Algebraic Expressions – Maharashtra Board Class 9 Solutions for Algebra

Algebraic Expressions – Maharashtra Board Class 9 Solutions for Algebra

Exercise – 3.1

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Let ‘x’ be the expression to be subtracted from 2a + 6b – 5 to get -3a + 2b + 3.
∴ 2a + 6b – 5 – x = -3a + 2b + 3.
∴ 2a + 6b – 5 – x + 3a – 2b – 3= 0
∴ 2a + 3a + 6b – 2b – 5 – 3 – x = 0
∴ 5a + 4b – 8 – x = 0
∴x = 5a + 4b – 8

Solution 5:

First let us add 3x – 2y + 7 and 5x – 3y – 8
(3x – 2y + 7) + (5x – 3y – 8)
= 3x – 2y + 7 + 5x – 3y – 8
= 3x + 5x – 2y – 3y + 7 – 8
= 8x – 5y – 1
Let us subtract 4x + y + 2 from 8x – 5y – 1
(8x – 5y – 1) – (4x + y + 2)
= 8x – 5y – 1 – 4x – y – 2
= 8x – 4x – 5y – y – 1 – 2
= 4x – 6y – 3.

Solution 6:

Solution 7:

Exercise – 3.2

Solution 1:

4x – 8y
= 4(x – 2y)

Solution 2:

5t + 25t²
= 5t(1 + 5t)

Solution 3:

x⁴y⁵ – 3x⁵y⁴
= x⁴y⁴(y – 3x)

Solution 4:

x² + xy – 3x – 3y
= x(x + y) – 3(x + y)
= (x + y) (x – 3)

Solution 5:

6ax – 6by – 4ay + 9bx
= 6ax – 4ay + 9bx – 6by
= 2a (3x – 2y) + 3b (3x – 2y)
= (3x – 2y) (2a + 3b)

Solution 6:

7x² – 21x + 2xy – 6y
= 7x (x – 3) + 2y(x – 3)
= (x – 3) (7x + 2y)

Solution 7:

2x² – 3xy – 8xy² + 12y³
= x(2x – 3y) – 4y²(2x – 3y)
= (2x – 3y) (x – 4y²)

Solution 8:

81x² – 64y²
= (9x)² – (8y)²
= (9x + 8y) (9x – 8y)

Solution 9:

27a² – 75b²
= 3(9a² – 25b²)
= 3[(3a)² – (5b)²]
= 3(3a + 5b) (3a – 5b)

Solution 10:

3a³ – 3a
= 3a (a² – 1)
= 3a(a + 1)(a – 1)

Solution 11:

x² – y² – 6x – 6y
= (x)² – (y)² – 6(x + y)
= (x + y)(x – y) – 6(x + y)
= (x + y) { (x – y) – 6}
= (x + y) (x – y – 6)

Solution 12:

(a + b) (c + d) – a² + b²
= (a + b) (c + d) – (a² – b²)
= (a + b) (c + d) – (a + b) (a – b)
= (a + b) {(c + d) – (a – b)}
= (a + b) ( c + d – a + b)
= (a + b) (-a + b + c + d)

Solution 13:

x² + 8x + 24y – 9y²
= x² – 9y² + 8x + 24y
= {(x²) – (3y)²} + 8(x + 3y)
= (x + 3y) (x – 3y) + 8(x +3y)
= (x + 3y) (x – 3y + 8)

Solution 14:

a² – 12ab + 36b² – 25
= (a – 6b)² – (5)²
= (a – 6b + 5) (a – 6b – 5)

Solution 15:

= x² + 6xy + 9y²– 25m² + 40mn – 16n²
= (x² + 6xy + 9y²) – (25m² – 40mn + 16n²)
= (x + 3y)² – (5m – 4n)²
= [(x + 3y) + (5m – 4n)][(x + 3y) – (5m – 4n)]
= (x + 3y + 5m – 4n) (x + 3y – 5m + 4n)

Exercise – 3.3

Solution 1:

8x³ + 125y³
= (2x)³ + (5y)³
= (2x + 5y) (4x² – 10xy + 25y²)

Solution 2:

2a³ – 54b³
= 2(a³ – 27b³)
= 2[(a)³ – (3b)³]
= 2(a – 3b) (a² + 3ab + 9b²)

Solution 3:

(a + b)³ – 8
= (a + b)³ – (2)³
= (a + b – 2) [(a + b)² + (a + b) × 2 + (2)²]
= (a + b – 2) (a² + 2ab + b² + 2a + 2b + 4)

Solution 4:

Solution 5:

Solution 6:

Let a + b = x and a – b = y,
(a – b)³ – (a – b)³ = x³ – y³ = (x – y) (x² + xy + y²)
Substituting the values of x and y,
[(a + b) – (a – b)][(a + b)² + (a + b)(a – b) + (a – b)²]
= (a + b – a + b) (a² + 2ab + b² + a² – b² + a² – 2ab + b²)
= (a – a + b + b) (a² + a² + a² + 2ab – 2ab + b² – b² + b²)
= 2b (3a² + b²)

Solution 7:

Let 2m + 3n = a and 3m + 2n = b
(2m + 3n)³ – (3m + 2n)³
= a³ – b³
= (a – b)(a² + ab + b²)
Substituting the values of a and b,
= [(2m + 3n) – (3m + 2n)] [(2m + 3n)² + (2m + 3n) (3m + 2n) + (3m + 2n)²]
= (2m + 3n – 3m – 2n)(4m² + 12mn + 9n² + 6m² + 13mn + 6n² + 9m² + 12mn + 4n²)
= (-m + n) (6m² + 9m² + 12mn + 13mn + 12mn + 9n² + 6n² + 4n²)
=(-m + n) (19m² + 37mn + 19n²)

Solution 8:

Substituting 3x + 5y = a and 2x – y = b,
(3x + 5y)³ – (2x – y)³
= a³ – b³
= (a – b) (a² + ab + b²)
Substituting the values of a and b,
= [(3x + 5y) – (2x – y)][(3x + 5y)² + (3x + 5y) (2x -y) + (2x – y)²]
= (3x + 5y – 2x + y)(9x² + 30xy + 25y² + 6x² + 7xy – 5y² + 4x² – 4xy + y²)
= (x + 6y)(9x² + 6x² + 4x² + 30xy + 7xy – 4xy + 25y² – 5y² + y²)
= (x + 6y)(19x² + 33xy + 21y²)

Solution 9:

Substituting x – 1 = a,
27(x – 1)³ + y³
= 27a³ + y³
= (3a)³ + y³
= (3a + y) (9a² – 3ay + y²)
Substituting the value of a,
= [3(x – 1) + y] [9(x – 1)² – 3(x – 1) (y) + y²]
= (3x – 3 + y) [9(x² – 2x + 1) – 3xy + 3y + y²]
= (3x + y – 3) (9x² – 18x + 9 – 3xy + 3y + y²)
= (3x + y – 3) (9x² – 18x + 9 – 3xy + 3y + y²)

Solution 10:

a⁶ – b⁶
= (a³)² – (b³)²
= (a³ + b³)(a³ – b³)
= (a + b) (a² – ab + b²)(a – b) (a² + ab + b²)

Exercise – 3.4

Solution 1:

2x² + 3x – 5
= 2x² + 5x – 2x – 5 …………….. ∵ (-5)  × 2 = -10
= x(2x + 5) – 1(2x + 5) ………. 5 × (-2) = -10
= (2x + 5) (x – 1) ………………… and 5 + (-2) = 3

Solution 2:

3x² – 14x + 8
= 3x² – 12x – 2x + 8 …………. ∵ 3 × 8 = 24
= 3x(x – 4) -2(x – 4) …………. (-12) × (- 2) = 24
= (x – 4) (3x – 2) ……………….. and – 12 – 2 = -14

Solution 3:

6x² + 11x – 10
= 6x² + 15x – 4x – 10 ………………… ∵ 6 × (-10) = -60
= 3x(2x + 5) -2(2x + 5) ……………. 15 × (- 4) = -60
= (2x + 5) (3x – 2) …………………….. and 15 – 4 = 11

Solution 4:

2x² – 7x – 15
= 2x²  – 10x + 3x – 15 ………………… ∵ 2 × (-15) = -30
= 2x(x – 5) + 3(x – 5) ………………… -10 × 3 = -30
= (x – 5) (2x + 3) ……………………….. and -10 + 3 = -7

Solution 5:

x² + 9xy + 18y²
= x² + 6xy + 3xy + 18y² ………………… ∵ 6 × 3 = 18
= x(x + 6y) + 3y(x + 6y) ………………. and 6 + 3 = 9
= (x + 6y) (x + 3y)

Solution 6:

a² – 5ab – 36b²
= a² + 4ab – 9ab – 36b² …………….. ∵ 4 × (-9) = -36
= a(a + 4b) – 9b(a + 4b) …………… and 4 – 9 = -5
= (a + 4b) (a – 9b)

Solution 7:

a² + 14ab – 51b²
= a² – 3ab + 17ab – 51b² ………. ∵ -3 × 17 = -51
= a (a – 3b) +17b(a – 3b) …….. and -3 + 17 = 14
= (a – 3b) (a + 17b)

Solution 8:

2m² + 19mn + 30n²
= 2m² + 4mn + 15mn + 30n² …….. ∵ 2 × 30 = 60
= 2m(m + 2n) + 15n (m + 2n) ….. 4 × 15 = 60
= (m + 2n) (2m + 15n) ………………. and 4 + 15 = 19

Solution 9:

3a² – 11ab + 6b²
= 3a² – 9ab – 2ab + 6b² ……………. ∵ 3 × 6 = 18
= 3a (a – 3b) – 2b(a – 3b) ………… -9 × -2 = 18
= (a – 3b) (3a – 2b) ………………….. and -9 – 2 = -11

Solution 10:

6x² – 7xy – 13y²
= 6x² + 6xy – 13xy – 13y² …………. ∵ 6 × (-13) = -78
= 6x (x + y) -13y(x + y) ……………. and 6 – 13 = -7
= (x + y) (6x – 13y)

Solution 11:

Exercise – 3.5

Solution 1:

x⁴ – 8x²y² + 12y⁴
Let x² = p and y² = q
Then x⁴ = p², = q² and x²y² = pq
∴ x⁴ – 8x²y² + 12y⁴
= p² – 8pq + 12q² ……………… ∵ -6 × -2 = 12
= p² – 6pq – 2pq + 12q² …….. and -6 – 2 = -8
= p(p – 6q) – 2q(p – 6q)
= (p – 6q) (p – 2q)
Re-substituting the values of p and q we get,
= (x² – 6y²) (x² – 2y²)

Solution 2:

2x⁴ – 13x²y² + 15y⁴
Let x² = a and y² = b.
Then x⁴ = a², y⁴ = b² and x²y² = ab
∴ 2x⁴ – 13x²y² + 15y⁴ ……………… ∵ 2 × 15 = 30
= 2a² – 13ab + 15ab² ………………. -10 × -3 = 30
= 2a² – 10ab – 3ab + 15b² ………. and -10 – 3 = -13
= 2a(a – 5b) -3b (a – 5b)
= (a – 5b) (2a – 3b)
Re-substituting the values of a and b we get,
= (x² – 5y²) (2x² – 3y²)

Solution 3:

6a⁴ + 11a²b² – 10b⁴
Let a² = m and b² = n.
Then a⁴ = m², b⁴ = n² and a²b² = mn
∴ 6a⁴ + 11a²b² – 10b⁴ …………….. ∵ 6 × (-10) = -60
= 6m² + 11mn – 10n² ……………… 15 × -4 = -60
= 6m² + 15mn – 4mn – 10n² ….. and 15 – 4 = 11
= 3m(2m + 5n) – 2n(2m + 5n)
= (2m + 5n) (3m – 2n)
Re-substituting the values of m and n we get,
= (2a² + 5b²) (3a² – 2b²)

Solution 4:

3(x² – 5x)² – 2(x² – 5x + 5) – 6
Substituting (x² – 5x) = m
= 3m² – 2(m + 5) – 6
= 3m² – 2m – 10 – 6
= 3m² – 2m – 16 ………………….. ∵ 3 × (-16) = -48
= 3m² – 8m + 6m – 16 ………… -2 × -10 = 20
= m(3m – 8) + 2 (3m – 8) ….. and -8 + 6 = -2
= m(3m – 8) + 2(3m – 8)
= (3m – 8) (m + 2)
Re-substituting the value of m we get,
= [3(x² – 5x) – 8] [(x² – 5x) + 2]
= (3x² – 15x – 8)(x² – 5x + 2)

Solution 5:

(y² + 5y)(y² + 5y – 2) – 24
= m(m – 2) – 24
= m² – 6m + 4m – 24
= m² – 2m – 24 …………………….. ∵ -6 × -4 = 24
= m² – 6m + 4m – 24 …………… and -6 + 4 = -2
= m(m – 6) + 4 (m – 6)
= (m – 6) (m + 4)
Re-substituting the value of m we get,
= (y² + 5y – 6) (y² + 5y + 4))
= (y² + 6y – y – 6) (y² + 4y + y + 4)
= [y(y + 6) – 1(y + 6)][y(y + 4) + 1(y + 4)]
= (y + 6)(y – 1)(y + 4) (y + 1)

Exercise – 3.6

Solution 1:

x³ – 27y³ +125 + 45xy
= (x)³ + (-3y)³ + (5)³ – 3(x)(-3y)(5)
= (x – 3y + 5)(x² + 9y² + 25 + 3xy + 15y – 5x)

Solution 2:

a³ – b³ + 8c³ + 6abc
= (a)³ + (-b)³ + (2c)³ – 3(a)(-b)(2c)
= (a – b + 2c)(a² + b² + 4c² + ab + 2bc -2ca)

Solution 3:

8a³ + 27b³ + 64c³ – 72abc
= (2a)³ + (3b)³ + (4c)³ – 3(2a)(3b)(4c)
= (2a + 3b + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 8ca)

Solution 4:

-27x³ + y³ – z³ – 9xyz
= (-3x)³ + (y)³ + (-z)³ -3(-3x)(y) (-z)
= (-3x + y – z) (9x² +y² + z² +3xy + yz – 3zx)

Solution 5:

y⁶ + 32y³ – 64
= y⁶ + 8y³ – 64 + 24y³
= (y²)³ + (2y)³ + (-4)³ – 3(y²) (2y) (-4)
= (y² + 2y -4) (y⁴ + 4y² + 16 – 2y³ + 8y + 4y²)
= (y² + 2y – 4)(y⁴ – 2y³ + 8y² + 8y + 16)

Solution 6:

x⁶ – 10x³ – 27
= x⁶ – x³ – 27 – 9x³
= (x²)³ + (-x)³ + (-3)³ – 3(x²)(-x)(-3)
= (x² – x – 3)(x⁴ + x² + 9 + x³ – 3x + 3x²)
= (x² – x – 3)(x⁴ + x³ + 4x² – 3x + 9)

Solution 7:

Solution 8:

(p – 3q)³ + (3q – 7r)³ + (7r – p)³
Let (p – 3q) = a, (3q – 7r) = b and (7r – p)=c.
∴ (p – 3q)³ + (3q – 7r)³ + (7r – p)³ = a³ + b³ + c³
Here, a + b + c = p – 3q + 3q – 7r + 7r – p = 0
∴ a + b + c = 0
If a + b + c = 0, a³ + b³ + c³ = 3abc
∴ (p – 3q)³ + (3q – 7r)³ + (7r – p)³ = 3(p – 3q)(3q – 7r)(7r – p)

Solution 9:

(5x – 6y)³ + (7z – 5x)³ + (6y – 7z)³
Let (5x – 6y) = p, (7z – 5x) = q and (6y – 7z) = r
∴ (5x – 6y)³ + (7z – 5x)³ + (6y – 7z)³ = p³ + q³ + r³
Now, p + q + r = 5x – 6y + 7z – 5x + 6y -7z = 0
∴ p + q + r = 0
If p + q + r = 0, p³ + q³ + r³ = 3pqr
∴ (5x – 6y)³ + (7z – 5x)³ + (6y -7z)³ = 3(5x – 6y)(7z – 5x)(6y – 7z)

Solution 10:

27(a – b)³ + (2a – b)³ + (4b – 5a)³
= (3)³(a – b)³ + (2a – b)³ + (4b – 5a)³
= [3(a – b)]³ + (2a – b)³ + (4b – 5a)³
= (3a – 3b)³ + (2a – b)³ + (4b – 5a)³
Let 3a – 3b = x, 2a – b = y and 4b – 5a = z
= x³ + y³ + z³
Here, x + y + z = 3a – 3b + 2a – b + 4b – 5a = 0
∴ x + y + z = 0
If x + y + z = 0, x³ + y³ +z³ = 3xyz
∴ 27(a – b)³ + (2a – b)³ + (4b – 5a)³ = 3(3a – 3b)(2a – b) (4b – 5a)
= 9 (a – b) (2a – b) (4b – 5a)

Exercise – 3.7

Solution 1:

Expressions (i), (ii), (iv), (v) and (vii) are polynomials.

Solution 2:

Solution 3:

1. The degree of the polynomial is 0.
2. The degree of the polynomial is 0.
3. The degree of the polynomial is 1.
4. The degree of the polynomial is 2.
5. The degree of the polynomial is 12.
6. The degree of the polynomial is 3.
7. The degree of the polynomial is 3.
8. The degree of the polynomial is 8.
9. The degree of the polynomial is 7.
10. The degree of the polynomial is 7.

Exercise – 3.8

Solution 1(i):

2x³ – 7x² + 3x + 4; 2x³ – 3x² + 4x + 1
= (2x³ – 7x² + 3x + 4) + (2x³ – 3x² + 4x + 1)
= 2x³ – 7x² + 3x + 4 + 2x³ – 3x² + 4x + 1
= 2x³ + 2x³ – 7x² – 3x² + 3x + 4x + 4 +1
(Arranging the like terms together)
= 4x² – 10x² + 7x + 5 and degree 3.

Solution 1(ii):

3x² + 5x – x⁷ ; – 3x² + 5x + 8
= (3x² + 5x – x⁷) + (- 3x² + 5x + 8)
= 3x² + 5x – x⁷ – 3x² + 5x + 8
= – x⁷ + 3x² – 3x² +5x + 5x + 8
(Arranging the like terms together)
= x⁷ + 10x + 8 and degree 7.

Solution 1(iii):

x⁴ + 5x³ + 7x; 4x³ – 3x² + 5
= (x⁴ + 5x³ + 7x) + (4x³ – 3x² + 5)
= x⁴ + 5x³ + 7x + 4x³ – 3x² + 5
= x⁴ + 5x³ + 4x³– 3x² + 7x + 5
(Arranging the like terms together)
= x⁴ + 9x³ – 3x² + 7x + 5 and degree 4.

Solution 1(iv):

y² + 2y – 5; y³ + 2y² + 3y + 4; y³ + 7y – 2
= (y² + 2y – 5) + (y³ + 2y² + 3y + 4) + (y³ + 7y – 2)
= y² + 2y – 5 + y³ + 2y² + 3y + 4+ y³ + 7y – 2
= y³ + y³ + y² + 2y² + 2y + 3y + 7y – 5 + 4 – 2
(Arranging the like terms together)
= 2y³ + 3y² + 12y – 3 and degree 3.

Solution 1(v):

5m² + 3m + 8; m³ – 6m² + 4m; m³ – m² – m + 5
= (5m² + 3m + 8) + (m³ – 6m² + 4m) + (m³ – m² – m + 5)
= 5m² + 3m + 8 + m³ – 6m² + 4m + m³ – m² – m + 5
= m³ + m³+ 5m² – 6m² – m² +3m + 4m – m + 8 + 5
(Arranging the like terms together)
= 2m³ – 2m² + 6m + 13 and degree 3.

Solution 2(i):

x⁴ + x² + x – 1; x⁴ – x³ – x² + 1
= (x⁴ + x² + x – 1) – (x⁴ – x³ – x² + 1)
= x⁴ + x² + x – 1 – x⁴ + x³ + x² – 1
= x⁴ – x⁴ + x³ + x² + x- 1 – 1
(Arranging the like terms together)
= x³ + 2x² + x – 2 and degree 3.

Solution 2(ii):

n³ – 5n² + 6; n³ – 3n + 8
= (n³ – 5n² + 6) – (n³ – 3n + 8)
= n³ – 5n² + 6 – n³ + 3n – 8
= n³ – n³ – 5n² + 3n + 6 – 8
(Arranging the like terms together)
= – 5n² + 3n – 2 and degree 2.

Solution 2(iii):

2a + 3a² – 7; 3a² – 12 + 2a
= (2a + 3a² – 7) – (3a² – 12 + 2a)
= 2a + 3a² – 7 – 3a² + 12 – 2a
= 3a² – 3a² + 2a – 2a – 7 + 12
(Arranging the like terms together)
= 5 and degree 0.

Solution 3(i):

(3x² – 2x + 1) + (x² + 5x -3) + (4x² + 8)
= 3x² – 2x + 1 + x² + 5x -3 + 4x² + 8
= 3x² + x² + 4x² – 2x + 5x + 1 – 3 + 8
(Arranging the like terms together)
= 8x² + 3x + 6.

Solution 3(ii):

(2y³ + 3y -7) – (8y – 6) + (4y³ – 2y + 1)
= 2y³ + 3y – 7 – 8y + 6 + 4y³ – 2y + 1
= 2y³ + 4y³+ 3y – 8y – 2y – 7 + 6 + 1
(Arranging the like terms together)
= 6y³ -7y.

Solution 3(iii):

5m³ – m + 6m² – (3m² – 2 + m)
(5m³ – m + 6m²) – (3m² – 2 + m)
= 5m³ – m + 6m² -3m² + 2 – m
= 5m³ + 6m² – 3m² – m – m + 2
= 5m³ + 3m² – 2m + 2.

Solution 4:

Let the polynomial to be added be p(x)
(2x⁴ – 3x² + 5x + 8) + p(x) = (2x² – 5x + 4)
p(x) = (2x² – 5x + 4) – (2x⁴ – 3x² + 5x + 8)
= 2x² -5x + 4 – 2x⁴ + 3x² – 5x – 8
= -2x⁴ + 2x² + 3x² -5x – 5x + 4 – 8
= -2x⁴ + 5x² – 10x – 4
Therefore, – 2x⁴ + 5x² – 10x – 4 should be added to 2x⁴ – 3x² + 5x + 8 to get 2x² – 5x + 4.

Solution 5:

Let the polynomial to be subtracted be p(y).
(y³ + 2y² + 5y -1) – p(y) = 2y² + 12
∴ (y³ + 2y² + 5y – 1) – (2y² + 12) = p(y)
∴ p (y) = y³ + 2y² + 5y – 1 – 2y² – 12
= y³ + 2y² – 2y² + 5y – 1 – 12
= y³ + 5y – 13
Therefore, y³ + 5y – 13 should be subtracted from y³ + 2y² + 5y – 1 to get 2y² + 12.

Solution 6:

= (z³ + 3z² + 5z + 8) + (4z³ + 2z² – 7z – 2) – (2z³ – 3z² + z – 4)
= z³ + 3z² + 5z + 8 + 4z³ + 2z² – 7z – 2 – 2z³ + 3z² – z + 4
= z³ + 4z³ – 2z³ + 3z² + 2z² + 3z² + 5z – 7z – z + 8 – 2 + 4
= 3z³ + 8z² – 3z + 10

Exercise – 3.9

Solution 1(i):
(x² + 3x + 1)(2x – 3)
= x²(2x – 3) + 3x(2x – 3) + 1(2x – 3)
= 2x³ – 3x² + 6x² – 9x + 2x – 3
= 2x³ + 3x² – 7x – 3 and degree 3.

Solution 1(ii):

(3x² + 5x)(x² + 2x + 1)
= 3x²(x² + 2x + 1) + 5x(x² + 2x + 1)
= 3x⁴ + 6x³ + 3x² + 5x³ + 10x² + 5x
= 3x⁴ + 6x³ + 5x³ + 3x² + 10x² + 5x
(Arranging the like terms together)
= 3x⁴ + 11x³ + 13x² + 5x and degree 4.

Solution 1(iii):

(x³ + 4x + 2) (x² + x + 5)
= x³(x² + x + 5) + 4x(x² + x + 5)+ 2(x² + x + 5)
= x⁵ + x⁴ + 5x³ + 4x³ + 4x² + 20x + 2x² + 2x + 10
= x⁵ + x⁴ + 9x³ + 4x² + 2x² + 20x + 2x + 2x + 10 (Arranging the like terms together)
= x⁵ + x⁴ + 9x³ + 6x² + 22x + 10 and degree 5.

Solution 1(iv):

(x³ – 1)(x² – x + 4)
= x³(x² – x + 4) – 1(x² – x + 4)
= x⁵ – x⁴ + 4x³ – x² + x – 4 and degree 5.

Solution 1(v):

(2y² + 3)(3y³ + 1)
= 2y²(3y³ + 1) + 3(3y³ + 1)
= 6y⁵ + 2y² + 9y³ + 3
= 6y⁵ + 9y³ + 2y² + 3 and degree 5.

Solution 2(i):

Solution 2(ii):

Solution 2(iii):

Solution 2(iv):

Solution 2(v):

Exercise – 3.10

Solution 1:

1. The coefficient form of the polynomial2x² + 5x + 12 is (2, 5, 12).
2. The index form of the polynomial y⁴ – 3y² + 2y – 7 = y⁴ + 0y³ – 3y² + 2y – 7.∴The coefficient form = (1, 0, -3, 2, -7).
3. The index form of the polynomial x⁵ + 3x² = x⁵ + 0x⁴ +0x³ + 3x² + 0x + 0.∴The coefficient form = (1, 0, 0, 3, 0, 0).
4. The index form of the polynomial y⁴ – 3 = y⁴ + 0y³ + 0y² + 0y – 3.∴The coefficient form = (1, 0, 0, 0, -3).
5. The index form of the polynomial9x = 9x + 0.  ∴ The coefficient form = (9, 0).

Solution 2:

1. Number of coefficients = 3.∴The degree of the polynomial = 3 – 1 = 2
   ∴The index form of the given polynomial is 3x² + 2x + 7.
2. The number of coefficients = 4.∴The degree of the polynomial = 4-1=3
   ∴The index form of the given polynomial is 2x³– 4.
3. Number of coefficients = 5.∴The degree of the polynomial = 5 – 1= 4
   ∴The index form of the given polynomial is x⁴ –  3x² + x + 5.
4. Number of coefficients = 4.∴The degree of the polynomial = 4 – 1= 3,
   ∴The index form of the given polynomial is -x³ + 3x² – 5x + 6.
5. Number of coefficients = 7.∴The degree of the polynomial = 7 – 1 = 6
   ∴The index form of the given polynomial is x⁶ + 64.

Solution 3(i):

Solution 3(ii):

Solution 3(iii):

Solution 3(iv):

Solution 3(v):

Solution 3(vi):

Solution 3(vii):

Solution 3(viii):

Exercise – 3.11

Solution 1:

1. p(x) = x² + 2x + 5= 0 + 0 + 5
2. = 5
3. ∴ p(0) = (0)² + 2(0) + 5
4. p(x) = x² + 2x + 5= 9 + 6 + 5
5. = 20
6. ∴ p(3) = (3)² + 2(3) + 5
7. p(x) = x² + 2x + 5= 1 – 2 + 5
8. = 4
9. ∴ p(-1) = (-1)² + 2(-1) + 5
10. p(x) = x² + 2x + 5= 9 – 6 + 5
11. = 8
12. ∴ p(-3) = (-3)² + 2(-3) + 5
13. p(x) = x² + 2x + 5= a² + 2a + 5
14. ∴ p(a) = (a)² + 2(a) + 5

Solution 2:

P(y) = y³ – 5y – 2y² + 3

= y³ – 2y² – 5y + 3 …(Standard form)

1. p(y) = y³ – 2y² – 5y + 3= 1 – 2 – 5 + 3
2. = -3
3. ∴ p(1) = (1)³ – 2(1)² – 5(1) + 3
4. p(y) = y³ – 2y² – 5y + 3= 8 – 2(4) – 10 + 3= -7
5. = 8 – 8 – 10 + 3
6. ∴ p(2) = (2)³ – 2(2)² – 5(2) + 3
7. p(y) = y³ – 2y² – 5y + 3= -8 – 2(4) + 10 + 3= -3
8. = -8 – 8 + 10 + 3
9. ∴ p(-2) = (-2)³ – 2(-2)² – 5(-2) + 3
10. p(y) = y³ – 2y² – 5y + 3= 64 – 2(16) – 20 + 3= 15
11. = 64 – 32 – 20 + 3
12. ∴ p(4) = (4)³ – 2(4)² – 5(4) + 3
13. p(y) = y³ – 2y² – 5y + 3= -b³ – 2b² + 5b + 3
14. ∴ p(-b) = (-b)³ – 2(-b)² – 5(-b) + 3

Solution 3:

p(x) = x² – mx + 7
∴ p(2) = (2)² – m(2) + 7
= 4 – 2m + 7
= 11 – 2m
But p(2) = 35 given
∴ 11 – 2 m = 35
∴ 2m = 11 – 35
∴ 2m = -24
∴ m = -12

Solution 4:

p(y) = ay² + 2y – 6
∴ p(-3) = a(-3 )² + 2(-3 ) – 6
= 9a – 6 – 6
= 9a – 12
But p(y) = 15
∴ 9a – 12 = 15
∴ 9a = 15 + 12
∴ 9a = 27
∴ a = 3

Exercise – 3.12

Solution 1:

Solution 2(i):

p(x) = x – 2
p(x) = 0
∴ x – 2 = 0
∴ x = 2
∴ 2 is the zero of the given polynomial.

Solution 2(ii):

p(x) = (x – 2)
p(x) = 0
∴ x – 2 = 0
∴ x =2
p(x) = (x – 9)
p(x) = 0
∴ x – 9 = 0
∴ x = 9
∴ 2 and 9 are the zero of the given polynomial.

Solution 2(iii):

Solution 3(i):

Solution 3(ii):

Solution 4(i):

Solution 4(ii):

Exercise – 3.13

Solution 1(i):

p(x) = 3x² + x + 7
Divisor is x + 2
∴ Put x = -2 in p(x)
∴ By the Remainder Theorem
Remainder = p(-2) = 3(-2)² + (-2) + 7
= 3(4) – 2 + 7
= 12 – 2 + 7
= 17
∴ Remainder = 17

Solution 1(ii):

p(x) = 4x³ + 5x – 10
Divisor is x – 3
∴ Put x = 3 in p(x)
∴ By the Remainder Theorem
Remainder = p(3) = 4(3)³ + 5(3) – 10
= 4(27) + 15 – 10
=108 + 15 – 10
= 113
∴ Remainder = 113

Solution 1(iii):

p(x) = x³ – ax² + 2x – a
Divisor is x – a.
∴ Put x = a in p(x)
∴ By the Remainder Theorem
Remainder = p(a) = (a)³ – a(a)² + 2 (a) – a
= a³ – a³ + 2a – a
= a
∴ Remainder = a

Solution 2(i):

p(x) = 2x³ – 3x² + 4x – 5
Divisor is x – 2
∴ Put x = 2 in p(x)
∴ By the Remainder Theorem
Remainder = p(2) = 2 (2)³ – 3(2)² + 4(2) – 5
=16 – 3(4) + 8 – 5
= 16 – 12 + 8 – 5
= 7
∴ Remainder = 7

Solution 2(ii):

p(x) = 2x³ – 3x² + 4x – 5
Divisor is x + 3
∴ Put x = -3 in p(x)
∴ By the Remainder Theorem
Remainder = p(-3) = 2 (-3)³ – 3(-3)² + 4(-3) – 5
= 2 (- 27) – 3(9) – 12 – 5
= – 54 – 27 – 12 – 5
= -98
∴ Remainder = -98

Solution 2(iii):

p(x) = 2x³ – 3x² + 4x – 5
Divisor is x – 1
∴ Put x = 1 in p(x)
∴ By the Remainder Theorem
Remainder = p(1) = 2 (1)³ – 3(1)² + 4(1) – 5
= 2 (1) – 3(1) + 4 – 5
= 2 – 3 + 4 – 5
= -2
∴ Remainder = -2.
Solution 3:
p(x) = x³ + ax² + 4x – 5
Divisor is x + 1
∴ Put x = 1 in p(x)
∴ By the Remainder Theorem
Remainder = p(-1) = (-1)³ + a(-1)² + 4(-1) – 5
= -1 +a – 4 – 5
= a – 10
But, remainder = 14
∴ a – 10 = 14
∴ a = 14 + 10
∴ a = 24

Exercise – 3.14

Solution 1(i):

p(x) = x² – 4
Put x = -2 in p(x), we get
p(-2) = (-2)² – 4
= 4 – 4
= 0
As p(-2) = 0,
∴ By the Factor Theorem (x + 2) is a factor of x² – 4.

Solution 1(ii):

p(x) = x³ – 27
Put x = 3 in p(x), we get
p(3) = (3)³ – 27
= 27 – 27
= 0
As p(3) = 0,
∴ By the Factor Theorem (x – 3) is a factor of x³ – 27.

Solution 1(iii):

p(x) = 2x⁴ + 9x³ + 6x² – 11x – 6
Put x = 1 in p(x), we get
p(1) = 2(1)⁴ + 9(1)³ + 6(1)² – 11(1) – 6
= 2 + 9 + 6 – 11 – 6
= 0
As p(1) = 0,
∴ By the Factor Theorem (x – 1) is a factor of 2x⁴ + 9x³ + 6x² – 11x – 6.

Solution 1(iv):

p(x) = x² + 10x + 24
Put x = -4 in p(x), we get
p(1) = (-4)² + 10(-4) + 24
= 16 – 40 + 24
= 0
As p(-4) = 0,
∴ By the Factor Theorem (x + 4) is a factor of x² + 10x + 24.

Solution 2:

p(x) = x³ – 3x² + 4x + 4
Put x = 2 in p(x), we get
p(1) = (2)³ – 3(2)² + 4(2) + 4
= 8 – 3(4) + 8 + 4
= 8 – 12 + 8 + 4
= 8 ≠ 0
As p(2) ≠ 0,
∴ By the Factor Theorem (x – 2) is not a factor of x³ – 3x² + 4x + 4.

Solution 3:

Let p(x) = 2x³ – 6x² + 5x + a
Put x = 2 in p(x), we get
p(2) = 2(2)³ – 6(2)² + 5(2) + a
= 2(8) – 6(4) + 10 + a
= 16 – 24 + 10 + a
= 2 + a
But p(2) must be 0, because (x – 2) is a factor
∴ 2 + a = 0
∴ a = -2