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:
- The degree of the polynomial is 0.
- The degree of the polynomial is 0.
- The degree of the polynomial is 1.
- The degree of the polynomial is 2.
- The degree of the polynomial is 12.
- The degree of the polynomial is 3.
- The degree of the polynomial is 3.
- The degree of the polynomial is 8.
- The degree of the polynomial is 7.
- 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:
- The coefficient form of the polynomial2x2 + 5x + 12 is (2, 5, 12).
- The index form of the polynomial y4 – 3y2 + 2y – 7 = y4 + 0y3 – 3y2 + 2y – 7.∴The coefficient form = (1, 0, -3, 2, -7).
- The index form of the polynomial x5 + 3x2 = x5 + 0x4 +0x3 + 3x2 + 0x + 0.∴The coefficient form = (1, 0, 0, 3, 0, 0).
- The index form of the polynomial y4 – 3 = y4 + 0y3 + 0y2 + 0y – 3.∴The coefficient form = (1, 0, 0, 0, -3).
- The index form of the polynomial9x = 9x + 0. ∴ The coefficient form = (9, 0).
Solution 2:
- Number of coefficients = 3.∴The degree of the polynomial = 3 – 1 = 2
∴The index form of the given polynomial is 3x2 + 2x + 7. - The number of coefficients = 4.∴The degree of the polynomial = 4-1=3
∴The index form of the given polynomial is 2x3– 4. - Number of coefficients = 5.∴The degree of the polynomial = 5 – 1= 4
∴The index form of the given polynomial is x4 – 3x2 + x + 5. - Number of coefficients = 4.∴The degree of the polynomial = 4 – 1= 3,
∴The index form of the given polynomial is -x3 + 3x2 – 5x + 6. - 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:
- p(x) = x2 + 2x + 5= 0 + 0 + 5
- = 5
- ∴ p(0) = (0)2 + 2(0) + 5
- p(x) = x2 + 2x + 5= 9 + 6 + 5
- = 20
- ∴ p(3) = (3)2 + 2(3) + 5
- p(x) = x2 + 2x + 5= 1 – 2 + 5
- = 4
- ∴ p(-1) = (-1)2 + 2(-1) + 5
- p(x) = x2 + 2x + 5= 9 – 6 + 5
- = 8
- ∴ p(-3) = (-3)2 + 2(-3) + 5
- p(x) = x2 + 2x + 5= a2 + 2a + 5
- ∴ p(a) = (a)2 + 2(a) + 5
Solution 2:
P(y) = y3 – 5y – 2y2 + 3
= y3 – 2y2 – 5y + 3 …(Standard form)
- p(y) = y3 – 2y2 – 5y + 3= 1 – 2 – 5 + 3
- = -3
- ∴ p(1) = (1)3 – 2(1)2 – 5(1) + 3
- p(y) = y3 – 2y2 – 5y + 3= 8 – 2(4) – 10 + 3= -7
- = 8 – 8 – 10 + 3
- ∴ p(2) = (2)3 – 2(2)2 – 5(2) + 3
- p(y) = y3 – 2y2 – 5y + 3= -8 – 2(4) + 10 + 3= -3
- = -8 – 8 + 10 + 3
- ∴ p(-2) = (-2)3 – 2(-2)2 – 5(-2) + 3
- p(y) = y3 – 2y2 – 5y + 3= 64 – 2(16) – 20 + 3= 15
- = 64 – 32 – 20 + 3
- ∴ p(4) = (4)3 – 2(4)2 – 5(4) + 3
- p(y) = y3 – 2y2 – 5y + 3= -b3 – 2b2 + 5b + 3
- ∴ 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