Free worked solutions · 2026 Edition

Chapter 9: Polynomials worked solutions

Worked solutions to the practice questions in Chapter 9 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[9 marks]

(a) Write down the zeros of f(x) = (x + 4)(2x - 3)(x - 1)^2, and state which zero is repeated.

[3]

(b) Find the y-intercept of the graph of y = f(x).

[1]

(c) A cubic polynomial g(x) has zeros -2, \frac{1}{3} and 5, and its leading coefficient is 3. Write g(x) as a product of linear factors with integer coefficients, and hence in the form ax^3 + bx^2 + cx + d.

[3]

(d) Solve the equation x(x^2 - 9)(x + 2) = 0.

[2]

Show solutionHide solution

(a) zeros -4 and \frac{3}{2}

zero 1

x = 1 is the repeated (double) zero because the factor (x - 1) is squared

(b) f(0) = (4)(-3)(-1)^2 = -12, so the y-intercept is (0, -12)

(c) g(x) = 3(x + 2)\left(x - \frac{1}{3}\right)(x - 5)

g(x) = (x + 2)(3x - 1)(x - 5)

(x + 2)(3x - 1) = 3x^2 + 5x - 2, so g(x) = 3x^3 - 10x^2 - 27x + 10

(d) x(x - 3)(x + 3)(x + 2) = 0 (M1)

x = 0,\ 3,\ -3,\ -2

Final answer

(a) -4, \frac32, 1 (repeated)
(b) (0,-12)
(c) (x+2)(3x-1)(x-5)=3x^3-10x^2-27x+10
(d) x=-3,-2,0,3

Question 2

[7 marks]

(a) Use long division to find the quotient and the remainder when 2x^3 + 3x^2 - 8x + 5 is divided by (x + 3).

[3]

(b) Find the quotient and the remainder when x^4 - 2x^2 + 3x - 1 is divided by (x - 2).

[3]

(c) Write your answer to part (b) in the form x^4 - 2x^2 + 3x - 1 \equiv (x - 2)Q(x) + R.

[1]

Show solutionHide solution

(a) 2x^3 \div x = 2x^2; subtract 2x^3 + 6x^2 to leave -3x^2 - 8x; then -3x; subtract -3x^2 - 9x to leave x + 5; then +1

quotient 2x^2 - 3x + 1

remainder 2

(b) writing the dividend as x^4 + 0x^3 - 2x^2 + 3x - 1 and dividing

quotient x^3 + 2x^2 + 2x + 7

remainder 13

(c) x^4 - 2x^2 + 3x - 1 \equiv (x - 2)(x^3 + 2x^2 + 2x + 7) + 13

Final answer

(a) 2x^2-3x+1, remainder 2
(b) x^3+2x^2+2x+7, remainder 13
(c) (x-2)(x^3+2x^2+2x+7)+13

Question 3

[9 marks]

In each part, find the value of the constant k.

(a) (x + 3) is a factor of x^3 + kx^2 + x + 21.

[3]

(b) When 4x^3 + kx^2 - 3x + 5 is divided by (2x + 1) the remainder is 8.

[3]

(c) (x - 2) is a factor of x^4 - 3x^3 + kx^2 + kx - 10.

[3]

Show solutionHide solution

(a) substituting x = -3: -27 + 9k - 3 + 21 = 0 M1A1

9k - 9 = 0, so k = 1

(b) substituting x = -\frac{1}{2}: 4\left(-\frac{1}{8}\right) + \frac{k}{4} + \frac{3}{2} + 5 = 8 M1A1

6 + \frac{k}{4} = 8, so k = 8

(c) substituting x = 2: 16 - 24 + 4k + 2k - 10 = 0 M1A1

6k - 18 = 0, so k = 3

Note: award M1 only for substituting the correct value (-3, -\frac{1}{2}, 2).

Final answer

(a) k=1
(b) k=8
(c) k=3

Question 4

[6 marks]

Let P(x) = 2x^3 + 5x^2 - 22x + 15.

(a) Show that P(1) = 0.

[1]

(b) Hence find a quadratic q(x) such that P(x) = (x - 1)q(x).

[2]

(c) Factorise P(x) completely and solve the equation P(x) = 0.

[3]

Show solutionHide solution

(a) P(1) = 2 + 5 - 22 + 15 = 0 A1AG

(b) dividing P(x) by (x - 1) (or comparing coefficients)

q(x) = 2x^2 + 7x - 15

(c) 2x^2 + 7x - 15 = (2x - 3)(x + 5)

P(x) = (x - 1)(2x - 3)(x + 5)

x = 1, x = \frac{3}{2}, x = -5

Final answer

(a) shown
(b) 2x^2+7x-15
(c) (x-1)(2x-3)(x+5); x=1,\frac32,-5

Question 5

[11 marks]

The roots of the equation 3x^2 - 4x - 2 = 0 are \alpha and \beta. Do not solve the equation.

(a) Write down the values of \alpha + \beta and \alpha\beta.

[2]

(b) Show that \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta, and hence find the value of \alpha^2 + \beta^2.

[3]

(c) Find the value of \dfrac{1}{\alpha} + \dfrac{1}{\beta}.

[2]

(d) Given that \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta), find the value of \alpha^3 + \beta^3.

[2]

(e) Find the value of (\alpha - \beta)^2, and hence the value of \lvert \alpha - \beta \rvert.

[2]

Show solutionHide solution

(a) \alpha + \beta = \frac{4}{3}

\alpha\beta = -\frac{2}{3}

(b) (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2, so \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta M1AG

\alpha^2 + \beta^2 = \frac{16}{9} - 2\left(-\frac{2}{3}\right)

= \frac{28}{9}

(c) \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}

= \frac{4/3}{-2/3} = -2

(d) \left(\frac{4}{3}\right)^3 - 3\left(-\frac{2}{3}\right)\left(\frac{4}{3}\right) = \frac{64}{27} + \frac{8}{3}

= \frac{136}{27}

(e) (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{16}{9} + \frac{8}{3} = \frac{40}{9}

\lvert \alpha - \beta \rvert = \frac{2\sqrt{10}}{3}

Final answer

(a) \frac43, -\frac23
(b) \frac{28}{9}
(c) -2
(d) \frac{136}{27}
(e) \frac{40}{9}, \frac{2\sqrt{10}}{3}

Question 6

[6 marks]

When a factor of the denominator is (px + q) with p \ne 1, its partial fraction is written \dfrac{A}{px + q}.

Express each of the following in partial fractions.

(a) \dfrac{5x - 4}{(2x - 1)(x + 1)}

[3]

(b) \dfrac{11x + 6}{(3x + 2)(x + 2)}

[3]

Show solutionHide solution

(a) 5x - 4 \equiv A(x + 1) + B(2x - 1)

x = \frac{1}{2}: -\frac{3}{2} = \frac{3}{2}A, A = -1; x = -1: -9 = -3B, B = 3

-\dfrac{1}{2x - 1} + \dfrac{3}{x + 1}

(b) 11x + 6 \equiv A(x + 2) + B(3x + 2)

x = -\frac{2}{3}: -\frac{4}{3} = \frac{4}{3}A, A = -1; x = -2: -16 = -4B, B = 4

-\dfrac{1}{3x + 2} + \dfrac{4}{x + 2}

Note: do not award the final A1 for a fraction written over x - \frac{1}{2} or x + \frac{2}{3} unless the constant is correct for that form (e.g. -\dfrac{1}{2\left(x - \frac{1}{2}\right)}).

Final answer

(a) \frac{3}{x+1}-\frac{1}{2x-1}
(b) \frac{4}{x+2}-\frac{1}{3x+2}

Question 7

[5 marks]

A student is asked to express \dfrac{4x + 5}{(2x + 1)(x + 2)} in partial fractions. The student writes the zero of 2x + 1 as x = -\frac{1}{2} and works as follows.

“Covering up (x + \frac{1}{2}) and putting x = -\frac{1}{2} gives \dfrac{4\left(-\frac{1}{2}\right) + 5}{-\frac{1}{2} + 2} = 2. Covering up (x + 2) and putting x = -2 gives \dfrac{4(-2) + 5}{2(-2) + 1} = 1. So the answer is \dfrac{2}{x + \frac{1}{2}} + \dfrac{1}{x + 2}."

(a) By substituting x = 0, show that the student's answer is not equal to the original fraction.

[2]

(b) Explain the student's error.

[1]

(c) Write down the correct partial fractions.

[2]

Show solutionHide solution

(a) original at x = 0: \dfrac{5}{(1)(2)} = \dfrac{5}{2}

student at x = 0: \dfrac{2}{\frac{1}{2}} + \dfrac{1}{2} = 4 + \dfrac{1}{2} = \dfrac{9}{2} \ne \dfrac{5}{2}

(b) (2x + 1) = 2\left(x + \frac{1}{2}\right), so covering up \left(x + \frac{1}{2}\right) leaves a factor 2 in the denominator that the student ignored (the first value should be \dfrac{3}{2 \times \frac{3}{2}} = 1, or the fraction should be over 2x + 1)

(c) 4x + 5 \equiv A(x + 2) + B(2x + 1): x = -\frac{1}{2} gives 3 = \frac{3}{2}A, A = 2

\dfrac{2}{2x + 1} + \dfrac{1}{x + 2}

Note: accept the equivalent \dfrac{1}{x + \frac{1}{2}} + \dfrac{1}{x + 2}.

Final answer

(a) \frac92\ne\frac52
(b) the factor 2 of 2x+1 is lost
(c) \frac{2}{2x+1}+\frac{1}{x+2}

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 8

[7 marks]

Let P(x) = 2x^3 - x^2 - 15x + 18.

(a) Determine which of (x + 3), (2x - 3) and (x - 3) are factors of P(x), showing your working.

[5]

(b) Hence factorise P(x) completely.

[2]

Show solutionHide solution

; (a) (x + 3) matches x = -3: P(-3) = -54 - 9 + 45 + 18 = 0, so (x + 3) is a factor.

[2]

(2x - 3) matches x = \tfrac32 (not x = 3): P(\tfrac32) = \tfrac{27}{4} - \tfrac94 - \tfrac{45}{2} + 18 = 0, so (2x - 3) is a factor.

[2]

(x - 3) matches x = 3: P(3) = 54 - 9 - 45 + 18 = 18 \ne 0, so (x - 3) is not a factor.

[1]

(b) Two factors are known and P has degree 3, so P(x) = (x + 3)(2x - 3)(x + k).

[1]

Compare constant terms: 3 \times (-3) \times k = 18, so k = -2 and P(x) = (x + 3)(2x - 3)(x - 2).

[1]

Final answer

(a) (x+3) and (2x-3) are factors, (x-3) is not
(b) (x+3)(2x-3)(x-2)

Question 9

[4 marks]

Let P(x) = x^4 - 2x^3 + 3x^2 - x + 5.

(a) Divide P(x) by (x^2 + x - 1), giving the quotient and the remainder.

[3]

(b) Check your result by substituting x = 1 into the division identity.

[1]

Show solutionHide solution

; (a) x^4 \div x^2 = x^2; subtract x^2(x^2 + x - 1) = x^4 + x^3 - x^2, leaving -3x^3 + 4x^2 - x.

[1]

-3x^3 \div x^2 = -3x; subtract -3x(x^2 + x - 1) = -3x^3 - 3x^2 + 3x, leaving 7x^2 - 4x + 5.

7x^2 \div x^2 = 7; subtract 7(x^2 + x - 1) = 7x^2 + 7x - 7, leaving -11x + 12, which has degree 1 < 2, so the division stops.

Quotient x^2 - 3x + 7.

[1]

Remainder -11x + 12.

[1]

(b) Left side: P(1) = 1 - 2 + 3 - 1 + 5 = 6. Right side: (1 + 1 - 1)(1 - 3 + 7) + (-11 + 12) = 5 + 1 = 6.

[1]

Final answer

(a) quotient x^2-3x+7, remainder 12-11x
(b) both sides equal 6

Question 10

[5 marks]

The polynomial P(x) = x^4 - 3x^2 + 5x + 2 is divided by (x - 1)(x + 2).

(a) Given that the remainder is mx + c, find m and c.

[4]

(b) Explain why the remainder must have this form.

[1]

Show solutionHide solution

; (a) Write P(x) = (x - 1)(x + 2)Q(x) + mx + c.

[1]

Put x = 1: P(1) = 1 - 3 + 5 + 2 = 5 = m + c. Put x = -2: P(-2) = 16 - 12 - 10 + 2 = -4 = -2m + c.

[2]

Subtract: 3m = 9, so m = 3 and c = 2. The remainder is 3x + 2.

[1]

(b) The remainder has degree less than the degree 2 of the divisor, so it is at most linear.

[1]

Final answer

(a) m=3, c=2
(b) the divisor has degree 2

Question 11

[5 marks]

Let \alpha and \beta be the two solutions of x^2 - 5x + 2 = 0. Without solving, form the quadratic equation with integer coefficients whose solutions are the ratios \dfrac{\alpha}{\beta} and \dfrac{\beta}{\alpha}.

[5]

Show solutionHide solution

; Vieta: \alpha + \beta = 5 and \alpha\beta = 2.

[1]

New sum: \dfrac\alpha\beta + \dfrac\beta\alpha = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}.

[1]

= \dfrac{25 - 4}{2} = \dfrac{21}{2}.

[1]

New product: \dfrac\alpha\beta \cdot \dfrac\beta\alpha = 1.

[1]

The equation is x^2 - \dfrac{21}{2}x + 1 = 0, i.e. 2x^2 - 21x + 2 = 0.

[1]

Final answer

2x^2-21x+2=0

Question 12

[5 marks]

Let \alpha, \beta and \gamma be the solutions of 2x^3 - 3x^2 + x + 4 = 0.

(a) Without solving the equation, find a cubic equation with integer coefficients whose solutions are \alpha + 2, \beta + 2 and \gamma + 2.

[3]

(b) Find a cubic equation with integer coefficients whose roots are 2\alpha, 2\beta and 2\gamma.

[2]

Show solutionHide solution

; (a) Let w = x + 2 be a new root, so x = w - 2. Substitute into the equation: 2(w - 2)^3 - 3(w - 2)^2 + (w - 2) + 4 = 0.

[1]

Expand: 2(w^3 - 6w^2 + 12w - 8) - 3(w^2 - 4w + 4) + w - 2 + 4 = 0.

[1]

2w^3 - 15w^2 + 37w - 26 = 0.

[1]

(b) Let w = 2x, so x = \dfrac w2: 2\left(\dfrac w2\right)^3 - 3\left(\dfrac w2\right)^2 + \dfrac w2 + 4 = 0.

[1]

Multiply by 4: w^3 - 3w^2 + 2w + 16 = 0.

[1]

Final answer

(a) 2x^3-15x^2+37x-26=0
(b) x^3-3x^2+2x+16=0

Question 13

[5 marks]

The quadratic equation x^2 + px + 27 = 0 has real roots \alpha and \alpha^2.

(a) Show that \alpha^3 = 27, and hence find \alpha.

[3]

(b) Find the value of p.

[2]

Show solutionHide solution

; (a) The product of the roots is \alpha \cdot \alpha^2 = 27.

[1]

So \alpha^3 = 27.

\alpha is real, so \alpha = 3 and the roots are 3 and 9.

[2]

(b) The sum of the roots is 3 + 9 = 12 = -p.

[1]

p = -12.

[1]

Final answer

(a) \alpha=3
(b) p=-12

Question 14

[5 marks]

Solve the inequality 2x^3 - x^2 - 8x + 4 < 0.

[5]

Show solutionHide solution

; Factorise by grouping: 2x^3 - x^2 - 8x + 4 = x^2(2x - 1) - 4(2x - 1).

[1]

= (2x - 1)(x^2 - 4) = (2x - 1)(x - 2)(x + 2).

[1]

Roots: x = -2,\ \tfrac12,\ 2, all simple.

[1]

The leading coefficient is positive, so the signs of the polynomial on the four intervals, from left to right, are -,\ +,\ -,\ +.

We need the negative intervals: x < -2 or \tfrac12 < x < 2.

[2]

Final answer

x<-2 or \frac12<x<2

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 15

[9 marks]

The polynomial P(x) = x^4 + ax^3 + bx^2 + cx + 6 is exactly divisible by (x - 1)^2, and leaves remainder 24 when divided by (x + 1).

(a) Find the values of a, b and c.

[6]

(b) Hence factorise P(x), and show that P(x) = 0 has exactly one real root.

[3]

Show solutionHide solution

; (a) Divisible by (x - 1)^2: P(1) = 0 and P'(1) = 0. Remainder 24: P(-1) = 24.

[1]

P(1) = 1 + a + b + c + 6 = 0, so a + b + c = -7.

[1]

P'(x) = 4x^3 + 3ax^2 + 2bx + c, so P'(1) = 4 + 3a + 2b + c = 0, giving 3a + 2b + c = -4.

[1]

P(-1) = 1 - a + b - c + 6 = 24, so -a + b - c = 17.

[1]

Add the first and third equations: 2b = 10, so b = 5 and a + c = -12. The second equation gives 3a + c = -14.

[1]

Subtract: 2a = -2, so a = -1, c = -11 and b = 5.

[1]

(b) P(x) = x^4 - x^3 + 5x^2 - 11x + 6. Divide by (x - 1)^2 = x^2 - 2x + 1: the quotient is x^2 + x + 6.

[1]

P(x) = (x - 1)^2(x^2 + x + 6).

[1]

For x^2 + x + 6: \Delta = 1 - 24 = -23 < 0, so it has no real roots. The only real root is x = 1 (repeated).

[1]

Final answer

(a) a=-1, b=5, c=-11
(b) (x-1)^2(x^2+x+6), and x^2+x+6 has \Delta=-23<0

Question 16

[8 marks]

When P(x) = x^4 + ax^3 + bx^2 - 2x + 3 is divided by x^2 - x - 2, the remainder is 3x + 1.

(a) By writing P(x) \equiv (x^2 - x - 2)Q(x) + 3x + 1, explain why P(2) = 7, and find P(-1).

[3]

(b) Hence find the values of a and b.

[3]

(c) Find Q(x).

[2]

Show solutionHide solution

(a) x^2 - x - 2 = (x - 2)(x + 1)

at x = 2 the term (x^2 - x - 2)Q(x) is zero, so P(2) = 3(2) + 1 = 7

P(-1) = 3(-1) + 1 = -2

(b) 16 + 8a + 4b - 4 + 3 = 7 and 1 - a + b + 2 + 3 = -2

2a + b = -2 and -a + b = -8

a = 2, b = -6

(c) dividing x^4 + 2x^3 - 6x^2 - 2x + 3 by x^2 - x - 2 (the remainder checks as 3x + 1)

Q(x) = x^2 + 3x - 1

Final answer

(a) P(2)=7, P(-1)=-2
(b) a=2, b=-6
(c) Q(x)=x^2+3x-1

Question 17

[12 marks]

(a) The equation 2x^3 - 3x^2 - 8x + k = 0 has roots \alpha, -\alpha and \gamma.

(i) Show that \gamma = \frac{3}{2}.

[2]

(ii) Find the value of k.

[2]

(iii) Find the values of \alpha.

[2]

(b) The equation 3x^3 + px^2 - 19x + 6 = 0 has three real roots, two of which have product 1.

(i) Show that the third root is -2.

[2]

(ii) Find the value of p.

[2]

(iii) Find the other two roots, giving your answers in exact form.

[2]

Show solutionHide solution

(a)(i) sum of roots \alpha + (-\alpha) + \gamma = \gamma

\gamma = \frac{-(-3)}{2} = \frac{3}{2} A1AG

(ii) x = \frac{3}{2} is a root: 2\left(\frac{27}{8}\right) - 3\left(\frac{9}{4}\right) - 8\left(\frac{3}{2}\right) + k = 0

\frac{27}{4} - \frac{27}{4} - 12 + k = 0, so k = 12

(iii) product of roots -\alpha^2 \cdot \frac{3}{2} = \frac{(-1)^3(12)}{2} = -6 (or 2x^3 - 3x^2 - 8x + 12 = (2x - 3)(x^2 - 4))

\alpha^2 = 4, so \alpha = \pm 2

(b)(i) let the roots be \beta, \delta with \beta\delta = 1, and \gamma: product of roots \beta\delta\gamma = \frac{(-1)^3(6)}{3}

\gamma = -2 A1AG

(ii) x = -2 is a root: 3(-8) + 4p - 19(-2) + 6 = 0

4p + 20 = 0, so p = -5

(iii) 3x^3 - 5x^2 - 19x + 6 = (x + 2)(3x^2 - 11x + 3)

x = \frac{11 \pm \sqrt{85}}{6}

Note: check: the product of these two roots is \frac{3}{3} = 1.

Final answer

(a)(i) shown (ii) k=12 (iii) \alpha=\pm2
(b)(i) shown (ii) p=-5 (iii) \frac{11\pm\sqrt{85}}{6}

Question 18

[10 marks]

The equation x^3 - 6x^2 - 7x + 13 = 0 has roots \alpha, \beta and \gamma.

(a) Write down the values of \alpha + \beta + \gamma and \alpha\beta\gamma.

[2]

(b) By substituting x = \dfrac{u}{2}, show that the equation whose roots are 2\alpha, 2\beta and 2\gamma is u^3 - 12u^2 - 28u + 104 = 0.

[3]

(c) Find a cubic equation, in the form u^3 + pu^2 + qu + r = 0, whose roots are \alpha + 1, \beta + 1 and \gamma + 1.

[3]

(d) Hence write down the value of (\alpha + 1)(\beta + 1)(\gamma + 1), and verify the sum of the roots of your equation in part (c) using part (a).

[2]

Show solutionHide solution

(a) \alpha + \beta + \gamma = 6

\alpha\beta\gamma = \frac{(-1)^3(13)}{1} = -13

(b) if u = 2\alpha then x = \frac{u}{2} is a root of the original equation

\frac{u^3}{8} - \frac{3u^2}{2} - \frac{7u}{2} + 13 = 0

multiplying by 8: u^3 - 12u^2 - 28u + 104 = 0 A1AG

(c) u = \alpha + 1 gives x = u - 1: (u - 1)^3 - 6(u - 1)^2 - 7(u - 1) + 13 = 0

u^3 - 3u^2 + 3u - 1 - 6u^2 + 12u - 6 - 7u + 7 + 13 = 0

u^3 - 9u^2 + 8u + 13 = 0

(d) (\alpha + 1)(\beta + 1)(\gamma + 1) = \frac{(-1)^3(13)}{1} = -13

sum of the new roots is 9, and (\alpha + \beta + \gamma) + 3 = 6 + 3 = 9

Final answer

(a) 6, -13
(b) shown
(c) u^3-9u^2+8u+13=0
(d) -13; the new roots sum to 9=6+3

Question 19

[11 marks]

Let h(x) = \dfrac{7 + kx}{(1 + x)(3 - 2x)}, where k is a constant.

(a) Show that h(x) \equiv \dfrac{A}{1 + x} + \dfrac{B}{3 - 2x}, where A = \dfrac{7 - k}{5} and B = \dfrac{14 + 3k}{5}.

[4]

(b) In the expansion of h(x) in ascending powers of x, the coefficient of x is -\dfrac{1}{9}. Find the value of k.

[4]

(c) For this value of k, find the coefficient of x^2 in the expansion of h(x).

[2]

(d) State the interval of validity of the expansion.

[1]

Show solutionHide solution

(a) 7 + kx \equiv A(3 - 2x) + B(1 + x)

x = -1: 7 - k = 5A

x = \frac{3}{2}: 7 + \frac{3}{2}k = \frac{5}{2}B

A = \dfrac{7 - k}{5}, B = \dfrac{2\left(7 + \frac{3}{2}k\right)}{5} = \dfrac{14 + 3k}{5}

A1AG

(b) \dfrac{A}{1 + x} = A(1 - x + x^2 - \ldots) and \dfrac{B}{3 - 2x} = \dfrac{B}{3}\left(1 - \dfrac{2x}{3}\right)^{-1} = \dfrac{B}{3}\left(1 + \dfrac{2}{3}x + \dfrac{4}{9}x^2 + \ldots\right) M1A1

coefficient of x: -A + \dfrac{2B}{9} = -\dfrac{7 - k}{5} + \dfrac{2(14 + 3k)}{45} = -\dfrac{1}{9}

-9(7 - k) + 2(14 + 3k) = -5 \Rightarrow 15k = 30, k = 2

(c) A = 1, B = 4: coefficient of x^2 = A + \dfrac{4B}{27} = 1 + \dfrac{16}{27} (M1)

= \dfrac{43}{27}

(d) \lvert x \rvert < 1

Note: \lvert x \rvert < 1 and \left\lvert \dfrac{2x}{3} \right\rvert < 1 both hold exactly when -1 < x < 1.

Final answer

(a) shown
(b) k=2
(c) \frac{43}{27}
(d) |x|<1

Question 20

[6 marks]

Let g(x)=3x^3+5x^2-26x+8.

(a) Show that (x-2) is a factor of g(x), and factorise g(x) completely.

[3]

(b) Hence solve 3e^{3x}+5e^{2x}-26e^{x}+8=0, giving a reason for each value you reject.

[3]

Show solutionHide solution

(a) g(2)=24+20-52+8=0, so (x-2) is a factor

M1A1

g(x)=(x-2)(3x^2+11x-4)=(x-2)(x+4)(3x-1)

(b) put x\to\mathrm{e}^x in the factors: \mathrm{e}^x=2, \mathrm{e}^x=\frac13 or \mathrm{e}^x=-4

\mathrm{e}^x>0 for every real x, so \mathrm{e}^x=-4 is rejected

x=\ln2 or x=-\ln3

Final answer

(a) (x-2)(x+4)(3x-1)
(b) x=\ln2, x=-\ln3

Question 21

[16 marks]

Let P(x) = 2x^4 + ax^3 + bx^2 + cx + 12, where a, b, c \in \mathbb{R}.

The sum of the roots of the equation P(x) = 0 is \frac{3}{2}. The polynomial P(x) has a factor (x + 1), and when P(x) is divided by (x - 1) the remainder is 10.

(a) Find the value of a.

[2]

(b) Find the values of b and c.

[4]

(c) Show that (x - 2) is a factor of P(x).

[1]

(d) Hence factorise P(x) completely, showing that (x - 2) is a repeated factor.

[4]

(e) Write down the product of the roots of P(x) = 0, and use your factorisation to verify it.

[2]

(f) Sketch the graph of y = P(x) for -2 \le x \le 3, labelling the intercepts with the axes.

[3]

Show solutionHide solution

(a) \frac{-a}{2} = \frac{3}{2}

a = -3

(b) using P(-1) = 0 and P(1) = 10

2 + 3 + b - c + 12 = 0 \Rightarrow b - c = -17

2 - 3 + b + c + 12 = 10 \Rightarrow b + c = -1

b = -9, c = 8

(c) P(2) = 32 - 24 - 36 + 16 + 12 = 0 A1AG

(d) (x + 1)(x - 2) = x^2 - x - 2 is a factor; dividing 2x^4 - 3x^3 - 9x^2 + 8x + 12 by x^2 - x - 2

quotient 2x^2 - x - 6

2x^2 - x - 6 = (2x + 3)(x - 2)

P(x) = (x + 1)(2x + 3)(x - 2)^2, so (x - 2) is a repeated factor

(e) product = \frac{(-1)^4(12)}{2} = 6

roots -1, -\frac{3}{2}, 2, 2: (-1)\left(-\frac{3}{2}\right)(2)(2) = 6

(f) positive quartic shape (y \to \infty at both ends)

crosses the x-axis at -\frac{3}{2} and -1, touches it at 2

y-intercept (0, 12)

Final answer

(a) a=-3
(b) b=-9, c=8
(c) shown
(d) (x+1)(2x+3)(x-2)^2
(e) 6
(f) crosses at -\frac32 and -1, touches at 2, meets the y-axis at 12

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Constant differences

G1 Worked example

[11 marks]

For a function u, the first difference is \Delta u(x)=u(x+1)-u(x); the second difference \Delta^2u is the first difference of \Delta u, and so on. A difference table lists u(0),u(1),u(2),\dots in a row, and each later row lists the differences of the row above.

(a) For p(x)=2x^3-3x^2+1, write down p(0),p(1),\dots,p(5) and the rows of first, second and third differences.

[2]

(b) Show that \Delta p(x)=6x^2-1 and \Delta^2p(x)=12x+6. Hence explain why every third difference is 12.

[3]

(c) Show that, for every cubic q(x)=ax^3+bx^2+cx+d, every third difference is 6a.

[3]

(d) The values of a cubic at x=0,1,2,3,4 are 4, 2, 6, 22, 56. Find its values at x=5 and x=6, and find the cubic.

[3]

Show solutionHide solution

(a) p: 1,0,5,28,81,176; ; first: -1,5,23,53,95; ; second: 6,18,30,42

third: 12,12,12

(b) \Delta p(x)=2\bigl((x+1)^3-x^3\bigr)-3\bigl((x+1)^2-x^2\bigr)=2(3x^2+3x+1)-3(2x+1)=6x^2-1

\Delta^2p(x)=6\bigl((x+1)^2-x^2\bigr)=6(2x+1)=12x+6

\Delta^3p(x)=12(x+1)+6-(12x+6)=12 for every x

(c) \Delta q(x)=a(3x^2+3x+1)+b(2x+1)+c=3ax^2+(3a+2b)x+(a+b+c)

M1A1

\Delta^2q(x)=3a(2x+1)+(3a+2b)=6ax+6a+2b, so \Delta^3q(x)=6a

(d) Differences: -2,4,16,34; then 6,12,18; then 6,6. Extend with third differences 6: second 24,30; first 58,88

values 114 and 202

6a=6 gives a=1. Subtracting x^3 leaves 4,1,-2,-5,-8, which is 4-3x, so the cubic is x^3-3x+4

Check: 5^3-15+4=114 and 6^3-18+4=202.

Final answer

(d) 114, 202; x^3-3x+4

G1 Investigation

[13 marks]

In this question \Delta u(x)=u(x+1)-u(x), \Delta^2u is the first difference of \Delta u, and so on.

(a) Write down the difference table of u(x)=x^4 for x=0,1,\dots,5, and show that the fourth differences are 24.

[2]

(b) Use the binomial theorem to show that, if p(x) is a polynomial of degree n\ge1 with leading coefficient a, then \Delta p(x) is a polynomial of degree n-1 with leading coefficient na.

[3]

(c) Conjecture a formula for the nth differences of a polynomial of degree n with leading coefficient a, and prove it by induction on n.

[3]

(d) For k\in\mathbb{Z}^+ let B_k(x)=\dfrac{x(x-1)(x-2)\cdots(x-k+1)}{k!}, and let B_0(x)=1. Show that \Delta B_k(x)=B_{k-1}(x).

[2]

(e) Let u(x)=c_0+c_1B_1(x)+c_2B_2(x)+c_3B_3(x). Explain why u(0), \Delta u(0), \Delta^2u(0) and \Delta^3u(0) are c_0, c_1, c_2 and c_3. Hence find the cubic whose values at x=0,1,2,3,4 are 2, 3, 8, 23, 54, giving it in the form px^3+qx^2+rx+s.

[3]

Show solutionHide solution

(a) 0,1,16,81,256,625; ; 1,15,65,175,369; ; 14,50,110,194; ; 36,60,84

fourth differences 24,24

(b) p(x)=ax^n+bx^{n-1}+\dots; by the binomial theorem (x+1)^n-x^n=nx^{n-1}+\binom n2x^{n-2}+\dots+1

and (x+1)^m-x^m has degree m-1 for each lower power x^m, so it adds nothing to the x^{n-1} term except from m=n

\Delta p(x)=nax^{n-1}+(\text{terms of lower degree}): degree n-1, leading coefficient na

(c) x^3: 6=3!; 2x^3: 12=2\times3!; x^4: 24=4!: conjecture \Delta^np=n!\,a

n=1: \Delta(ax+b)=a=1!\,a. Assume true for degree k. For degree k+1, by (b) \Delta p has degree k and leading coefficient (k+1)a

so \Delta^{k+1}p=\Delta^k(\Delta p)=k!\,(k+1)a=(k+1)!\,a; true for n=1 and true for k+1 whenever true for k, so true for every n\in\mathbb{Z}^+

(d) B_k(x+1)-B_k(x)=\dfrac{(x+1)x\cdots(x-k+2)-x(x-1)\cdots(x-k+1)}{k!}=\dfrac{x(x-1)\cdots(x-k+2)\bigl((x+1)-(x-k+1)\bigr)}{k!}

=\dfrac{k\,x(x-1)\cdots(x-k+2)}{k!}=B_{k-1}(x)

A1AG

(e) By (d), \Delta^ju=c_j+c_{j+1}B_1+\dots, and B_m(0)=0 for m\ge1, so \Delta^ju(0)=c_j

Table: 2,3,8,23,54; 1,5,15,31; 4,10,16; 6,6: so c_0=2, c_1=1, c_2=4, c_3=6

u(x)=2+B_1(x)+4B_2(x)+6B_3(x); expanding, u(x)=x^3-x^2+x+2

Final answer

(a) 24
(c) \Delta^np=n!\,a
(e) x^3-x^2+x+2

G2 A fence for the roots

G2 Worked example

[10 marks]

Let p(x)=x^3+bx^2+cx+d, where b,c,d\in\mathbb{R}, and let M be the largest of |b|, |c| and |d|, with M>0. You may use the fact that |u+v+w|\le|u|+|v|+|w| for any real or complex u, v, w.

(a) Suppose that p(t)=0, where t is real or complex and |t|>1. Show that |t|^3\le M\left(|t|^2+|t|+1\right).

[2]

(b) Hence show that |t|<1+M. Explain why every root of p, whatever its modulus, satisfies |t|<1+M.

[3]

(c) For q(x)=x^3-4x^2+x+6, write down the bound given by (b). Find the roots of q and comment on the bound.

[3]

(d) For r(x)=x^3-10x^2-10x-10, write down the bound given by (b). Use your GDC to find the real root of r, and comment.

[2]

Show solutionHide solution

(a) t^3=-\left(bt^2+ct+d\right), so |t|^3\le|b||t|^2+|c||t|+|d|

\le M\left(|t|^2+|t|+1\right)

(b) With |t|>1: |t|^2+|t|+1=\dfrac{|t|^3-1}{|t|-1}<\dfrac{|t|^3}{|t|-1}

so |t|^3<\dfrac{M|t|^3}{|t|-1}, which gives |t|-1<M

A root with |t|\le1 already has |t|\le1<1+M

(c) M=6, so every root has |x|<7

q(x)=(x+1)(x-2)(x-3): roots -1, 2, 3

Every root is inside the bound, but the largest modulus is 3, far below 7: here the bound is safe but weak

(d) M=10: every root has |x|<11

GDC: x=10.99 (4 s.f.), just inside the bound: for this polynomial the bound is very nearly reached

Final answer

(c) |x|<7; roots -1,2,3
(d) |x|<11; x=10.99

G2 Investigation

[11 marks]

For M\ge1 and an integer n\ge2, let g_n(x)=x^n-M\left(x^{n-1}+x^{n-2}+\dots+x+1\right). The bound |x|<1+M holds for every root of a monic polynomial whose other coefficients have largest modulus M; this investigation asks how close a root can come to it.

(a) For n=2, find the positive root r of g_2 in terms of M. Find (1+M)-r when M=1, 10 and 100, and suggest its limit as M\to\infty.

[3]

(b) Show that, for x\ne1, g_n(x)=0 if and only if x^{n+1}-(M+1)x^n+M=0.

[2]

(c) Show that g_n(M)<0, and deduce that g_n has a root r with r>M.

[2]

(d) Show that this root satisfies r=1+M-\dfrac{M}{r^n}.

[2]

(e) Hence show that 0<(1+M)-r<M^{1-n}. What does this tell you about the bound 1+M?

[2]

Show solutionHide solution

(a) x^2-Mx-M=0, so r=\dfrac{M+\sqrt{M^2+4M}}{2}

M=1: 0.382; ; M=10: 0.0839; ; M=100: 0.00980

The gap appears to tend to 0

(b) For x\ne1, x^{n-1}+\dots+x+1=\dfrac{x^n-1}{x-1}, so g_n(x)=0\iff x^n(x-1)=M(x^n-1)

\iff x^{n+1}-(M+1)x^n+M=0

A1AG

(c) g_n(M)=M^n-M\cdot M^{n-1}-M\left(M^{n-2}+\dots+1\right)=-M\left(M^{n-2}+\dots+1\right)<0

g_n(x)>0 for large x, because x^n dominates; g_n is continuous, so it has a root r>M

(d) r>M\ge1, so r\ne1 and (b) applies: r^{n+1}-(M+1)r^n+M=0

Divide by r^n: r-(M+1)+\dfrac{M}{r^n}=0, so r=1+M-\dfrac{M}{r^n}

A1AG

(e) (1+M)-r=\dfrac{M}{r^n}>0, and r>M gives \dfrac{M}{r^n}<\dfrac{M}{M^n}=M^{1-n}

For large M (or large n with M>1) the gap is tiny, so the bound cannot be lowered by any fixed amount: no bound 1+M-\varepsilon with \varepsilon>0 works for every polynomial

Note: the values in (a) satisfy (e): 0.382<1, 0.0839<0.1, 0.00980<0.01.

Final answer

(a) r=\frac{M+\sqrt{M^2+4M}}{2}; 0.382, 0.0839, 0.00980; limit 0
(e) the bound 1+M cannot be lowered by a fixed amount

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Polynomial roots	`APPS, PlySmlt2 (or graph and 2nd TRACE, 2:zero, for real roots)`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 9 teaches polynomials from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book