CHAPTER 5

Polynomials

5.1 Revision 232

5.2 Cubic polynomials 238

5.3 Remainder theorem 246

5.4 Factor theorem 250

5.5 Solving cubic equations 252

5.6 Summary 254

5 Polynomials

• Proofs of theorems are not examinable.

• Be careful of spending too much time on revision. Learners should be familiar with thesefactorisation methods from Grade 10 and 11.

• Explain terminology carefully. It is very important for learners to understand what a factoris and how it relates to the graphs of quadratic and cubic functions (covered in DifferentialCalculus chapter).

• Long division and synthetic division are included as an introduction to the remaindertheorem. Do not spend too much time on these techniques.

• It is important for learners to understand that a remainder of zero means that divisor is afactor.

• Encourage learners to practise factorising by inspection.

5.1 Revision

Identifying polynomials

Exercise 5 – 1: Identifying polynomials

1. Given f(x) = 2x3 +3x2−1, determine whether the following statements are true or false.If false, provide the correct statement.

a) f(x) is a trinomial.

Solution: True

b) The coefficient of the x is zero.

Solution: True

c) f�

12

)= 1

12.

Solution:

f(x) = 2x3 + 3x2 − 1

f

(1

2

)= 2

(1

2

)3

+ 3

(1

2

)2

− 1

= 2

(1

8

)+ 3

(1

4

)− 1

=1

4+

3

4− 1

= 1− 1

= 0

∴ False

d) f(x) is of degree 3.

Solution: True

e) The constant term is 1.

Solution: False: −1

f) f(x) will have 3 real roots.

Solution: True

2. Given g(x) = 2x3 − 9x2 + 7x+ 6, determine the following:

a) the number of terms in g(x).

232 5.1. Revision

Solution: 4

b) the degree of g(x).Solution: 3

c) the coefficient of the x2 term.Solution: −9

d) the constant term.Solution: 6

3. Determine which of the following expressions are polynomials and which are not. Forthose that are not polynomials, give reasons.

a) y3 +√

5

Solution: Cubic polynomialb) −x2 − x− 1

Solution: Quadratic polynomial

c) 4√k − 9

Solution: Not a polynomial; in k12 the exponent is not a natural number.

d) 2p

+ p+ 3

Solution: Not a polynomial; in p−1 the exponent is not a natural number.e) x(x− 1)(x− 2)− 2

Solution: Cubic polynomialf) (√m− 1) (

√m+ 1)

Solution:

�√m− 1

) �√m+ 1

)=�√m)2

+√m−

√m− 1

= m− 1

Linear polynomialg) t0 − 1

Solution:

t0 − 1 = 1− 1

= 0

Zero polynomialh) 16y7

Solution: Polynomial; degree 7

i) −x3

2+ 5x2 + x

3− 11

Solution: Cubic polynomialj) 4b0 + 3b−1 + 5b2 − b3

Solution: Not a polynomial; in b−1 the exponent is not a natural number.

4. Peter’s mathematics homework is shown below. Find and correct his mistakes.Homework:

Given p(x) = x+4

x− 5, answer the following questions:

a) Simplify the expression.b) Is p(x) a polynomial?c) What is the coefficient of the x term?

Peter’s answers:

a)p(x) = x+ 4

x− 5 (restrictions: x 6= 0)

= x2 + 4− 5x (multiply through by x)= x2 − 5x+ 4 (write in descending order)= (x− 1)(x+ 4) (factorise, quadratic has two roots)

233Chapter 5. Polynomials

b) Yes, because it can be simplified to have exponents that are all natural numbers. Itis a quadratic binomial because the highest exponent is two and there are only twoterms; (x− 1) and (x+ 4).

c) Before I simplified, the coefficient of the x term was nothing and after I simplified itbecame 5.

Solution:Peter made the following mistakes:

a) This is not an equation, therefore Peter must not multiply through by x. Peter did notfactorise correctly, he made an error with the sign in the second bracket .

p(x) = x+ 4x− 5 (restrictions: x 6= 0)

= x2+4−5xx

(common denominator is x)

= x2−5x+4x

(write in standard form)

= (x−1)(x−4)x

(factorise and be careful of the signs)

b) The original expression is not a polynomial because the exponent of the second term�4x

= 4x−1)

is not a natural number. Peter’s simplified expression is a quadratictrinomial because it has three terms and two factors.

c) In the original expression, the coefficient of the x term is 1. After Peter wronglysimplified the expression, the coefficient of the x term was −5 and not +5. Al-ways remember that the coefficient includes the sign and the number in front of thevariable.

Check answers online with the exercise code below or click on ’show me the answer’.1. 28R3 2. 28R4 3a. 28R5 3b. 28R6 3c. 28R7 3d. 28R8

3e. 28R9 3f. 28RB 3g. 28RC 3h. 28RD 3i. 28RF 3j. 28RG4. 28RH

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Quadratic polynomials

Exercise 5 – 2: Quadratic polynomials

1. Solve the following quadratic equations by factorisation. Answers may be left in surd form,where applicable.

a) 7p2 + 14p = 0

Solution:

7p2 + 14p = 0

7p(p+ 2) = 0

p(p+ 2) = 0

p = 0 or p = −2

b) k2 + 5k − 36 = 0

Solution:

k2 + 5k − 36 = 0

(k − 4)(k + 9) = 0

k = 4 or k = −9

234 5.1. Revision

c) 400 = 16h2

Solution:

16h2 − 400 = 0

16(h2 − 25) = 0

(h− 5)(h+ 5) = 0

h = ±5

d) (x− 1)(x+ 10) + 24 = 0

Solution:

(x− 1)(x+ 10) + 24 = 0

x2 + 9x− 10 + 24 = 0

x2 + 9x+ 14 = 0

(x+ 7)(x+ 2) = 0

x = −7 or x = −2

e) y2 − 5ky + 4k2 = 0

Solution:

y2 − 5ky + 4k2 = 0

(y − 4k)(y − k) = 0

y = 4k or y = k

2. Solve the following equations by completing the square:

a) p2 + 10p− 2 = 0

Solution:

p2 + 10p− 2 = 0

p2 + 10p = 2

p2 + 10p+ 25 = 2 + 25

(p+ 5)2 − 27 = 0

(p+ 5)2 = 27

∴ (p+ 5) = ±√

27

∴ (p+ 5) = −√

27 or (p+ 5) =√

27

∴ p = −5− 3√

3 or p = −5 + 3√

3

b) 2(6y + y2) = −4

Solution:

y2 + 6y = −2

y2 + 6y + 9 = −2 + 9

(y + 3)2 = 7

y + 3 = ±√

7

y = −3±√

7

c) x2 + 5x+ 9 = 0

235Chapter 5. Polynomials

Solution:

x2 + 5x+ 9 = 0

x2 + 5x = −9

x2 + 5x+25

4= −9 +

25

4(x+

5

2

)2

= −11

4

x+5

2= ±

√−11

4

No real solution

d) f2 + 30 = 2(10− 8f)

Solution:

f2 + 30 = 2(10− 8f)

f2 + 16f + 10 = 0

f2 + 16f = −10

f2 + 16f + 64 = −10 + 64

(f + 8)2 = 54

f + 8 = ±√

54

f = −8±√

9× 6

∴ f = −8± 3√

6

e) 3x2 + 6x− 2 = 0

Solution:

3x2 + 6x− 2 = 0

Divide through by 3 to get leading coefficient of 1

x2 + 2x =2

3

x2 + 2x+ 1 =2

3+ 1

(x+ 1)2 =5

3

x+ 1 = ±√

5

3

x = −1±√

5

3

3. Solve the following using the quadratic formula.

a) 3m2 +m− 4 = 0

Solution:

a = 3; b = 1; c = −4

236 5.1. Revision

3m2 +m− 4 = 0

m =−(1)±

√12 − 4(3)(−4)

2(3)

=−1±

√1 + 48

6

=−1±

√49

6

=−1± 7

6

m =−1 + 7

6=

6

6= 1 or m =

−1− 7

6=−8

6= −4

3

Note: it would be easier and faster to solve by factorisation rather than using thequadratic formula. If the question does not specify which method to use, first try tosolve by factorisation.

b) 2t2 + 6t+ 5 = 0

Solution:

2t2 + 6t+ 5 = 0

t =−6±

√(6)2 − 4(2)(5)

2(2)

=−6±

√36− 40

4

=−6±

√−4

4No real solution

c) y2 − 4y + 2 = 0

Solution:

y2 − 4y + 2 = 0

y =−(−4)±

√(−4)2 − 4(1)(2)

2(1)

=4±√

16− 8

2

=4±√

8

2

= 2±√

2

∴ y = 2 +√

2 or y = 2−√

2

d) 3f − 2 = −2f2

Solution:

2f2 + 3f − 2 = 0

f =−3±

√32 − 4(2)(−2)

2(2)

=−3±

√9 + 16

4

=−3±

√25

4

=−3± 5

4

therefore f =−3 + 5

4=

1

2or f =

−3− 5

4= −2

237Chapter 5. Polynomials

4. Factorise the following:

a) 27p3 − 1

Solution:

27p3 − 1 = (3p− 1)(9p2 + 3p+ 1)

b) 16 + 2x3

Solution:

16 +2

x3= 2

(8 +

1

x3

)= 2

(2 +

1

x

)(4− 2

x+

1

x2

)

Check answers online with the exercise code below or click on ’show me the answer’.1a. 28RJ 1b. 28RK 1c. 28RM 1d. 28RN 1e. 28RP 2a. 28RQ2b. 28RR 2c. 28RS 2d. 28RT 2e. 28RV 3a. 28RW 3b. 28RX3c. 28RY 3d. 28RZ 4a. 28S2 4b. 28S3

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5.2 Cubic polynomials

Exercise 5 – 3: Cubic polynomials

1. Factorise the following:

a) p3 − 1

Solution:

p3 − 1 = (p− 1)(p2 + p+ 1)

b) t3 + 27

Solution:

t3 + 27 = (t+ 3)(t2 + 3t+ 9)

c) 64−m3

Solution:

64−m3 = (4−m)(16 + 4m+m2)

d) k − 125k4

Solution:

k − 125k4 = k(1− 125k3)

= k(1− 5k)(1 + 5k + 25k2)

e) 8a6 − b9

238 5.2. Cubic polynomials

Solution:

8a6 − b9 =�2a2)3 − �b3)3

=�2a2 − b3

) �4a4 + 2a2b3 + b6

)f) 8− (p+ q)3

Solution:

8− (p+ q)3 = [2− (p+ q)][4 + 2(p+ q) + (p+ q)2]

= (2− p− q)(4 + 2p+ 2q + p2 + 2pq + q2)

= (2− p− q)(4 + 2p+ 2q + p2 + 2pq + q2)

2. For each of the following:

a) Use long division to determine the quotient Q(x) and the remainder R(x).b) Write a(x) in the form a(x) = b(x) . Q(x) +R(x).

c) Check your answer by expanding the brackets to get back to the original cubic poly-nomial.

a) a(x) = x3 + 2x2 + 3x+ 7 is divided by (x+ 1)

Solution:

x2 + x+ 2

x+ 1|x3 + 2x2 + 3x+ 7

−�x3 + x2)

0 + x2 + 3x

−�x2 + x

)0 + 2x+ 7

− (2x+ 2)

0 + 5

Q(x) = x2 + x+ 2

R(x) = 5

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x+ 1)(x2 + x+ 2) + 5

Check:

(x+ 1)(x2 + x+ 2) + 5 = x3 + x2 + 2x+ x2 + x+ 2 + 5

= x3 + 2x2 + 3x+ 7

b) a(x) = 1 + 4x2 − 5x− x3 and b(x) = x+ 2

Solution:a(x) = −x3 + 4x2 − 5x+ 1 and b(x) = x+ 2

− x2 + 6x− 17

x+ 2|−x3 + 4x2 − 5x+ 1

−�−x3 − 2x2)

0 + 6x2 − 5x

−�6x2 + 12x

)0− 17x+ 1

− (−17x− 34)

0 + 35

239Chapter 5. Polynomials

Q(x) = −x2 + 6x− 17

R(x) = 35

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x+ 2)(−x2 + 6x− 17) + 35

Check:

(x+ 2)(−x2 + 6x− 17) + 35 = −x3 + 6x2 − 17x− 2x2 + 12x− 34 + 35

= −x3 + 4x2 − 5x+ 1

c) a(x) = 2x3 + 3x2 + x− 6 and b(x) = x− 1

Solution:

2x2 + 5x+ 6

x− 1|2x3 + 3x2 + x− 6

−�2x3 − 2x2)

0 + 5x2 + x

−�5x2 − 5x

)0 + 6x− 6

− (6x− 6)

0 + 0

The remainder is equal to zero, therefore b(x) is a factor of a(x).

Q(x) = 2x2 + 5x+ 6

R(x) = 0

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x− 1)(2x2 + 5x+ 6)

Check:

(x− 1)(2x2 + 5x+ 6) = 2x3 + 5x2 + 6x− 2x2 − 5x− 6

= 2x3 + 3x2 + x− 6

d) a(x) = x3 + 2x2 + 5 and b(x) = x− 1

Solution:

x2 + 3x+ 3

x− 1|x3 + 2x2 + 5

−�x3 − x2)

0 + 3x2 + 0x

−�3x2 − 3x

)0 + 3x+ 5

− (3x− 3)

0 + 8

Q(x) = x2 + 3x+ 3

R(x) = 8

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x− 1)(x2 + 3x+ 3) + 8

240 5.2. Cubic polynomials

Check:

(x− 1)(x2 + 3x+ 3) + 8 = x3 + 3x2 + 3x− x2 − 3x− 3 + 8

= x3 + 2x2 + 5

e) (x− 1) is divided into a(x) = x4 + 2x3 − 3x2 + 5x+ 4

Solution:

x3 + 3x2 + 0x+ 5

x− 1|x4 + 2x3 − 3x2 + 5x+ 4

−�x4 − x3)

0 + 3x3 − 3x2

−�3x3 − 3x2)0 + 0 + 5x+ 4

− (5x− 5)

0 + 9

Q(x) = x3 + 3x2 + 5

R(x) = 9

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x− 1)(x3 + 3x2 + 5) + 9

Check:

(x− 1)(x3 + 3x2 + 5) + 9 = x4 + 3x3 + 5x− x3 − 3x2 − 5 + 9

= x4 + 2x3 − 3x2 + 5x+ 4

f) a(x)b(x)

= 5x4+3x3+6x2+x+2x2−2

Solution:

5x2 + 3x+ 16

x2 + 0x− 2|5x4 + 3x3 + 6x2 + x+ 2

−�5x4 − 0x3 − 10x2)

0 + 3x3 + 16x2 + x

−�3x3 + 0x2 − 6x

)0 + 16x2 + 7x+ 2

−�16x2 + 0x− 32

)7x+ 34

Q(x) = 5x2 + 3x+ 16

R(x) = 7x+ 34

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x2 − 2)(5x2 + 3x+ 16) + 7x+ 34

Check:

(x2 − 2)(5x2 + 3x+ 16) + 7x+ 34

= 5x4 + 3x3 + 16x2 − 10x2 − 6x− 32 + 7x+ 34

= 5x4 + 3x3 + 6x2 + x+ 2

241Chapter 5. Polynomials

g) a(x) = 3x3 − x2 + 2x+ 1 is divided by (3x− 1)

Solution:

x2 +2

3

3x− 1|3x3 − x2 + 2x+ 1

−�3x3 − x2)

0 + 0 + 2x+ 1

−(

2x− 2

3

)0 +

5

3

Q(x) = x2 +2

3

R(x) =5

3and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (3x− 1)

(x2 +

2

3

)+

5

3

Check:

(3x− 1)

(x2 +

2

3

)+

5

3= 3x3 + 2x− x2 − 2

3+

5

3

= 3x3 − x2 + 2x+ 1

h) a(x) = 2x5 + x3 + 3x2 − 4 and b(x) = x+ 2

Solution:

2x4 − 4x3 + 9x2 − 15x+ 30

x+ 2|2x5 + 0x4 + x3 + 3x2 + 0x− 4

−�2x5 + 4x4)

0− 4x4 + x3

−�−4x4 − 8x3)0 + 9x3 + 3x2

−�9x3 + 18x2)0− 15x2 + 0x

−�−15x2 − 30x

)0 + 30x− 4

− (30x+ 60)

0− 64

Q(x) = 2x4 − 4x3 + 9x2 − 15x+ 30

R(x) = −64

and a(x) = b(x) . Q(x) +R(x)

∴ a(x) = (x+ 2)(2x4 − 4x3 + 9x2 − 15x+ 30)− 64

Check:

(x+ 2)(2x4 − 4x3 + 9x2 − 15x+ 30)− 64

= 2x5 − 4x4 + 9x3 − 15x2 + 30x+ 4x4 − 8x3 + 18x2 − 30x+ 60− 64

= 2x5 + x3 + 3x2 − 4

242 5.2. Cubic polynomials

3. Use synthetic division to determine the quotient Q(x) and the remainder R(x) when f(x)is divided by g(x):

a)

f(x) = x2 + 5x+ 1

g(x) = x+ 2

Solution:

1 3 − 5

−2|1 5 1

q1 = 1

q0 = 5 + (−2)(1) = 3

R = 1 + (−2)(3) = −5

Q(x) = x+ 3

R(x) = −5

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x+ 2)(x+ 3)− 5

b)

f(x) = x2 − 5x− 7

g(x) = x− 1

Solution:

1 − 4 − 11

1|1 − 5 − 7

q1 = 1

q0 = −5 + (1)(1) = −4

R = −7 + (−4)(1) = −11

Q(x) = x− 4

R(x) = −11

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x− 1)(x− 4)− 11

c)

f(x) = 2x3 + 5x− 4

g(x) = x− 1

Solution:

2 2 7 3

1|2 0 5 − 4

q2 = 2

q1 = 0 + (2)(1) = 2

q0 = 5 + (2)(1) = 7

R = −4 + (7)(1) = 3

Q(x) = 2x2 + 2x+ 7

R(x) = 3

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x− 1)(2x2 + 2x+ 7) + 3

243Chapter 5. Polynomials

d)

f(x) = 19 + x2 + 8x

g(x) = x+ 3

Solution:

1 5 4

−3|1 8 19

q1 = 1

q0 = 8 + (−3)(1) = 5

R = 19 + (5)(−3) = 4

Q(x) = x+ 5

R(x) = 4

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x+ 3)(x+ 5) + 4

e)

f(x) = x3 + 2x2 + x− 10

g(x) = x− 1

Solution:

1 3 4 − 6

1|1 2 1 − 10

q2 = 1

q1 = 2 + (1)(1) = 3

q0 = 1 + (1)(3) = 4

R = −10 + (1)(4) = −6

Q(x) = x2 + 3x+ 4

R(x) = −6

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x− 1)(x2 + 3x+ 4)− 6

f)

f(x) = 2x3 + 7x2 + 2x− 3

g(x) = x+ 3

Solution:

2 1 − 1 + 0

−3|2 7 2 − 3

q2 = 2

q1 = 7 + (−3)(2) = 1

q0 = 2 + (−3)(1) = −1

R = −3 + (−3)(−1) = 0

Q(x) = 2x2 + x− 1

R(x) = 0

and f(x) = g(x) . Q(x) +R(x)

∴ f(x) = (x+ 3)(2x2 + x− 1)

244 5.2. Cubic polynomials

g)

f(x) = 4x3 + 4x2 − x− 2

g(x) = 2x− 1

= 2

(x− 1

2

)

Solution:

4 6 2 − 1

1

2|4 4 − 1 − 2

q2 = 4

q1 = 4 +

(1

2

)(4) = 6

q0 = −1 +

(1

2

)(6) = 2

R = −2 +

(1

2

)(2) = −1

Q(x) = 4x2 + 6x+ 2

R(x) = −1

and f(x) =1

2g(x) . Q(x) +R(x)

=1

2. 2

(x− 1

2

)(4x2 + 6x+ 2)− 1

=

(x− 1

2

)(2)(2x2 + 3x+ 1)− 1

∴ f(x) = (2x− 1) (2x2 + 3x+ 1)− 1

h)

f(x) = 5x+ 22 + 2x3 + x2

g(x) = 2x+ 3

= 2

(x+

3

2

)

Solution:

f(x) = 2x3 + x2 + 5x+ 22

g(x) = 2x+ 3

2 − 2 8 10

−3

2|2 1 5 22

q2 = 2

q1 = 1 +

(−3

2

)(2) = −2

q0 = 5 +

(−3

2

)(−2) = 8

R = 22 +

(−3

2

)(8) = 10

245Chapter 5. Polynomials

Q(x) = 2x2 − 2x+ 8

R(x) = 10

and f(x) =1

2g(x) . Q(x) +R(x)

=1

2. 2

(x+

3

2

)(2x2 − 2x+ 8) + 10

= 2

(x+

3

2

)(x2 − x+ 4) + 10

∴ f(x) = (2x+ 3) (x2 − x+ 4) + 10

Check answers online with the exercise code below or click on ’show me the answer’.1a. 28S7 1b. 28S8 1c. 28S9 1d. 28SB 1e. 28SC 1f. 28SD2a. 28SF 2b. 28SG 2c. 28SH 2d. 28SJ 2e. 28SK 2f. 28SM2g. 28SN 2h. 28SP 3a. 28SQ 3b. 28SR 3c. 28SS 3d. 28ST3e. 28SV 3f. 28SW 3g. 28SX 3h. 28SY

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5.3 Remainder theorem

Exercise 5 – 4: Remainder theorem

1. Use the remainder theorem to determine the remainder R when g(x) is divided by h(x):

a)

g(x) = x3 + 4x2 + 11x− 5

h(x) = x− 1

Solution:

g(x) = x3 + 4x2 + 11x− 5

g(1) = (1)3 + 4(1)2 + 11(1)− 5

= 1 + 4 + 11− 5

∴ R = 11

b)

g(x) = 2x3 − 5x2 + 8

h(x) = 2x− 1

Solution:

g(x) = 2x3 − 5x2 + 8

g

(1

2

)= 2

(1

2

)3

− 5

(1

2

)2

+ 8

= 2

(1

8

)− 5

(1

4

)+ 8

=1

4− 5

4+ 8

∴ R = 7

246 5.3. Remainder theorem

c)

g(x) = 4x3 + 5x2 + 6x− 1

h(x) = x+ 2

Solution:

g(x) = 4x3 + 5x2 + 6x− 1

g(−2) = 4(−2)3 + 5(−2)2 + 6(−2)− 1

= −32 + 20− 12− 1

∴ R = −25

d)

g(x) = −5x3 − x2 − 10x+ 9

h(x) = 5x+ 1

Solution:

g(x) = −5x3 − x2 − 10x+ 9

g

(−1

5

)= −5

(−1

5

)3

−(−1

5

)2

− 10

(−1

5

)+ 9

= −5

(− 1

125

)−(

1

25

)+ 2 + 9

=1

25− 1

25+ 11

∴ R = 11

e)

g(x) = x4 + 5x2 + 2x− 8

h(x) = x+ 1

Solution:

g(x) = x4 + 5x2 + 2x− 8

g(−1) = (−1)4 + 5(−1)2 + 2(−1)− 8

= 1 + 5− 2− 8

= 6− 10

∴ R = −4

f)

g(x) = 3x5 − 8x4 + x2 + 2

h(x) = 2− x

Solution:

h(x) = 2− xIf x = 2 : h(2) = 2− 2 = 0

g(x) = h(x) . Q(x) +R

g(2) = h(2) . Q(2) +R

∴ g(2) = 0 . Q(2) +R

g(2) = 0 +R

∴ R = 3(2)5 − 8(2)4 + (2)2 + 2

= 3(32)− 8(16) + 6

= 96− 128 + 6

∴ R = −26

247Chapter 5. Polynomials

g)

g(x) = 2x100 − x− 1

h(x) = x+ 1

Solution:

g(x) = 2x100 − x− 1

g(−1) = 2(−1)100 − (−1)− 1

= 2 + 1− 1

∴ R = 2

2. Determine the value of t if x3 + tx2 + 8x+ 21 divided by x+ 1 gives a remainder of 16.

Solution:

a(x) = x3 + tx2 + 8x+ 21

b(x) = x+ 1

R = 16

a(−1) = (−1)3 + t(−1)2 + 8(−1) + 21

∴ 16 = −1 + t− 8 + 21

∴ t = 4

3. Calculate the value of m if 2x3−7x2 +mx−26 is divided by x−2 and gives a remainderof −24.

Solution:

a(x) = 2x3 − 7x2 +mx− 26

b(x) = x− 2

R = −24

a(2) = 2(2)3 − 7(2)2 +m(2)− 26

∴ −24 = 16− 28 + 2m− 26

14 = 2m

∴ m = 7

4. If x5 − 2x3 − kx− 1 is divided by x− 1 and the remainder is − 12, find the value of k .

Solution:

a(x) = x5 − 2x3 − kx− 1

b(x) = x− 1

R = −1

2

a(1) = (1)5 − 2(1)3 − k(1)− 1

∴ −1

2= 1− 2− k − 1

3

2= −k

∴ k = −3

2

5. Determine the value of p if 18x3 +px2−8x+9 is divided by 2x−1 and gives a remainderof 6.

248 5.3. Remainder theorem

Solution:

a(x) = 18x3 + px2 − 8x+ 9

b(x) = 2x− 1

R = 6

a

(1

2

)= 18

(1

2

)3

+ p

(1

2

)2

− 8

(1

2

)+ 9

∴ 6 = 18

(1

8

)+ p

(1

4

)− 4 + 9

6 =18

8+p

4+ 5

1 =18

8+p

44 = 9 + p

∴ p = −5

6. If x3 + x2 − x+ b is divided by x− 2 and the remainder is 2 12, calculate the value of b.

Solution:

Let f(x) = x3 + x2 − x+ b

R =5

2

f(2) = (2)3 + (2)2 − (2) + b

∴5

2= 8 + 4− 2 + b

5

2− 10 = b

∴ b = −15

2

7. Calculate the value of h if 3x5 + hx4 + 10x2 − 21x + 12 is divided by x − 2 and gives aremainder of 10.

Solution:

a(x) = 3x5 + hx4 + 10x2 − 21x+ 12

b(x) = x− 2

R = 10

a(2) = 3(2)5 + h(2)4 + 10(2)2 − 21(2) + 12

∴ 10 = 3(32) + 16h+ 40− 42 + 12

10 = 96 + 16h+ 10

−96 = 16h

∴ h = −6

8. If x3 + 8x2 +mx− 5 is divided by x+ 1 and the remainder is n, express m in terms of n.

Solution:

a(x) = x3 + 8x2 +mx− 5

b(x) = x+ 1

R = n

a(−1) = (−1)3 + 8(−1)2 +m(−1)− 5

∴ n = −1 + 8−m− 5

n = 2−m∴ m = 2− n

249Chapter 5. Polynomials

9. When the polynomial 2x3 + px2 + qx+ 1 is divided by x+ 1 or x− 4, the remainder is5. Determine the values of p and q.

Solution:

a(x) = 2x3 + px2 + qx+ 1

a(−1) = 2(−1)3 + p(−1)2 + q(−1) + 1

∴ 5 = −2 + p− q + 1

6 = p− q∴ q = p− 6 . . . . . . (1)

a(4) = 2(4)3 + p(4)2 + q(4) + 1

∴ 5 = 128 + 16p+ 4q + 1

−124 = 16p+ 4q

∴ −31 = 4p+ q . . . . . . (2)

Substitute eqn (1) into eqn (2)

∴ −31 = 4p+ (p− 6)

−25 = 5p

∴ −5 = p

And q = −5− 6

= −11

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5.4 Factor theorem

Exercise 5 – 5: Factorising cubic polynomials

1. Find the remainder when 4x3 − 4x2 + x− 5 is divided by x+ 1.

Solution:

Let a(x) = 4x3 − 4x2 + x− 5

a(−1) = 4(−1)3 − 4(−1)2 + (−1)− 5

= −4− 4− 1− 5

= −14

2. Use the factor theorem to factorise x3 − 3x2 + 4 completely.

Solution:

250 5.4. Factor theorem

Let a(x) = x3 − 3x2 + 4

a(−1) = (−1)3 − 3(−1)2 + 4

= −1− 3 + 4

= 0

∴ (x+ 1) is a factor

a(x) = (x+ 1)(x2 − 4x+ 4)

= (x+ 1)(x− 2)(x− 2)

= (x+ 1)(x− 2)2

3. f (x) = 2x3 + x2 − 5x+ 2

a) Find f (1).

Solution:

f(x) = 2x3 + x2 − 5x+ 2

f(1) = 2(1)3 + (1)2 − 5(1) + 2

= 2 + 1− 5 + 2

= 0

b) Factorise f (x) completely.

Solution:

f(1) = 0

∴ (x− 1) is a factor of f(x)

f(x) = (x− 1)(2x2 + 3x− 2)

= (x− 1)(2x− 1)(x+ 2)

4. Use the factor theorem to determine all the factors of the following expression:

x3 + x2 − 17x+ 15

Solution:

Let a(x) = x3 + x2 − 17x+ 15

a(1) = (1)3 + (1)2 − 17(1) + 15

= 1 + 1− 17 + 15

= 0

∴ a(x) = (x− 1)(x2 + 2x− 15)

= (x− 1)(x+ 5)(x− 3)

5. Complete: If f (x) is a polynomial and p is a number such that f (p) = 0, then (x− p)is...

Solution:a factor of f(x)

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251Chapter 5. Polynomials

5.5 Solving cubic equations

Exercise 5 – 6: Solving cubic equations

Solve the following cubic equations:

1. x3 + x2 − 16x = 16

Solution:

x3 + x2 − 16x = 16

x3 + x2 − 16x− 16 = 0

Let a(x) = x3 + x2 − 16x− 16

a(−1) = (−1)3 + (−1)2 − 16(−1)− 16

= −1 + 1 + 16− 16

= 0

∴ a(x) = (x+ 1)(x2 − 16)

= (x+ 1)(x− 4)(x+ 4)

∴ 0 = (x+ 1)(x− 4)(x+ 4)

∴ x = −1 or x = 4 or x = −4

2. −n3 − n2 + 22n+ 40 = 0

Solution:

n3 + n2 − 22n− 40 = 0

Let a(n) = n3 + n2 − 22n− 40

a(−2) = (−2)3 + (−2)2 − 22(−2)− 40

= −8 + 4 + 44− 40

= 0

∴ a(n) = (n+ 2)(n2 − n− 20)

= (n+ 2)(n− 5)(n+ 4)

∴ 0 = (n+ 2)(n− 5)(n+ 4)

∴ n = −2 or n = −4 or n = 5

3. y(y2 + 2y) = 19y + 20

Solution:

y(y2 + 2y) = 19y + 20

y3 + 2y2 − 19y − 20 = 0

Let a(y) = y3 + 2y2 − 19y − 20

a(−1) = (−1)3 + 2(−1)2 − 19(−1)− 20

= −1 + 2 + 19− 20

= 0

∴ a(y) = (y + 1)(y2 + y − 20)

= (y + 1)(y + 5)(y − 4)

∴ 0 = (y + 1)(y + 5)(y − 4)

∴ y = −1 or y = 4 or y = −5

4. k3 + 9k2 + 26k + 24 = 0

252 5.5. Solving cubic equations

Solution:

Let a(k) = k3 + 9k2 + 26k + 24

a(−2) = (−2)3 + 9(−2)2 + 26(−2) + 24

= −8 + 36− 52 + 24

= 0

∴ a(k) = (k + 2)(k2 + 7k + 12)

= (k + 2)(k + 3)(k + 4)

∴ 0 = (k + 2)(k + 3)(k + 4)

∴ k = −2 or k = −3 or k = −4

5. x3 + 2x2 − 50 = 25x

Solution:

x3 + 2x2 − 50 = 25x

x3 + 2x2 − 25x− 50 = 0

Let a(x) = x3 + 2x2 − 25x− 50

a(−2) = (−2)3 + 2(−2)2 − 25(−2)− 50

= −8 + 8 + 50− 50

= 0

∴ a(x) = (x+ 2)(x2 − 25)

= (x+ 2)(x− 5)(x+ 5)

∴ 0 = (x+ 2)(x− 5)(x+ 5)

∴ x = −2 or x = 5 or x = −5

6. −p3 + 19p = 30

Solution:

−p3 + 19p− 30 = 0

p3 − 19p+ 30 = 0

Let a(p) = p3 − 19p+ 30

a(3) = (3)3 − 19(3) + 30

= 27− 57 + 30

= 0

∴ a(p) = (p− 3)(p2 + 3p− 10)

= (p− 3)(p− 2)(p+ 5)

∴ 0 = (p− 3)(p− 2)(p+ 5)

∴ p = 3 or p = 2 or p = −5

7. 6x2 − x3 = 5x+ 12

Solution:

0 = x3 − 6x2 + 5x+ 12

Let a(x) = x3 − 6x2 + 5x+ 12

a(−1) = (−1)3 − 6(−1)2 + 5(−1) + 12

= −1− 6− 5 + 12

= 0

∴ a(x) = (x+ 1)(x2 − 7x+ 12)

= (x+ 1)(x− 3)(x− 4)

∴ 0 = (x+ 1)(x− 3)(x− 4)

∴ x = −1 or x = 3 or x = 4

253Chapter 5. Polynomials

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5.6 Summary

Exercise 5 – 7: End of chapter exercises

1. Solve for x: x3 + x2 − 5x+ 3 = 0

Solution:

Let a(x) = x3 + x2 − 5x+ 3

a(1) = (1)3 + (1)2 − 5(1) + 3

= 0

∴ a(x) = (x− 1)(x2 + 2x− 3)

= (x− 1)(x+ 3)(x− 1)

= (x− 1)2(x+ 3)

∴ 0 = (x− 1)2(x+ 3)

∴ x = 1 or x = −3

2. Solve for y: y3 = 3y2 + 16y + 12

Solution:

Let a(y) = y3 − 3y2 − 16y − 12

a(−1) = (−1)3 − 3(−1)2 − 16(−1)− 12

= −1− 3 + 16− 12

= 0

∴ a(y) = (y + 1)(y2 − 4y − 12)

= (y + 1)(y − 6)(y + 2)

∴ 0 = (y + 1)(y − 6)(y + 2)

∴ y = −1 or y = 6 or y = −2

3. Solve for m: m(m2 −m− 4) = −4

Solution:

Let a(m) = m3 −m2 − 4m+ 4

a(1) = (1)3 − (1)2 − 4(1) + 4

= 1− 1− 4 + 4

= 0

∴ a(m) = (m− 1)(m2 − 4)

= (m− 1)(m+ 2)(m− 2)

∴ 0 = (m− 1)(m+ 2)(m− 2)

∴ m = 1 or m = 2 or m = −2

254 5.6. Summary

4. Solve for x: x3 − x2 = 3 (3x+ 2)

Solution:

x3 − x2 = 3 (3x+ 2)

x3 − x2 = 9x+ 6

x3 − x2 − 9x− 6 = 0

Let x = −2 : (−2)3 − (−2)2 − 9(−2)− 6

= −8− 4 + 18− 6

= 0

∴ (x+ 2) is a factor

(x+ 2)(x2 − 3x− 3) = 0

Using quadratic formula to solve for x : x2 − 3x− 3 = 0

a = 1; b = −3; c = −3

x =−b±

√b2 − 4ac

2a

=−(−3)±

√(−3)2 − 4(1)(−3)

2(1)

=3±√

9 + 12

2

=3±√

21

2

∴ x = −2 or x =3 +√

21

2or x =

3−√

21

2

5. Solve for x if 2x3 − 3x2 − 8x = 3.

Solution:

2x3 − 3x2 − 8x = 3

2x3 − 3x2 − 8x− 3 = 0

Let a(x) = 2x3 − 3x2 − 8x− 3

a(−1) = 2(−1)3 − 3(−1)2 − 8(−1)− 3

= −2− 3 + 8− 3

= 0

∴ a(x) = (x+ 1)(2x2 − 5x− 3)

= (x+ 1)(2x+ 1)(x− 3)

∴ 0 = (x+ 1)(2x+ 1)(x− 3)

∴ x = −1 or x = −1

2or x = 3

6. Solve for x: 16 (x+ 1) = x2 (x+ 1)

Solution:

255Chapter 5. Polynomials

16 (x+ 1) = x2 (x+ 1)

16x+ 16 = x3 + x2

0 = x3 + x2 − 16x− 16

Let a(x) = x3 + x2 − 16x− 16

a(−1) = (−1)3 + (−1)2 − 16(−1)− 16

= −1 + 1 + 16− 16

= 0

∴ a(x) = (x+ 1)(x2 − 16)

= (x+ 1)(x− 4)(x+ 4)

∴ 0 = (x+ 1)(x− 4)(x+ 4)

∴ x = −1 or x = 4 or x = −4

7. a) Show that x− 2 is a factor of 3x3 − 11x2 + 12x− 4.

Solution:

Let a(x) = 3x3 − 11x2 + 12x− 4

a(2) = 3(2)3 − 11(2)2 + 12(2)− 4

= 24− 44 + 24− 4

= 0

∴ (x− 2) is a factor of a(x)

b) Hence, by factorising completely, solve the equation:3x3 − 11x2 + 12x− 4 = 0

Solution:

3x3 − 11x2 + 12x− 4 = 0

(x− 2)(3x2 − 5x+ 2) = 0

∴ (x− 2)(3x− 2)(x− 1) = 0

∴ x = 2 or x =2

3or x = 1

8. 2x3 − x2 − 2x+ 2 = Q (x) . (2x− 1) +R for all values of x. What is the value of R?

Solution:

Let a(x) = 2x3 − x2 − 2x+ 2

R = a

(1

2

)= 2

(1

2

)3

−(

1

2

)2

− 2

(1

2

)+ 2

= 2

(1

8

)−(

1

4

)− 1 + 2

=1

4− 1

4+ 1

= 1

∴ R = 1

9. a) Use the factor theorem to solve the following equation for m:8m3 + 7m2 − 17m+ 2 = 0

Solution:

256 5.6. Summary

Let a(m) = 8m3 + 7m2 − 17m+ 2

a(1) = 8(1)3 + 7(1)2 − 17(1) + 2

= 8 + 7− 17 + 2

= 0

∴ a(m) = (m− 1)(8m2 + 15m− 2)

= (m− 1)(8m− 1)(m+ 2)

∴ 0 = (m− 1)(8m− 1)(m+ 2)

∴ m = 1 or m =1

8or m = −2

b) Hence, or otherwise, solve for x:23x+3 + 7.22x + 2 = 17.2x

Solution:

23x+3 + 7 . 22x + 2 = 17 . 2x

23x . 23 + 7 . 22x + 2 = 17 . 2x

8 . (2x)3 + 7 . (2x)2 − 17 . 2x + 2 = 0

which we can compare with a(m) = 8m3 + 7m2 − 17m+ 2

Let 2x = m

and from part (a) we know that m = 1 or m =1

8or m = −2

So 2x = 1

2x = 20

∴ x = 0

Or 2x =1

8

2x = 2−3

∴ x = −3

Or 2x = −2

∴ no solution

10. Find the value of R if x− 1 is a factor of h(x) = (x− 6) . Q(x) +R and Q(x) divided byx− 1 gives a remainder of 8.

Solution:

h(x) = (x− 6) . Q(x) +R

h(1) = (1− 6) . Q(1) +R

∴ 0 = −5 . Q(1) +R

And Q(1) = 8

0 = −5(8) +R

∴ R = 40

11. Determine the values of p for which the function

f (x) = 3p3 − (3p− 7)x2 + 5x− 3

leaves a remainder of 9 when it is divided by (x− p).

Solution:

257Chapter 5. Polynomials

f (x) = 3p3 − (3p− 7)x2 + 5x− 3

∴ f(p) = 3p3 − (3p− 7) p2 + 5p− 3

= 3p3 − 3p3 + 7p2 + 5p− 3

= 7p2 + 5p− 3

f(p) = 9

∴ 9 = 7p2 + 5p− 3

0 = 7p2 + 5p− 12

0 = (7p+ 12)(p− 1)

∴ p = −12

7or p = 1

Alternative (long) method:

We first take out the factor using long division:

(7− 3p)x+ (5 + 7p− 3p2)

(x− p)|(7− 3p)x2 + 5x+ (3p3 − 3)

− {(7− 3p)x2 − p(7− 3p)x}

0 + 5x+ p(7− 3p)x+ (3p3 − 3)

5x+ 7px− 3p2x+ 3p3x

[5 + 7p− 3p2]x+ 3p3 − 3

− {[5 + 7p− 3p2]x− p(5 + 7p− 3p2)}

0 + 3p3 − 3 + 5p+ 7p2 − 3p3

We take the remainder and set it equal to 9:

−3 + 5p+ 7p2 = 9

7p2 + 5p− 12 = 0

(7p+ 12)(p− 1) = 0

∴ p = −12

7or p = 1

12. Calculate t and Q(x) if x2 + tx+ 3 = (x+ 4) . Q(x)− 17.

Solution:

x2 + tx+ 20 = (x+ 4) . Q(x)

Let f(x) = x2 + tx+ 20

f(−4) = (−4)2 + t(−4) + 20

0 = 16− 4t+ 20

4t = 36

∴ t = 9

x2 + 9x+ 20 = (x+ 4) . Q(x)

(x+ 4)(x+ 5) = (x+ 4) . Q(x)

∴ Q(x) = x+ 5

Check answers online with the exercise code below or click on ’show me the answer’.1. 28V2 2. 28V3 3. 28V4 4. 28V5 5. 28V6 6. 28V77. 28V8 8. 28V9 9. 28VB 10. 28VC 11. 28VD 12. 28VF

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258 5.6. Summary