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ESAT Maths 2 Polynomials questions

10 original ESAT Maths 2 polynomials questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Factor and remainder theorem equationsCount roots using turning points

Question 1ESAT Maths 2Warm-up

The polynomial p(x)=x3+ax2+bx−7p(x) = x^3 + ax^2 + bx - 7 has (x−1)(x - 1) as a factor and leaves a remainder of 7474 when divided by (x−3)(x - 3). What is a+ba + b?
  1. A44
  2. B55
  3. C66
  4. D77
  5. E88
  6. F99
Show the answer and worked solution

Answer: C · Method: Factor and remainder theorem equations

  1. Factor theorem: p(1)=0p(1) = 0, so a+b=6a + b = 6.
  2. Remainder theorem: p(3)=74p(3) = 74, so 9a+3b=549a + 3b = 54.
  3. Solving these together gives a=6a = 6 and b=0b = 0, so a+b=6a + b = 6.

Why the tempting wrong answers are wrong

  • B: A factor of (x−1)(x - 1) means p(1)=0p(1) = 0, not p(−1)=0p(-1) = 0.

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Question 2ESAT Maths 2Warm-up

The polynomial p(x)=x3+ax2+bx−3p(x) = x^3 + ax^2 + bx - 3 has (x+1)(x + 1) as a factor and leaves a remainder of 99 when divided by (x+2)(x + 2). What is a+ba + b?
  1. A00
  2. B22
  3. C44
  4. D77
  5. E88
  6. F99
  7. G1010
Show the answer and worked solution

Answer: E · Method: Factor and remainder theorem equations

  1. Factor theorem: p(−1)=0p(-1) = 0, so a−b=4a - b = 4.
  2. Remainder theorem: p(−2)=9p(-2) = 9, so 4a−2b=204a - 2b = 20.
  3. Solving these together gives a=6a = 6 and b=2b = 2, so a+b=8a + b = 8.

Why the tempting wrong answers are wrong

  • A: Dividing by (x+2)(x + 2) leaves remainder p(−2)p(-2), not p(2)p(2).
  • B: A factor of (x+1)(x + 1) means p(−1)=0p(-1) = 0, not p(1)=0p(1) = 0.

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Question 3ESAT Maths 2Warm-up

The polynomial p(x)=x3+ax2+bx+5p(x) = x^3 + ax^2 + bx + 5 has (x+1)(x + 1) as a factor and leaves a remainder of 99 when divided by (x−2)(x - 2). What is a+ba + b?
  1. A−6-6
  2. B−4-4
  3. C−2-2
  4. D−1-1
  5. E00
  6. F11
  7. G22
  8. H2424
Show the answer and worked solution

Answer: E · Method: Factor and remainder theorem equations

  1. Factor theorem: p(−1)=0p(-1) = 0, so a−b=−4a - b = -4.
  2. Remainder theorem: p(2)=9p(2) = 9, so 4a+2b=−44a + 2b = -4.
  3. Solving these together gives a=−2a = -2 and b=2b = 2, so a+b=0a + b = 0.

Why the tempting wrong answers are wrong

  • A: A factor of (x+1)(x + 1) means p(−1)=0p(-1) = 0, not p(1)=0p(1) = 0.
  • H: Dividing by (x−2)(x - 2) leaves remainder p(2)p(2), not p(−2)p(-2).

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Question 4ESAT Maths 2Warm-up

The polynomial p(x)=x3+ax2+bx+24p(x) = x^3 + ax^2 + bx + 24 has (x−3)(x - 3) as a factor and leaves a remainder of 1010 when divided by (x−2)(x - 2). What is a+ba + b?
  1. A−9-9
  2. B−7-7
  3. C−6-6
  4. D−5-5
  5. E−4-4
  6. F−3-3
Show the answer and worked solution

Answer: D · Method: Factor and remainder theorem equations

  1. Factor theorem: p(3)=0p(3) = 0, so 9a+3b=−519a + 3b = -51.
  2. Remainder theorem: p(2)=10p(2) = 10, so 4a+2b=−224a + 2b = -22.
  3. Solving these together gives a=−6a = -6 and b=1b = 1, so a+b=−5a + b = -5.

Why the tempting wrong answers are wrong

  • A: A factor of (x−3)(x - 3) means p(3)=0p(3) = 0, not p(−3)=0p(-3) = 0.

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Question 5ESAT Maths 2Standard

For how many integer values of kk does the equation x3+6x2+5=kx^3 + 6x^2 + 5 = k have three distinct real solutions?
  1. A2727
  2. B2929
  3. C3030
  4. D3131
  5. E3232
  6. F3333
  7. G3434
  8. H3535
Show the answer and worked solution

Answer: D · Method: Count roots using turning points

  1. Let f(x)=x3+6x2+5f(x) = x^3 + 6x^2 + 5. Then f′(x)=3x2+12x=3(x+4)xf'(x) = 3x^2 + 12x = 3(x + 4)x, so the turning points are at x=−4x = -4 and x=0x = 0.
  2. The local maximum value is f(−4)=37f(-4) = 37 and the local minimum value is f(0)=5f(0) = 5.
  3. A horizontal line y=ky = k crosses three times exactly when 5<k<375 < k < 37: that's 3131 integers.

Why the tempting wrong answers are wrong

  • E: Count the integers strictly between the two turning values.
  • F: At k=37k = 37 and k=5k = 5 the line touches a turning point, giving only two distinct solutions.

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Question 6ESAT Maths 2Standard

For how many integer values of kk does the equation x3+6x2−6=kx^3 + 6x^2 - 6 = k have three distinct real solutions?
  1. A2929
  2. B3030
  3. C3131
  4. D3232
  5. E3333
  6. F3434
Show the answer and worked solution

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3+6x2−6f(x) = x^3 + 6x^2 - 6. Then f′(x)=3x2+12x=3(x+4)xf'(x) = 3x^2 + 12x = 3(x + 4)x, so the turning points are at x=−4x = -4 and x=0x = 0.
  2. The local maximum value is f(−4)=26f(-4) = 26 and the local minimum value is f(0)=−6f(0) = -6.
  3. A horizontal line y=ky = k crosses three times exactly when −6<k<26-6 < k < 26: that's 3131 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=26k = 26 and k=−6k = -6 the line touches a turning point, giving only two distinct solutions.

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Question 7ESAT Maths 2Standard

For how many integer values of kk does the equation x3−3x2+2=kx^3 - 3x^2 + 2 = k have three distinct real solutions?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G77
  8. H88
Show the answer and worked solution

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2. Then f′(x)=3x2−6x=3x(x−2)f'(x) = 3x^2 - 6x = 3x(x - 2), so the turning points are at x=0x = 0 and x=2x = 2.
  2. The local maximum value is f(0)=2f(0) = 2 and the local minimum value is f(2)=−2f(2) = -2.
  3. A horizontal line y=ky = k crosses three times exactly when −2<k<2-2 < k < 2: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=2k = 2 and k=−2k = -2 the line touches a turning point, giving only two distinct solutions.

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Question 8ESAT Maths 2Standard

For how many integer values of kk does the equation x3+3x2−9x+5=kx^3 + 3x^2 - 9x + 5 = k have three distinct real solutions?
  1. A2727
  2. B2929
  3. C3030
  4. D3131
  5. E3232
  6. F3333
  7. G3434
  8. H3535
Show the answer and worked solution

Answer: D · Method: Count roots using turning points

  1. Let f(x)=x3+3x2−9x+5f(x) = x^3 + 3x^2 - 9x + 5. Then f′(x)=3x2+6x−9=3(x+3)(x−1)f'(x) = 3x^2 + 6x - 9 = 3(x + 3)(x - 1), so the turning points are at x=−3x = -3 and x=1x = 1.
  2. The local maximum value is f(−3)=32f(-3) = 32 and the local minimum value is f(1)=0f(1) = 0.
  3. A horizontal line y=ky = k crosses three times exactly when 0<k<320 < k < 32: that's 3131 integers.

Why the tempting wrong answers are wrong

  • E: Count the integers strictly between the two turning values.
  • F: At k=32k = 32 and k=0k = 0 the line touches a turning point, giving only two distinct solutions.

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Question 9ESAT Maths 2Standard

For how many integer values of kk does the equation x3−3x+5=kx^3 - 3x + 5 = k have three distinct real solutions?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G77
  8. H88
Show the answer and worked solution

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3−3x+5f(x) = x^3 - 3x + 5. Then f′(x)=3x2−3=3(x+1)(x−1)f'(x) = 3x^2 - 3 = 3(x + 1)(x - 1), so the turning points are at x=−1x = -1 and x=1x = 1.
  2. The local maximum value is f(−1)=7f(-1) = 7 and the local minimum value is f(1)=3f(1) = 3.
  3. A horizontal line y=ky = k crosses three times exactly when 3<k<73 < k < 7: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=7k = 7 and k=3k = 3 the line touches a turning point, giving only two distinct solutions.

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Question 10ESAT Maths 2Standard

For how many integer values of kk does the equation x3−12x2+36x=kx^3 - 12x^2 + 36x = k have three distinct real solutions?
  1. A2727
  2. B2929
  3. C3030
  4. D3131
  5. E3232
  6. F3333
  7. G3434
  8. H3535
Show the answer and worked solution

Answer: D · Method: Count roots using turning points

  1. Let f(x)=x3−12x2+36xf(x) = x^3 - 12x^2 + 36x. Then f′(x)=3x2−24x+36=3(x−2)(x−6)f'(x) = 3x^2 - 24x + 36 = 3(x - 2)(x - 6), so the turning points are at x=2x = 2 and x=6x = 6.
  2. The local maximum value is f(2)=32f(2) = 32 and the local minimum value is f(6)=0f(6) = 0.
  3. A horizontal line y=ky = k crosses three times exactly when 0<k<320 < k < 32: that's 3131 integers.

Why the tempting wrong answers are wrong

  • E: Count the integers strictly between the two turning values.
  • F: At k=32k = 32 and k=0k = 0 the line touches a turning point, giving only two distinct solutions.

Try it interactively →

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