Std 9 Mathematics Part 1 · Chapter 3

3. Polynomials

Solutions to Practice Sets 3.1 to 3.6 and Problem Set 3, with synthetic division worked out. Maharashtra State Board (Balbharati), Standard Nine Mathematics Part 1 (Algebra).

Practice Set 3.1

1State whether the given algebraic expressions are polynomials? Justify.

An algebraic expression is a polynomial only if every power of the variable is a whole number (0, 1, 2, …).

(i) y + 1y — y + 1y = y + y⁻¹. The power −1 is not a whole number. Not a polynomial

(ii) 2 − 5√x — 2 − 5√x = 2 − 5x¹². The power 12 is not a whole number. Not a polynomial

(iii) x² + 7x + 9 — the powers 2, 1 and 0 are whole numbers. Polynomial

(iv) 2m⁻² + 7m − 5 — the power −2 is not a whole number. Not a polynomial

(v) 10 — 10 = 10x⁰, and 0 is a whole number. Polynomial (a constant polynomial)

2Write the coefficient of m³ in each of the given polynomial.

(i) m³ — 1

(ii) −32 + m − √3m³ — −√3

(iii) −23m³ − 5m² + 7m − 1 — −23

3Write the polynomial in x using the given information.

Sample answers (any polynomial of the right type and degree is correct):

(i) Monomial with degree 7 — 2x⁷

(ii) Binomial with degree 35 — x³⁵ + 4

(iii) Trinomial with degree 8 — x⁸ + 3x² − 1

4Write the degree of the given polynomials.

(i) √5 — a non-zero constant. 0

(ii) x⁰ — x⁰ = 1, a non-zero constant. 0

(iii) x² — 2

(iv) √2m¹⁰ − 7 — 10

(v) 2p − √7 — 1

(vi) 7y − y³ + y⁵ — 5

(vii) xyz + xy − z — sums of powers in the terms are 3, 2 and 1. 3

(viii) m³n⁷ − 3m⁵n + mn — sums of powers are 3 + 7 = 10, 5 + 1 = 6 and 1 + 1 = 2. 10

5Classify the following polynomials as linear, quadratic and cubic polynomial.

Degree 1 → linear, degree 2 → quadratic, degree 3 → cubic.

(i) 2x² + 3x + 1 — Quadratic
(ii) 5p — Linear
(iii) √2y − 12 — Linear
(iv) m³ + 7m² + 52m − √7 — Cubic
(v) a² — Quadratic
(vi) 3r³ — Cubic

6Write the following polynomials in standard form.

Standard form: terms in descending powers of the variable.

(i) m³ + 3 + 5m = m³ + 5m + 3

(ii) −7y + y⁵ + 3y³ − 12 + 2y⁴ − y² = y⁵ + 2y⁴ + 3y³ − y² − 7y − 12

7Write the following polynomials in coefficient form.

(i) x³ − 2 = x³ + 0x² + 0x − 2 ⇒ (1, 0, 0, −2)

(ii) 5y = 5y + 0 ⇒ (5, 0)

(iii) 2m⁴ − 3m² + 7 = 2m⁴ + 0m³ − 3m² + 0m + 7 ⇒ (2, 0, −3, 0, 7)

(iv) −23 — a constant polynomial. (−23)

8Write the polynomials in standard form.

(i) (1, 2, 3) — x² + 2x + 3

(ii) (5, 0, 0, 0, −1) — 5x⁴ + 0x³ + 0x² + 0x − 1 = 5x⁴ − 1

(iii) (−2, 2, −2, 2) — −2x³ + 2x² − 2x + 2

9Write the appropriate polynomials in the boxes. (x + 7, x², x³ + x² + x + 5, 2x² + 5x + 10, x³ + 9, 3x² + 5x)

Box	Polynomials
Quadratic polynomial	x², 2x² + 5x + 10, 3x² + 5x
Cubic polynomial	x³ + x² + x + 5, x³ + 9
Linear polynomial	x + 7
Binomial	x + 7, x³ + 9, 3x² + 5x
Trinomial	2x² + 5x + 10
Monomial	x²

x³ + x² + x + 5 has four terms, so it goes only in the Cubic box, not in the Binomial, Trinomial or Monomial boxes.

Practice Set 3.2

1Use the given letters to write the answer.

(i) There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?
Increase in x years = b × x = bx. a + bx trees

(ii) For the parade there are y students in each row and x such rows are formed. How many students are there in all?
xy students

(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Value = 10 × (tens digit) + (units digit). 10m + n

2Add the given polynomials.

(i) x³ − 2x² − 9 ; 5x³ + 2x + 9
= (x³ + 5x³) − 2x² + 2x + (−9 + 9) = 6x³ − 2x² + 2x

(ii) −7m⁴ + 5m³ + √2 ; 5m⁴ − 3m³ + 2m² + 3m − 6
= (−7m⁴ + 5m⁴) + (5m³ − 3m³) + 2m² + 3m + (√2 − 6)
= −2m⁴ + 2m³ + 2m² + 3m + √2 − 6

(iii) 2y² + 7y + 5 ; 3y + 9 ; 3y² − 4y − 3
= (2y² + 3y²) + (7y + 3y − 4y) + (5 + 9 − 3) = 5y² + 6y + 11

3Subtract the second polynomial from the first.

(i) x² − 9x + √3 ; −19x + √3 + 7x²
(x² − 9x + √3) − (7x² − 19x + √3)
= x² − 9x + √3 − 7x² + 19x − √3
= −6x² + 10x

(ii) 2ab² + 3a²b − 4ab ; 3ab − 8ab² + 2a²b
(2ab² + 3a²b − 4ab) − (3ab − 8ab² + 2a²b)
= 2ab² + 8ab² + 3a²b − 2a²b − 4ab − 3ab
= 10ab² + a²b − 7ab

4Multiply the given polynomials.

(i) 2x ; x² − 2x − 1
2x(x² − 2x − 1) = 2x³ − 4x² − 2x

(ii) x⁵ − 1 ; x³ + 2x² + 2
= x⁵(x³ + 2x² + 2) − 1(x³ + 2x² + 2)
= x⁸ + 2x⁷ + 2x⁵ − x³ − 2x² − 2

(iii) 2y + 1 ; y² − 2y³ + 3y
= 2y(y² − 2y³ + 3y) + 1(y² − 2y³ + 3y)
= 2y³ − 4y⁴ + 6y² + y² − 2y³ + 3y
= −4y⁴ + 7y² + 3y

5Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor × Quotient + Remainder’.

(i) x³ − 64 ; x − 4

x³ − 64 = x²(x − 4) + 4x² − 64

= x²(x − 4) + 4x(x − 4) + 16x − 64

= x²(x − 4) + 4x(x − 4) + 16(x − 4)

Quotient = x² + 4x + 16 Remainder = 0

Dividend = Divisor × Quotient + Remainder:
x³ − 64 = (x − 4)(x² + 4x + 16)

(ii) 5x⁵ + 4x⁴ − 3x³ + 2x² + 2 ; x² − x

5x⁵ + 4x⁴ − 3x³ + 2x² + 2 = 5x³(x² − x) + 9x⁴ − 3x³ + 2x² + 2

= 5x³(x² − x) + 9x²(x² − x) + 6x³ + 2x² + 2

= 5x³(x² − x) + 9x²(x² − x) + 6x(x² − x) + 8x² + 2

= 5x³(x² − x) + 9x²(x² − x) + 6x(x² − x) + 8(x² − x) + 8x + 2

Quotient = 5x³ + 9x² + 6x + 8 Remainder = 8x + 2

Dividend = Divisor × Quotient + Remainder:
5x⁵ + 4x⁴ − 3x³ + 2x² + 2 = (x² − x)(5x³ + 9x² + 6x + 8) + 8x + 2

6*There is a rectangular farm with length (2a² + 3b²) metre and breadth (a² + b²) metre. The farmer used a square plot of side (a² − b²) metre to build a house. What is the area of the remaining part of the farm?

Area of farm = (2a² + 3b²)(a² + b²) = 2a⁴ + 2a²b² + 3a²b² + 3b⁴ = 2a⁴ + 5a²b² + 3b⁴

Area of plot = (a² − b²)² = a⁴ − 2a²b² + b⁴

Remaining area = (2a⁴ + 5a²b² + 3b⁴) − (a⁴ − 2a²b² + b⁴) = a⁴ + 7a²b² + 2b⁴

(a⁴ + 7a²b² + 2b⁴) sq. metre

Practice Set 3.3

1Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.

(i) (2m² − 3m + 10) ÷ (m − 5)

Synthetic division: standard form 2m² − 3m + 10 ⇒ coefficient form (2, −3, 10). Divisor m − 5, so write 5.

5	2	−3	10
		10	35
	2	7	45

Quotient = 2m + 7 Remainder = 45

Linear method:

2m² − 3m + 10 = 2m(m − 5) + 7m + 10

= 2m(m − 5) + 7(m − 5) + 45

∴ 2m² − 3m + 10 = (m − 5)(2m + 7) + 45

(ii) (x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)

Synthetic division: standard form x⁴ + 2x³ + 3x² + 4x + 5 ⇒ coefficient form (1, 2, 3, 4, 5). Divisor x + 2, so write −2.

−2	1	2	3	4	5
		−2	0	−6	4
	1	0	3	−2	9

Quotient = x³ + 3x − 2 Remainder = 9

Linear method:

x⁴ + 2x³ + 3x² + 4x + 5 = x³(x + 2) + 3x² + 4x + 5

= x³(x + 2) + 3x(x + 2) − 2x + 5

= x³(x + 2) + 3x(x + 2) − 2(x + 2) + 9

∴ x⁴ + 2x³ + 3x² + 4x + 5 = (x + 2)(x³ + 3x − 2) + 9

(iii) (y³ − 216) ÷ (y − 6)

Synthetic division: standard form y³ − 216 ⇒ coefficient form (1, 0, 0, −216). Divisor y − 6, so write 6.

6	1	0	0	−216
		6	36	216
	1	6	36	0

Quotient = y² + 6y + 36 Remainder = 0

Linear method:

y³ − 216 = y²(y − 6) + 6y² − 216

= y²(y − 6) + 6y(y − 6) + 36y − 216

= y²(y − 6) + 6y(y − 6) + 36(y − 6)

∴ y³ − 216 = (y − 6)(y² + 6y + 36)

(iv) (2x⁴ + 3x³ + 4x − 2x²) ÷ (x + 3)

Synthetic division: standard form 2x⁴ + 3x³ − 2x² + 4x ⇒ coefficient form (2, 3, −2, 4, 0). Divisor x + 3, so write −3.

−3	2	3	−2	4	0
		−6	9	−21	51
	2	−3	7	−17	51

Quotient = 2x³ − 3x² + 7x − 17 Remainder = 51

Linear method:

2x⁴ + 3x³ − 2x² + 4x = 2x³(x + 3) − 3x³ − 2x² + 4x

= 2x³(x + 3) − 3x²(x + 3) + 7x² + 4x

= 2x³(x + 3) − 3x²(x + 3) + 7x(x + 3) − 17x

= 2x³(x + 3) − 3x²(x + 3) + 7x(x + 3) − 17(x + 3) + 51

∴ 2x⁴ + 3x³ − 2x² + 4x = (x + 3)(2x³ − 3x² + 7x − 17) + 51

(v) (x⁴ − 3x² − 8) ÷ (x + 4)

Synthetic division: standard form x⁴ − 3x² − 8 ⇒ coefficient form (1, 0, −3, 0, −8). Divisor x + 4, so write −4.

−4	1	0	−3	0	−8
		−4	16	−52	208
	1	−4	13	−52	200

Quotient = x³ − 4x² + 13x − 52 Remainder = 200

Linear method:

x⁴ − 3x² − 8 = x³(x + 4) − 4x³ − 3x² − 8

= x³(x + 4) − 4x²(x + 4) + 13x² − 8

= x³(x + 4) − 4x²(x + 4) + 13x(x + 4) − 52x − 8

= x³(x + 4) − 4x²(x + 4) + 13x(x + 4) − 52(x + 4) + 200

∴ x⁴ − 3x² − 8 = (x + 4)(x³ − 4x² + 13x − 52) + 200

(vi) (y³ − 3y² + 5y − 1) ÷ (y − 1)

Synthetic division: standard form y³ − 3y² + 5y − 1 ⇒ coefficient form (1, −3, 5, −1). Divisor y − 1, so write 1.

1	1	−3	5	−1
		1	−2	3
	1	−2	3	2

Quotient = y² − 2y + 3 Remainder = 2

Linear method:

y³ − 3y² + 5y − 1 = y²(y − 1) − 2y² + 5y − 1

= y²(y − 1) − 2y(y − 1) + 3y − 1

= y²(y − 1) − 2y(y − 1) + 3(y − 1) + 2

∴ y³ − 3y² + 5y − 1 = (y − 1)(y² − 2y + 3) + 2

Practice Set 3.4

1For x = 0 find the value of the polynomial x² − 5x + 5.

p(0) = 0² − 5 × 0 + 5 = 0 − 0 + 5 ∴ p(0) = 5

2If p(y) = y² − 3√2y + 1 then find p(3√2).

p(3√2) = (3√2)² − 3√2 × 3√2 + 1 = 18 − 18 + 1 ∴ p(3√2) = 1

3If p(m) = m³ + 2m² − m + 10 then p(a) + p(−a) = ?

p(a) = a³ + 2a² − a + 10

p(−a) = (−a)³ + 2(−a)² − (−a) + 10 = −a³ + 2a² + a + 10

p(a) + p(−a) = a³ − a³ + 2a² + 2a² − a + a + 10 + 10

p(a) + p(−a) = 4a² + 20

4If p(y) = 2y³ − 6y² − 5y + 7 then find p(2).

p(2) = 2 × 2³ − 6 × 2² − 5 × 2 + 7 = 16 − 24 − 10 + 7 ∴ p(2) = −11

Practice Set 3.5

1Find the value of the polynomial 2x − 2x³ + 7 using given values for x.

(i) x = 3
p(3) = 2 × 3 − 2 × 3³ + 7 = 6 − 54 + 7 = −41

(ii) x = −1
p(−1) = 2(−1) − 2(−1)³ + 7 = −2 + 2 + 7 = 7

(iii) x = 0
p(0) = 0 − 0 + 7 = 7

2For each of the following polynomial, find p(1), p(0) and p(−2).

(i) p(x) = x³
p(1) = 1, p(0) = 0, p(−2) = (−2)³ = −8

(ii) p(y) = y² − 2y + 5
p(1) = 1 − 2 + 5 = 4
p(0) = 0 − 0 + 5 = 5
p(−2) = 4 + 4 + 5 = 13

(iii) p(x) = x⁴ − 2x² − x
p(1) = 1 − 2 − 1 = −2
p(0) = 0
p(−2) = 16 − 8 + 2 = 10

3If the value of the polynomial m³ + 2m + a is 12 for m = 2, then find the value of a.

p(2) = 2³ + 2 × 2 + a = 8 + 4 + a = 12 + a
But p(2) = 12 ∴ 12 + a = 12 ∴ a = 0

4For the polynomial mx² − 2x + 3 if p(−1) = 7 then find m.

p(−1) = m(−1)² − 2(−1) + 3 = m + 2 + 3 = m + 5
m + 5 = 7 ∴ m = 2

5Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.

By the remainder theorem, when p(x) is divided by (x + a) the remainder is p(−a); when divided by (x − a) it is p(a).

(i) (x² − 7x + 9) ; (x + 1)
Remainder = p(−1) = 1 + 7 + 9 = 17

(ii) (2x³ − 2x² + ax − a) ; (x − a)
Remainder = p(a) = 2a³ − 2a² + a × a − a = 2a³ − 2a² + a² − a = 2a³ − a² − a

(iii) (54m³ + 18m² − 27m + 5) ; (m − 3)
Remainder = p(3) = 54 × 27 + 18 × 9 − 27 × 3 + 5 = 1458 + 162 − 81 + 5 = 1544

6If the polynomial y³ − 5y² + 7y + m is divided by y + 2 and the remainder is 50 then find the value of m.

Remainder = p(−2) = (−2)³ − 5(−2)² + 7(−2) + m = −8 − 20 − 14 + m = m − 42
m − 42 = 50 ∴ m = 92

7Use factor theorem to determine whether x + 3 is a factor of x² + 2x − 3 or not.

p(−3) = (−3)² + 2(−3) − 3 = 9 − 6 − 3 = 0
Since p(−3) = 0, (x + 3) is a factor of x² + 2x − 3.

8If (x − 2) is a factor of x³ − mx² + 10x − 20 then find the value of m.

(x − 2) is a factor, so p(2) = 0.
2³ − m × 2² + 10 × 2 − 20 = 8 − 4m + 20 − 20 = 8 − 4m = 0 ∴ m = 2

9By using factor theorem, determine whether q(x) is a factor of p(x) or not.

(i) p(x) = x³ − x² − x − 1, q(x) = x − 1
p(1) = 1 − 1 − 1 − 1 = −2 ≠ 0 ∴ q(x) is not a factor of p(x)

(ii) p(x) = 2x³ − x² − 45, q(x) = x − 3
p(3) = 2 × 27 − 9 − 45 = 54 − 54 = 0 ∴ q(x) is a factor of p(x)

10If (x³¹ + 31) is divided by (x + 1) then find the remainder.

Remainder = p(−1) = (−1)³¹ + 31 = −1 + 31 (an odd power of −1 is −1)
Remainder = 30

11Show that m − 1 is a factor of m²¹ − 1 and m²² − 1.

Let p(m) = m²¹ − 1. Then p(1) = 1²¹ − 1 = 1 − 1 = 0.
Let q(m) = m²² − 1. Then q(1) = 1²² − 1 = 1 − 1 = 0.

Both values are 0, so by the factor theorem (m − 1) is a factor of m²¹ − 1 and of m²² − 1.

12*If x − 2 and x − 12 both are the factors of the polynomial nx² − 5x + m, then show that m = n = 2.

(x − 2) is a factor ⇒ p(2) = 0 ⇒ 4n − 10 + m = 0 ⇒ m + 4n = 10 …(I)

(x − 12) is a factor ⇒ p(12) = 0 ⇒ n4 − 52 + m = 0 ⇒ (multiply by 4) n − 10 + 4m = 0 ⇒ 4m + n = 10 …(II)

From (I), m = 10 − 4n. Put in (II): 4(10 − 4n) + n = 10 ⇒ 40 − 15n = 10 ⇒ 15n = 30 ⇒ n = 2

m = 10 − 4 × 2 = 2

m = n = 2 — shown.

13(i) If p(x) = 2 + 5x then find p(2) + p(−2) − p(1). (ii) If p(x) = 2x² − 5√3x + 5 then find p(5√3).

(i) p(2) = 12, p(−2) = 2 − 10 = −8, p(1) = 7
p(2) + p(−2) − p(1) = 12 − 8 − 7 = −3

(ii) p(5√3) = 2(5√3)² − 5√3 × 5√3 + 5 = 2 × 75 − 75 + 5 = 150 − 75 + 5 = 80

Practice Set 3.6

1Find the factors of the polynomials given below.

(i) 2x² + x − 1
= 2x² + 2x − x − 1 = 2x(x + 1) − 1(x + 1) = (x + 1)(2x − 1)

(ii) 2m² + 5m − 3
= 2m² + 6m − m − 3 = 2m(m + 3) − 1(m + 3) = (m + 3)(2m − 1)

(iii) 12x² + 61x + 77
12 × 77 = 924 = 28 × 33 and 28 + 33 = 61
= 12x² + 28x + 33x + 77 = 4x(3x + 7) + 11(3x + 7) = (3x + 7)(4x + 11)

(iv) 3y² − 2y − 1
= 3y² − 3y + y − 1 = 3y(y − 1) + 1(y − 1) = (y − 1)(3y + 1)

(v) √3x² + 4x + √3
√3 × √3 = 3 = 3 × 1 and 3 + 1 = 4
= √3x² + 3x + x + √3 = √3x(x + √3) + 1(x + √3) = (x + √3)(√3x + 1)
(here 3x is written as √3 × √3x)

(vi) 12x² − 3x + 4
= 12(x² − 6x + 8) = 12(x² − 4x − 2x + 8) = 12[x(x − 4) − 2(x − 4)] = 12(x − 4)(x − 2)

2Factorize the following polynomials.

(i) (x² − x)² − 8(x² − x) + 12
Put x² − x = m: m² − 8m + 12 = (m − 6)(m − 2)
= (x² − x − 6)(x² − x − 2)
= (x − 3)(x + 2)(x − 2)(x + 1)

(ii) (x − 5)² − (5x − 25) − 24
5x − 25 = 5(x − 5). Put x − 5 = m: m² − 5m − 24 = (m − 8)(m + 3)
= (x − 5 − 8)(x − 5 + 3) = (x − 13)(x − 2)

(iii) (x² − 6x)² − 8(x² − 6x + 8) − 64
Put x² − 6x = m: m² − 8(m + 8) − 64 = m² − 8m − 128 = (m − 16)(m + 8)
= (x² − 6x − 16)(x² − 6x + 8)
= (x − 8)(x + 2)(x − 4)(x − 2)

(iv) (x² − 2x + 3)(x² − 2x + 5) − 35
Put x² − 2x = m: (m + 3)(m + 5) − 35 = m² + 8m + 15 − 35 = m² + 8m − 20 = (m + 10)(m − 2)
= (x² − 2x + 10)(x² − 2x − 2)

(v) (y + 2)(y − 3)(y + 8)(y + 3) + 56
Pair so that the y-terms match: (y + 2)(y + 3) = y² + 5y + 6 and (y − 3)(y + 8) = y² + 5y − 24
Put y² + 5y = m: (m + 6)(m − 24) + 56 = m² − 18m − 144 + 56 = m² − 18m − 88 = (m − 22)(m + 4)
= (y² + 5y − 22)(y² + 5y + 4) = (y² + 5y − 22)(y + 1)(y + 4)

(vi) (y² + 5y)(y² + 5y − 2) − 24
Put y² + 5y = m: m(m − 2) − 24 = m² − 2m − 24 = (m − 6)(m + 4)
= (y² + 5y − 6)(y² + 5y + 4) = (y + 6)(y − 1)(y + 1)(y + 4)

(vii) (x − 3)(x − 4)²(x − 5) − 6
(x − 3)(x − 5) = x² − 8x + 15 and (x − 4)² = x² − 8x + 16
Put x² − 8x = m: (m + 15)(m + 16) − 6 = m² + 31m + 234 = (m + 13)(m + 18)
= (x² − 8x + 13)(x² − 8x + 18)

In (iv) and (vii) the quadratic factors cannot be split further into factors with integer coefficients, so the answers stop there.

Problem Set 3

1Write the correct alternative answer for each of the following questions.

(i) Which of the following is a polynomial? — (D) √2x² + 12
The other options have powers √2, −2 or a variable in the denominator.
(ii) What is the degree of the polynomial √7? — (D) 0
√7 is a non-zero constant.
(iii) What is the degree of the 0 polynomial? — (C) undefined
(iv) What is the degree of the polynomial 2x² + 5x³ + 7? — (A) 3
(v) What is the coefficient form of x³ − 1? — (C) (1, 0, 0, −1)
(vi) p(x) = x² − 7√7x + 3, then p(7√7) = ? — (A) 3
343 − 343 + 3 = 3
(vii) When x = −1, what is the value of 2x³ + 2x? — (D) −4
2(−1) + 2(−1) = −4
(viii) If x − 1 is a factor of 3x² + mx, find m. — (C) −3
p(1) = 3 + m = 0
(ix) Degree of the product (x² − 3)(2x − 7x³ + 4)? — (A) 5
2 + 3 = 5
(x) Which of the following is a linear polynomial? — (A) x + 5

2Write the degree of the polynomial for each of the following.

(i) 5 + 3x⁴ — 4 (ii) 7 — 0 (iii) ax⁷ + bx⁹ — 9 (taking b ≠ 0)

3Write the following polynomials in standard form.

(i) 4x² + 7x⁴ − x³ − x + 9 = 7x⁴ − x³ + 4x² − x + 9

(ii) p + 2p³ + 10p² + 5p⁴ − 8 = 5p⁴ + 2p³ + 10p² + p − 8

4Write the following polynomial in coefficient form.

(i) x⁴ + 16 = x⁴ + 0x³ + 0x² + 0x + 16 ⇒ (1, 0, 0, 0, 16)

(ii) m⁵ + 2m² + 3m + 15 = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15 ⇒ (1, 0, 0, 2, 3, 15)

5Write the index form of the polynomial using variable x from its coefficient form.

(i) (3, −2, 0, 7, 18) — 3x⁴ − 2x³ + 0x² + 7x + 18

(ii) (6, 1, 0, 7) — 6x³ + x² + 0x + 7

(iii) (4, 5, −3, 0) — 4x³ + 5x² − 3x + 0

6Add the following polynomials.

(i) 7x⁴ − 2x³ + x + 10 ; 3x⁴ + 15x³ + 9x² − 8x + 2
= (7 + 3)x⁴ + (−2 + 15)x³ + 9x² + (1 − 8)x + (10 + 2) = 10x⁴ + 13x³ + 9x² − 7x + 12

(ii) 3p³q + 2p²q + 7 ; 2p²q + 4pq − 2p³q
= (3 − 2)p³q + (2 + 2)p²q + 4pq + 7 = p³q + 4p²q + 4pq + 7

7Subtract the second polynomial from the first.

(i) 5x² − 2y + 9 ; 3x² + 5y − 7
= 5x² − 2y + 9 − 3x² − 5y + 7 = 2x² − 7y + 16

(ii) 2x² + 3x + 5 ; x² − 2x + 3
= 2x² + 3x + 5 − x² + 2x − 3 = x² + 5x + 2

8Multiply the following polynomials.

(i) (m³ − 2m + 3)(m⁴ − 2m² + 3m + 2)
= m³(m⁴ − 2m² + 3m + 2) − 2m(m⁴ − 2m² + 3m + 2) + 3(m⁴ − 2m² + 3m + 2)
= m⁷ − 2m⁵ + 3m⁴ + 2m³ − 2m⁵ + 4m³ − 6m² − 4m + 3m⁴ − 6m² + 9m + 6
= m⁷ − 4m⁵ + 6m⁴ + 6m³ − 12m² + 5m + 6

(ii) (5m³ − 2)(m² − m + 3)
= 5m⁵ − 5m⁴ + 15m³ − 2m² + 2m − 6
= 5m⁵ − 5m⁴ + 15m³ − 2m² + 2m − 6

9Divide polynomial 3x³ − 8x² + x + 7 by x − 3 using synthetic method and write the quotient and remainder.

Synthetic division: standard form 3x³ − 8x² + x + 7 ⇒ coefficient form (3, −8, 1, 7). Divisor x − 3, so write 3.

3	3	−8	1	7
		9	3	12
	3	1	4	19

Quotient = 3x² + x + 4 Remainder = 19

∴ 3x³ − 8x² + x + 7 = (x − 3)(3x² + x + 4) + 19

10For which value of m, x + 3 is the factor of the polynomial x³ − 2mx + 21?

(x + 3) is a factor, so p(−3) = 0.
(−3)³ − 2m(−3) + 21 = −27 + 6m + 21 = 6m − 6 = 0 ∴ m = 1

11At the end of 2016, the population of villages Kovad, Varud, Chikhali is 5x² − 3y², 7y² + 2xy and 9x² + 4xy respectively. At the beginning of 2017, x² + xy − y², 5xy and 3x² + xy persons from these villages respectively went to another village for education. What is the remaining total population of these three villages?

Total population = (5x² − 3y²) + (7y² + 2xy) + (9x² + 4xy) = 14x² + 6xy + 4y²

Persons who left = (x² + xy − y²) + 5xy + (3x² + xy) = 4x² + 7xy − y²

Remaining = (14x² + 6xy + 4y²) − (4x² + 7xy − y²) = 10x² − xy + 5y²

10x² − xy + 5y²

12Polynomials bx² + x + 5 and bx³ − 2x + 5 are divided by polynomial x − 3 and the remainders are m and n respectively. If m − n = 0 then find the value of b.

m = remainder of bx² + x + 5 = 9b + 3 + 5 = 9b + 8

n = remainder of bx³ − 2x + 5 = 27b − 6 + 5 = 27b − 1

m − n = 0 ⇒ 9b + 8 = 27b − 1 ⇒ 18b = 9 ⇒ b = 12

b = 12

13Simplify. (8m² + 3m − 6) − (9m − 7) + (3m² − 2m + 4)

= 8m² + 3m − 6 − 9m + 7 + 3m² − 2m + 4
= (8 + 3)m² + (3 − 9 − 2)m + (−6 + 7 + 4) = 11m² − 8m + 5

14Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x² + 5x − 4?

Required polynomial = (x² + 13x + 7) − (3x² + 5x − 4) = x² − 3x² + 13x − 5x + 7 + 4
−2x² + 8x + 11

15Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?

Required polynomial = (6m + 3n + 10) − (4m + 2n + 3)
2m + n + 7

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