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TMUA Polynomials questions

12 original TMUA 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 1TMUA Paper 1Warm-up

When p(x)=x3+ax2+bx+6p(x) = x^3 + ax^2 + bx + 6 is divided by (x−1)(x - 1) the remainder is 44, and when it is divided by (x+2)(x + 2) the remainder is −20-20. What is abab?
  1. A−12-12
  2. B−4-4
  3. C−3-3
  4. D33
  5. E44
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Answer: B · Method: Factor and remainder theorem equations

  1. By the remainder theorem, p(1)=7+a+b=4p(1) = 7 + a + b = 4, so a+b=−3a + b = -3.
  2. And p(−2)=−2+4a−2b=−20p(-2) = -2 + 4a - 2b = -20, so 2a−b=−92a - b = -9.
  3. Adding: 3a=−123a = -12, so a=−4a = -4, b=1b = 1 and ab=−4ab = -4.

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Question 2TMUA Paper 1Warm-up

The polynomial p(x)=x3+ax2+bx−8p(x) = x^3 + ax^2 + bx - 8 has (x−2)(x - 2) as a factor, and leaves remainder −30-30 when divided by (x+1)(x + 1). What is a+ba + b?
  1. A−21-21
  2. B−7-7
  3. C77
  4. D1414
  5. E2121
  6. F00
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Answer: C · Method: Factor and remainder theorem equations

  1. Factor theorem: p(2)=8+4a+2b−8=0p(2) = 8 + 4a + 2b - 8 = 0, so b=−2ab = -2a.
  2. Remainder theorem: p(−1)=−1+a−b−8=−30p(-1) = -1 + a - b - 8 = -30, so a−b=−21a - b = -21.
  3. Then 3a=−213a = -21, so a=−7a = -7, b=14b = 14 and a+b=7a + b = 7.

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Question 3TMUA Paper 1Warm-up

The polynomial p(x)=x3+ax2+bx+1p(x) = x^3 + ax^2 + bx + 1 has (x−1)(x - 1) as a factor and leaves a remainder of 1010 when divided by (x+1)(x + 1). What is abab?
  1. A−25-25
  2. B−24-24
  3. C−23-23
  4. D−22-22
  5. E−2-2
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Answer: B · Method: Factor and remainder theorem equations

  1. Factor theorem: p(1)=0p(1) = 0, so a+b=−2a + b = -2.
  2. Remainder theorem: p(−1)=10p(-1) = 10, so a−b=10a - b = 10.
  3. Solving these together gives a=4a = 4 and b=−6b = -6, so ab=−24ab = -24.

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Question 4TMUA Paper 1Warm-up

The polynomial p(x)=x3+ax2+bx+20p(x) = x^3 + ax^2 + bx + 20 has (x−2)(x - 2) as a factor and leaves a remainder of 88 when divided by (x+2)(x + 2). What is a+ba + b?
  1. A−11-11
  2. B−10-10
  3. C−9-9
  4. D−8-8
  5. E22
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Answer: B · Method: Factor and remainder theorem equations

  1. Factor theorem: p(2)=0p(2) = 0, so 4a+2b=−284a + 2b = -28.
  2. Remainder theorem: p(−2)=8p(-2) = 8, so 4a−2b=−44a - 2b = -4.
  3. Solving these together gives a=−4a = -4 and b=−6b = -6, so a+b=−10a + b = -10.

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Question 5TMUA Paper 1Warm-up

The polynomial p(x)=x3+ax2+bx+6p(x) = x^3 + ax^2 + bx + 6 has (x+2)(x + 2) as a factor and leaves a remainder of 33 when divided by (x−1)(x - 1). What is a+ba + b?
  1. A−5-5
  2. B−4-4
  3. C−3-3
  4. D22
  5. E88
Show the answer and worked solution

Answer: B · Method: Factor and remainder theorem equations

  1. Factor theorem: p(−2)=0p(-2) = 0, so 4a−2b=24a - 2b = 2.
  2. Remainder theorem: p(1)=3p(1) = 3, so a+b=−4a + b = -4.
  3. Solving these together gives a=−1a = -1 and b=−3b = -3, so a+b=−4a + b = -4.

Why the tempting wrong answers are wrong

  • E: Dividing by (x−1)(x - 1) leaves remainder p(1)p(1), not p(−1)p(-1).

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Question 6TMUA Paper 1Warm-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 −12-12 when divided by (x+3)(x + 3). What is a+ba + b?
  1. A−2-2
  2. B11
  3. C22
  4. D33
  5. E44
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=2a + b = 2.
  2. Remainder theorem: p(−3)=−12p(-3) = -12, so 9a−3b=189a - 3b = 18.
  3. Solving these together gives a=2a = 2 and b=0b = 0, so a+b=2a + b = 2.

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.

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Question 7TMUA Paper 1Standard

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

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3+3x2+4f(x) = x^3 + 3x^2 + 4. Then f′(x)=3x2+6x=3(x+2)xf'(x) = 3x^2 + 6x = 3(x + 2)x, so the turning points are at x=−2x = -2 and x=0x = 0.
  2. The local maximum value is f(−2)=8f(-2) = 8 and the local minimum value is f(0)=4f(0) = 4.
  3. A horizontal line y=ky = k crosses three times exactly when 4<k<84 < k < 8: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=8k = 8 and k=4k = 4 the line touches a turning point, giving only two distinct solutions.

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Question 8TMUA Paper 1Standard

For how many integer values of kk does the equation x3−12x−4=kx^3 - 12x - 4 = k have three distinct real solutions?
  1. A2929
  2. B3030
  3. C3131
  4. D3232
  5. E3333
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Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3−12x−4f(x) = x^3 - 12x - 4. Then f′(x)=3x2−12=3(x+2)(x−2)f'(x) = 3x^2 - 12 = 3(x + 2)(x - 2), so the turning points are at x=−2x = -2 and x=2x = 2.
  2. The local maximum value is f(−2)=12f(-2) = 12 and the local minimum value is f(2)=−20f(2) = -20.
  3. A horizontal line y=ky = k crosses three times exactly when −20<k<12-20 < k < 12: that's 3131 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=12k = 12 and k=−20k = -20 the line touches a turning point, giving only two distinct solutions.

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Question 9TMUA Paper 1Standard

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

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3−3x+4f(x) = x^3 - 3x + 4. 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)=6f(-1) = 6 and the local minimum value is f(1)=2f(1) = 2.
  3. A horizontal line y=ky = k crosses three times exactly when 2<k<62 < k < 6: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=6k = 6 and k=2k = 2 the line touches a turning point, giving only two distinct solutions.

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Question 10TMUA Paper 1Standard

For how many integer values of kk does the equation x3+9x2+24x−3=kx^3 + 9x^2 + 24x - 3 = k have three distinct real solutions?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
Show the answer and worked solution

Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3+9x2+24x−3f(x) = x^3 + 9x^2 + 24x - 3. Then f′(x)=3x2+18x+24=3(x+4)(x+2)f'(x) = 3x^2 + 18x + 24 = 3(x + 4)(x + 2), so the turning points are at x=−4x = -4 and x=−2x = -2.
  2. The local maximum value is f(−4)=−19f(-4) = -19 and the local minimum value is f(−2)=−23f(-2) = -23.
  3. A horizontal line y=ky = k crosses three times exactly when −23<k<−19-23 < k < -19: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=−19k = -19 and k=−23k = -23 the line touches a turning point, giving only two distinct solutions.

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Question 11TMUA Paper 1Standard

For how many integer values of kk does the equation x3+3x2−5=kx^3 + 3x^2 - 5 = k have three distinct real solutions?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
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Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3+3x2−5f(x) = x^3 + 3x^2 - 5. Then f′(x)=3x2+6x=3(x+2)xf'(x) = 3x^2 + 6x = 3(x + 2)x, so the turning points are at x=−2x = -2 and x=0x = 0.
  2. The local maximum value is f(−2)=−1f(-2) = -1 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<−1-5 < k < -1: that's 33 integers.

Why the tempting wrong answers are wrong

  • D: Count the integers strictly between the two turning values.
  • E: At k=−1k = -1 and k=−5k = -5 the line touches a turning point, giving only two distinct solutions.

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Question 12TMUA Paper 1Challenging

For how many integers kk does the equation x3−3x2−9x+k=0x^3 - 3x^2 - 9x + k = 0 have three distinct real roots, exactly two of which are positive?
  1. A2222
  2. B2525
  3. C2626
  4. D2727
  5. E3131
  6. F3232
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Answer: C · Method: Count roots using turning points

  1. Let f(x)=x3−3x2−9x+kf(x) = x^3 - 3x^2 - 9x + k. Then f′(x)=3(x−3)(x+1)f'(x) = 3(x - 3)(x + 1), so there is a maximum at x=−1x = -1 with f(−1)=k+5f(-1) = k + 5, and a minimum at x=3x = 3 with f(3)=k−27f(3) = k - 27.
  2. Three distinct real roots need f(−1)>0>f(3)f(-1) > 0 > f(3), so −5<k<27-5 < k < 27. The roots then lie in (−∞,−1)(-\infty, -1), (−1,3)(-1, 3) and (3,∞)(3, \infty).
  3. The first is negative and the last positive. The middle one is positive exactly when f(0)=k>0f(0) = k > 0, because ff is decreasing on (−1,3)(-1, 3). So 0<k<270 < k < 27: 2626 integers.

Why the tempting wrong answers are wrong

  • D: At k=0k = 0 one root is exactly 00, which isn't positive.
  • E: −5<k<27-5 < k < 27 only guarantees three distinct roots. You also need the middle root to be positive.

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