives x = 0, x = -4 or x = 4. In increasing order: -4, 0, 4.
💡 How to solve: Solving Polynomial Equations by Factoring
Get 0 on one side and factor completely: take out a common factor, factor by grouping, or treat x4 + bx2 + c like a quadratic in x2. Then set each factor equal to 0 and solve.
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Solve: x³ - 4x = 0
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💡 3! means multiply every whole number from 3 down to 1: 3 × 2 × 1 = 6.
✓ Correct!
Not quite. The answer is
💡 3! means multiply every whole number from 3 down to 1: 3 × 2 × 1 = 6.
💡 How to solve: Counting: Permutations & Combinations
Counting principle: multiply the number of choices at each step. n! = n × (n - 1) × ... × 1 counts ways to arrange n things. When order matters (1st, 2nd, 3rd), use P(n, r) = n! ÷ (n - r)!. When order does not matter (a group or committee), use C(n, r) = n! ÷ (r! × (n - r)!).
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You have 3 shirts, 3 pairs of pants and 2 pairs of shoes. How many different outfits can you make?
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💡 The denominator 2 means take the square root, and the numerator 1 is the power: 91/2 = (square root of 9)1. The square root of 9 is 3, because 32 = 9. So the answer is 3.
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💡 The denominator 2 means take the square root, and the numerator 1 is the power: 91/2 = (square root of 9)1. The square root of 9 is 3, because 32 = 9. So the answer is 3.
💡 How to solve: Rational Exponents
am/n means take the n-th root of a, then raise the result to the m-th power. Find the root first so the numbers stay small.
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Evaluate: 272/3
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💡 Add 4 to both sides: √(x + 11) = 4. Square both sides: x + 11 = 16. Subtract 11 from both sides: x = 5. Check: with x = 5, x + 11 = 16 and √16 = 4, so it works. The answer is x = 5.
✓ Correct!
Not quite. The answer is
💡 Add 4 to both sides: √(x + 11) = 4. Square both sides: x + 11 = 16. Subtract 11 from both sides: x = 5. Check: with x = 5, x + 11 = 16 and √16 = 4, so it works. The answer is x = 5.
💡 How to solve: Solving Radical Equations
Isolate the radical, then raise both sides to the index (square both sides for √, cube both sides for ∛) and solve. Substitute every answer back into the original equation: a value that does not check is extraneous. A square root can never equal a negative number.
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Solve: √(3x + 9) = 3
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Grade 11 Review Quiz — Easy Version 2
Answer every question. The lesson name before each question tells you which skill it checks — open that lesson if you need a reminder.