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Polynomials: Division, Factor and Remainder Theorems — Answers

Answer sheet with worked solutions for the 11th Grade functions worksheet.

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Success TutoringWorksheetsAnswer sheet11th Grade · Mathematics · Functions

Polynomials: Division, Factor and Remainder Theorems — Answers

Part A: The language of polynomials

A polynomial in x is a sum of terms of the form axn, where each n is a whole number (0, 1, 2, …). Its degree is the highest power of x, the leading coefficient is the coefficient of that term, and the constant term is the term without x. A polynomial is monic if its leading coefficient is 1.

Worked example. P(x) = 4x3 − 2x5 + x − 7 has degree 5, leading coefficient −2 and constant term −7; it is not monic. P(1) = 4 − 2 + 1 − 7 = −4.

1.For P(x) = 3x4 − x2 + 5x − 9, the degree is (answer space) and the leading coefficient is (answer space).

Answer: 4, 3

The highest power of x is x4, so the degree is 4, and the coefficient of that term (the leading coefficient) is 3.

2.If P(x) = x3 − 2x2 + 4, find P(−2).

Answer: −12

  1. P(−2) = (−2)3 − 2(−2)2 + 4
  2. = −8 − 8 + 4 = −12
3.Which of these is not a polynomial?
  1. Ax3 + 2x
  2. B√x + 1
  3. C5
  4. Dx2 − x

Answer: √x + 1

√x = x1/2, and a polynomial may only contain whole-number powers of x. A constant such as 5 is a polynomial of degree 0, and fractional coefficients are allowed.

4.The sum of two cubic polynomials is always a cubic polynomial.

Answer: False

The cubic terms can cancel: x3 + (−x3 + x) = x, which has degree 1.

Part B: Dividing polynomials

Dividing P(x) by D(x) gives a quotient Q(x) and a remainder R(x) with P(x) = D(x)Q(x) + R(x), where the degree of R(x) is less than the degree of D(x).

Long division: divide the leading term of what is left by the leading term of the divisor, multiply the divisor by that term, subtract, and repeat. Write in a 0 for any missing power of x.

Worked example. (x3 + 2x2 − 5x + 6) ÷ (x − 1): x2 gives x3 − x2, leaving 3x2 − 5x + 6; 3x gives 3x2 − 3x, leaving −2x + 6; −2 gives −2x + 2, leaving 4. So Q(x) = x[sup 2] + 3x − 2 and R = 4.

5.Divide x3 − 4x2 + x + 6 by x − 2. State the quotient and the remainder.

Answer: x2 − 2x − 3, remainder 0

  1. x3 ÷ x = x2; x2(x − 2) = x3 − 2x2; subtract: −2x2 + x
  2. −2x2 ÷ x = −2x; −2x(x − 2) = −2x2 + 4x; subtract: −3x + 6
  3. −3x ÷ x = −3; −3(x − 2) = −3x + 6; subtract: 0
  4. Quotient x[sup 2] − 2x − 3, remainder 0 (so x − 2 is a factor)
6.Divide 2x3 + 3x2 − 4x + 5 by x + 2. State the quotient and the remainder.

Answer: 2x2 − x − 2, remainder 9

  1. 2x3 ÷ x = 2x2; 2x2(x + 2) = 2x3 + 4x2; subtract: −x2 − 4x
  2. −x2 ÷ x = −x; −x(x + 2) = −x2 − 2x; subtract: −2x + 5
  3. −2x ÷ x = −2; −2(x + 2) = −2x − 4; subtract: 9
  4. Quotient 2x[sup 2] − x − 2, remainder 9
7.Divide x4 − 3x2 + 2x − 1 by x2 + 1. Write the answer in the form P(x) = D(x)Q(x) + R(x).

Answer: (x2 + 1)(x2 − 4) + 2x + 3

  1. x4 ÷ x2 = x2; x2(x2 + 1) = x4 + x2; subtract: −4x2 + 2x − 1
  2. −4x2 ÷ x2 = −4; −4(x2 + 1) = −4x2 − 4; subtract: 2x + 3
  3. The remainder 2x + 3 has degree less than 2, so stop.
  4. x[sup 4] − 3x[sup 2] + 2x − 1 = (x[sup 2] + 1)(x[sup 2] − 4) + 2x + 3
8.When a polynomial is divided by (x − 3)(x + 1), what is the highest possible degree of the remainder?
  1. A0
  2. B1
  3. C2
  4. D3

Answer: 1

The divisor (x − 3)(x + 1) = x2 − 2x − 3 has degree 2, and the remainder must have a lower degree than the divisor, so it is at most degree 1 (of the form ax + b).

Part C: The remainder and factor theorems

Remainder theorem: when P(x) is divided by x − a, the remainder is P(a). (Put x = a in P(x) = (x − a)Q(x) + R: the first term becomes 0.)

Factor theorem: x − a is a factor of P(x) exactly when P(a) = 0. To divide by x + 2, use a = −2.

Worked example. The remainder when P(x) = x3 − 3x2 + 4x − 2 is divided by x − 3 is P(3) = 27 − 27 + 12 − 2 = 10. Since P(1) = 1 − 3 + 4 − 2 = 0, x − 1 is a factor.

9.Find the remainder when P(x) = 2x3 − x2 + 3x − 4 is divided by x + 1.

Answer: −10

  1. By the remainder theorem the remainder is P(−1).
  2. P(−1) = 2(−1) − 1 + 3(−1) − 4 = −2 − 1 − 3 − 4 = −10
10.Find the remainder when x5 − 1 is divided by x + 1.

Answer: −2

  1. Remainder = P(−1) = (−1)5 − 1 = −1 − 1 = −2
11.x − 2 is a factor of x3 − 3x2 − 4x + 12.

Answer: True

P(2) = 8 − 12 − 8 + 12 = 0, so by the factor theorem x − 2 is a factor. (In fact x3 − 3x2 − 4x + 12 = (x − 2)(x + 2)(x − 3).)

12.When x3 + kx2 − 2x + 5 is divided by x − 2, the remainder is 13. Find k.

Answer: 1

  1. P(2) = 13: 8 + 4k − 4 + 5 = 13
  2. 4k + 9 = 13, so 4k = 4 and k = 1
13.P(x) = x3 + ax2 + bx − 6 has x − 1 as a factor and leaves a remainder of 16 when divided by x − 2. Then a = (answer space) and b = (answer space).

Answer: 2, 3

P(1) = 0: 1 + a + b − 6 = 0, so a + b = 5. P(2) = 16: 8 + 4a + 2b − 6 = 16, so 2a + b = 7. Subtract: a = 2, then b = 3. Check: P(x) = x3 + 2x2 + 3x − 6 gives P(1) = 0 and P(2) = 16.

Part D: Factoring and solving cubics

To factor a cubic with integer coefficients, test small factors of the constant term (±1, ±2, ±3, …) until P(a) = 0. Then divide by x − a and factor the quadratic quotient.

If the leading coefficient is not 1, also try fractions such as ±: any rational root has p dividing the constant term and q dividing the leading coefficient.

Worked example. P(x) = x3 − 7x + 6. P(1) = 0, so P(x) = (x − 1)(x2 + x − 6) = (x − 1)(x + 3)(x − 2), and P(x) = 0 when x = −3, 1, 2.

14.Factor P(x) = x3 − 2x2 − 5x + 6 fully, and hence solve P(x) = 0.

Answer: −2, 1, 3

  1. Test factors of 6: P(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor.
  2. Divide: P(x) = (x − 1)(x2 − x − 6)
  3. Factor the quadratic: P(x) = (x − 1)(x − 3)(x + 2)
  4. x = −2, 1, 3
15.Solve 2x3 + 3x2 − 11x − 6 = 0.

Answer: −3, −, 2

  1. Try x = 2: 16 + 12 − 22 − 6 = 0, so x − 2 is a factor.
  2. Divide: 2x3 + 3x2 − 11x − 6 = (x − 2)(2x2 + 7x + 3)
  3. Factor: (x − 2)(2x + 1)(x + 3) = 0
  4. x = −3, −[frac 1/2], 2

Part E: Extended problem

Let P(x) = x3 − 4x2 + x + 6.

(a) Show that x + 1 is a factor of P(x).

(b) Factor P(x) fully.

(c) Find the intercepts of y = P(x) with the axes and sketch its graph.

(d) Hence solve P(x) < 0.

(e) Find the remainder when P(x) is divided by x − 4.

16.Answer parts (a) to (e), showing full working.[10 marks]

Model answer: (a) P(−1) = −1 − 4 − 1 + 6 = 0, so by the factor theorem x + 1 is a factor. (b) Dividing gives P(x) = (x + 1)(x2 − 5x + 6) = (x + 1)(x − 2)(x − 3). (c) y-intercept P(0) = 6; x-intercepts −1, 2 and 3. The leading coefficient is positive, so the curve rises from bottom left, crosses at x = −1, turns between −1 and 2, crosses at x = 2, turns between 2 and 3, crosses at x = 3 and rises to the top right. (d) From the graph, P(x) < 0 below the x-axis: x < −1 or 2 < x < 3. (e) Remainder = P(4) = 64 − 64 + 4 + 6 = 10.

Success criteria:

  • (a) shows P(−1) = 0 and quotes the factor theorem (2 marks)
  • (b) correct quotient and full factorization (2 marks)
  • (c) intercepts and correct cubic shape for a positive leading coefficient (2 marks)
  • (d) x < −1 or 2 < x < 3 (2 marks)
  • (e) remainder 10 by the remainder theorem (2 marks)
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