GSEB Solutions for Class 9 Mathematics – Polynomials

GSEB Solutions for Class 9 Mathematics – Polynomials (English Medium)

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Exercise 3:

Solution 1:

Solution 2(1):

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Solution 2(2):

Solution 2(3):

Solution 3:

If we divide the total number of chocolates that the student had (x⁴ – 3x³ + 5x² + 8x + 5) by the number of chocolates received by each friend (x² – 1), the quotient will give the number of friends and the remainder will give the number of chocolates left over for his teacher.

Thus, the number of friends = x² – 3x + 6
The number of chocolates left over for his teacher = 5x + 11.
Given that 26 chocolates are left with him.
∴ 5x + 11 = 26
∴ 5x = 15
∴ x = 3
The total number of chocolates the student had
= x⁴ – 3x³ + 5x² + 8x + 5
= (3)⁴ – 3(3)³ + 5(3)² + 8(3) + 5
= 81 – 81 + 45 + 24 + 5
= 74
Thus, the student had 74 chocolates.
The number of chocolates received by each friend = x² – 1
= (3)² – 1
= 9 – 1
= 8
Thus, each friend received 8 chocolates.
The number of friends
= x² – 3x + 6
= (3)² – 3(3) + 6
= 9 – 9 + 6
= 6
Thus, the boy has 6 friends. 

Solution 4:

If we divide the total sum collected Rs.(2x³ + x² – 5x – 3) by the contribution received from each student  Rs.(2x + 3), the quotient will give the number of students and the remainder must be zero.

Solution 5:

Solution 6:

Since (x – 4) is a factor of x³ – 6x² + 4x + 16, split the term other than the first and the last to obtain the other factor.
x³ – 6x² + 4x + 16
= x³ – 4x² – 2x² + 8x – 4x + 16
 = x²(x – 4) – 2x(x – 4) – 4(x – 4)
= (x – 4)(x² – 2x – 4)
Thus, the other factor is (x² – 2x – 4).

Solution 7:

(107)²
= (100 + 7)²
= (100)² + 2(100)(7) + (7)²
= 10000 + 1400 + 49
= 11449

Solution 8:

If a + b + c = 0, then a³ + b³ + c³ = 3abc
Let a = (-7), b = 12 and c = (-5)
∴ a + b + c = (-7) + 12 + (-5) = -7 + 12 + -5 = 0
Now, a³ + b³ + c³ = 3abc
∴ (-7)³ + (12)³ + (-5)³
= 3(-7)(12)(-5)
= (-21)(-60)
= 1260

Solution 9:

4x² + 9y² + 25z² + 12xy – 30yz – 20zx
= (2x)² + (3y)² + (-5z)² + 2(2x)(3y) + 2(3y)(-5z) + 2(-5z)(2x)
= (2x + 3y – 5z)²

Solution 10(1) :

Solution 10(2) :

Solution 10(3) :

Solution 10(4) :

Solution 11 :

a + b + c = 6
∴ (a + b + c)² = (6)²
∴ a² + b² + c² + 2ab + 2bc + 2ca = 36
∴ 60 + 2(ab + bc + ca) = 36
∴ 2(ab + bc + ca) = 36 – 60
∴ 2(ab + bc + ca) = -24
∴ ab + bc + ca = -12

a³ + b³ + c³ – 3abc
= (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c)[(a² + b² + c²) – (ab + bc + ca)]
= (6)[(60) – (-12)]
= 6(60 + 12)
= 6(72)
= 432

Solution 12(1) :

a. (x – 3)
If p(3) = 0, then factor of p(x) is (x – 3).

Solution 12(2) :

b. 13
Remainder = p(2) = (2)³ + 2(2)² – 6(2) + 9 = 13
If x³ + 2x² – 6x + 9 is divided by (x – 2), then 13 is the remainder

Solution 12(3) :

d. 5
The degree of the polynomial x⁵ + 3x³ – 7x² + 9x + 11 is 5.

Solution 12(4) :

d. -18
(x – 2) is a factor of p(x) = 3x⁴ – 2x³ + 7x² – 21x + k.
∴ p(2) = 0
∴ 3(2)⁴ – 2(2)³ + 7(2)² – 21(2) + k = 0
∴ 48 – 16 + 28 – 42 + k = 0
∴ 18 + k = 0
∴ k = -18

Solution 12(5) :

Solution 12(6) :

b. 2
When polynomial p(x) is divided by x + 1, the remainder is p(-1).
∴  Remainder = p(-1) = (-1)² + 6(-1) + 7 = 2

Solution 12(7) :

d. (y + 3)(y + 7)
y² + 10y +21
= y² + 3y + 7y + 21
= y(y + 3) + 7(y + 3)
= (y + 3)(y + 7)
Factors of y² + 10y +21 are (y + 3)(y + 7).

Solution 12(8) :

c. 26
a³ – b³ = (a – b)³ + 3ab(a – b) = (2)³ + 3(3)(2) = 26

Solution 12(9) :

d. 0
a³ + b³ + c³ – 3abc
= a³ + a³ + a³ – 3a(a)(a)
= 3a³ – 3a³
= 0

Solution 12(10) :

c. x² + x – 6

Solution 12(11) :

b. 1
Remainder = p(-3) = (-3)³ + 28 = (-27) + 28 = 1

Solution 12(12) :

c. 12
Opposite of the remainder should be added to make the polynomial divisible by x – 4.
Remainder = p(4) = (4)³ – 76 = 64 – 76 = -12
Thus, 12 should be added.

Solution 12(13) :

c. 5x + 7y
25x² – 49y² = (5x)² – (7y)² = (5x + 7y)(5x – 7y)

Solution 12(14) :

b. 1
p(0) = -6 ≠ 0, p(1) = 0
p(2) = 8 ≠ 0 and p(3) = 24 ≠ 0.
Hence, 1 is a zero of p(x).

Solution 12(15) :

Solution 12(16) :

c. 64x³ – 343y³ – 336x²y + 588xy²
(4x – 7y)³
=(4x)³ – (7y)³ – 3(4x)(7y)(4x – 7y)
= 64x³ – 343y³ – 84xy(4x – 7y)
= 64x³ – 343y³ – 336x²y + 588xy²

Exercise 3.1:

Solution 1:

1. p(x) = 3x⁷ – 6x⁵ + 4x³ – x² – 5
   The degree of the polynomial is 7.
2. p(x) = x¹⁰⁰ – (x¹⁰)²⁰ + 3x⁵⁰ + x²⁵ + x⁵ – 7
   ∴p(x) = x¹⁰⁰ – x²⁰⁰ + 3x⁵⁰ + x²⁵ + x⁵ – 7
   The degree of polynomial is 200.
3. p(x) = 7x – 3x² + 4x³ – x⁴
   Thus, the degree of the polynomial is 4.
4. p(x) = 3.14x² + 1.57x + 1
   The degree of the polynomial is 2.

Solution 2:

1. p(x) = 4x³ + 3x² + 2x + 1
   In the polynomial p(x), the coefficient of x³ is 4.
2. p(x) = x² + 2x + 1
   In the polynomial p(x) the term with x³ is absent.
   Hence, the coefficient of x³ in the polynomial p(x) is 0.

Solution 3:

1. (x) = x²+ 27
   The degree of the polynomial p(x) is 2.
   Hence the given polynomial is a quadratic polynomial.
2. p(x) = 2010x + 2009
   The degree of the polynomial p(x) is 1.
   Hence the given polynomial is a linear polynomial.
3. p(x) = 4x² + 7x³ + 3
   The degree of the polynomial p(x) is 3.
   Hence the given polynomial is a cubic polynomial.
4. p(x) = (x – 1)(x + 1) = x² – 1
   The degree of the polynomial p(x) is 2.
   Hence, the given polynomial is a quadratic polynomial.

Solution 4:

1. p(x) = x⁷ + 10x⁵ + 4x³ + 3x + 1
   The index in each term of the given algebraic expression is a whole number.
   Thus, p(x) is a polynomial.

Solution 5:

8x¹⁰ is a monomial of degree 10.
4x²⁰ + 9x¹⁵ is a binomial of degree 20.

Exercise 3.2:

Solution 1:

Solution 2:

p(x) = x⁴ + 2x³ – x + 5
At x = 2,
p(2) = (2)⁴ + 2(2)³ – 2 + 5 = 16 + 16 – 2 + 5
∴p(2) = 35
p(x) = 3x³ – 5x² + 6x – 9

At x = 0,
p(0) = 3(0)³ – 5(0)² + 6(0) – 9 = 0 – 0 + 0 – 9
∴p(0) = -9
At x = -1
p(-1)
= 3(-1)³ – 5(-1)² + 6(-1) – 9
= 3(-1) – 5(1) + 6(-1) – 9
= -3 – 5 – 6 – 9
∴p(-1) = -23

p(x) = 5x³ + 11x² + 10
At x = -2,
p(-2)
= 5(-2)³ + 11(-2)² + 10
= 5(-8) + 11(4) + 10
= -40 + 44 + 10
∴p(-2) = 14

1. p(x) = x⁴ + 2x³ – x + 5
   At x = 2,
   p(2) = (2)⁴ + 2(2)³ – 2 + 5 = 16 + 16 – 2 + 5
   ∴p(2) = 35
2. p(x) = 3x³ – 5x² + 6x – 9
   At x = 0,
   p(0) = 3(0)³ – 5(0)² + 6(0) – 9 = 0 – 0 + 0 – 9
   ∴p(0) = -9
   At x = -1
   p(-1)
   = 3(-1)³ – 5(-1)² + 6(-1) – 9
   = 3(-1) – 5(1) + 6(-1) – 9
   = -3 – 5 – 6 – 9
   ∴p(-1) = -23
3. p(x) = 5x³ + 11x² + 10
   At x = -2,
   p(-2)
   = 5(-2)³ + 11(-2)² + 10
   = 5(-8) + 11(4) + 10
   = -40 + 44 + 10
   ∴p(-2) = 14

Solution 3:

1. p(x) = x⁷
   ∴ p(0) = 0⁷ = 0
   ∴ p(1) = (1)⁷ = 1
   ∴ p(2) = (2)⁷ = 128
2. p(x) = (x – 1)(x + 3)
   ∴ p(0) = (0 – 1)(0 + 3) = -1 × 3 = -3
   ∴ p(1) = (1 – 1)(1 + 3) = 0 × 4 = 0
   ∴ p(2) = (2 – 1)(2 + 3) = 1 × 5 = 5
3. p(x) = x²– 2x
   ∴ p(0) = (0)² – 2(0) = 0 – 0 = 0
   ∴ p(1) = (1)² – 2(1) = 1 – 2 = -1
   ∴ p(2) = (2)² – 2(2) = 4 – 4 = 0

Solution 4:

Exercise 3.3:

Solution 1(1):

Solution 1(2):

Solution 2(1):

Solution 2(2):

Solution 2(3):

Solution 2(4):

Solution 2(5):

Solution 3:

When p(x) = x⁴ – 4x³ + 3x – 1 is divided by d (x) = x + 2, the remainder is given by p(-2).
Remainder(R) = p(-2)
p(x) = x⁴ – 4x³ + 3x – 1
∴ p(-2)
x⁴ – 4x³ + 3x – 1
= (-2)⁴ – 4(-2)³ + 3(-2) – 1
= 16 – 4(-8) – 6 – 1
= 16 + 32 – 6 – 1
= 41
Thus, the remainder is 41.

Solution 4:

The quantity to be added to polynomial p(y) so that the resulting polynomial becomes divisible by y + 1 is the opposite of the remainder obtained on dividing polynomial p(y) by y + 1.

So, 4 should be added to polynomial p(y) = 12y³ – 39y² + 50y + 97 so that the resulting polynomial is divisible by y + 1.

Solution 5:

Solution 6:

Solution 7(1):

Solution 7(2):

Solution 7(3):

Solution 8:

When the polynomial p(x) = ax⁵ – 23x³ + 47x + 1 is divided by (x – 2), the remainder is given by p(2).
∴ Remainder = p(2)
p(x) = ax⁵ – 23x³ + 47x + 1
∴ p(2)
= a(2)⁵ – 23(2)³ + 47(2) + 1
= 32a – 184 + 94 + 1
= 32a – 89

But, the remainder is 7. (given)
∴ 32a – 89 = 7
∴ 32a = 7 + 89
∴ 32a = 96
∴ a = 3

Exercise 3.4:

Solution 1:

(x – 1) is a factor of the polynomial p(x), if and only if the sum of all the coefficients of p(x) is zero.
Also, if p(1) = 0, then (x – 1) is a factor of p(x).

1. 2x³ – 3x² + 3x – 2
   The sum of all the coefficients
   = 2 + (-3) + 3 + (-2)
   = 0
   Thus, (x – 1) is a factor of 2x³ – 3x² + 3x – 2.
2. 4x³ + x⁴ – x + 1
   The sum of all the coefficients
   = 4 + 1 + (-1) + 1
   ≠ 0
   Thus, (x – 1) is not a factor of 4x³ + x⁴ – x + 1.
3. 5x⁴ – 4x³ – 2x + 1
   The sum of all the coefficients
   = 5 + (-4) + (-2) + 1
   = 0
   Thus, (x – 1) is a factor of 5x⁴ – 4x³ – 2x + 1.
4. 3x³ + x² + x + 11
   The sum of all the coefficients
   = 3 + 1 + 1 + 11
   = 16
   ≠ 0
   Thus, (x – 1) is not a factor of 3x³ + x² + x + 11.

Solution 2:

1. p(x) = 21x² + 26x + 8
   d(x) = 3x + 2 is a factor of p(x).
   21x² + 26 + 8
   = 21x² + 14x + 12x + 8
   (Splitting the middle term to get 3x + 2 as a factor)
   = 7x(3x + 2) + 4(3x + 2)
   = (3x + 2)(7x + 4)
   Thus, (7x + 4) is other factor of p(x).
2. p(x) = x³ + 10x² + 23x + 14
   d(x) = x + 1 is a factor of p(x).
   x³ + 10x² + 23x + 14
   = x³ + x² + 9x² + 9x + 14x + 14
   (Splitting the terms to get x + 1 as a factor)
   = x²(x + 1) + 9x(x + 1) + 14(x + 1)
   = (x + 1)(x² + 9x +14)
   Thus, (x² + 9x +14) is other factor of p(x).
3. p(x) = x³ – 9x² + 20x – 12
   d(x) = x – 6 is a factor of p(x).
   x³ – 9x² + 20x – 12
   = x³ – 6x²– 3x²+ 18x + 2x – 12
   = (x – 6)(x² + 3x + 2)
   Thus, (x² + 3x + 2) is other factor of p(x).

Solution 3:

If p(x) = ax³ + 3x² + 7x + 13 is divided by (x + 3),
the remainder is given by p(-3).
Remainder = p(-3)
p(-3)
= a(-3)³ + 3(-3)² + 7(-3) + 13
= 27a + 27 – 21 + 13
= 19 – 27a
But, the remainder is -8. (given)
∴ 19 – 27a = -8
∴ -27a = -8 – 19
∴ -27a = -27
∴ a = 1

Solution 4:

1. 3x²+ 7x + 4
   = 3x² + 3x + 4x + 4
   =3x(x + 1) + 4(x + 1)
   = (x + 1)(3x + 4)
2. 15x²+ 16x + 4
   = 15x² + 10x + 6x + 4
   = 5x(3x + 2) + 2(3x + 2)
   = (3x + 2)(5x + 2)
3. -21x²+ 16x + 5
   = -(21x² – 16x – 5)
   = -(21x² – 21x + 5x – 5)
   = -[21x(x – 1) + 5(x – 1)]
   = -(x – 1)(21x + 5)

Solution 5:

1. p(x) = 3x³ – 7x² + 5x – 1
   The sum of all the coefficients
   = 3 + (-7) + 5 + (-1) = 0
   The sum of the coefficients of the odd powers of x
   = 3 + 5 = 8
   The sum of the coefficients of the even powers of x
   = (-7) + (-1) = (-8) ≠ 8
   Thus, (x – 1) is a factor of p(x), but (x + 1) is not a
   actor of p(x).
2. p(x) = 21x³ + 16x² + 4x + 9
   The sum of all the coefficients
   = 21 + 16 + 4 + 9 = 50 ≠ 0
   The sum of the coefficients of the odd powers of x
   = 21 + 4 = 25
   The sum of the coefficients of the even powers of x
   = 16 + 9 = 25
   Thus, (x + 1) is a factor of p(x), but (x – 1) is not a
   actor of p(x).
3. p(x) = 2x⁴ – 3x³ + 4x² – 5x + 2
   The sum of all the coefficients
   = 2 + (-3) + 4 + (-5) + 2
   = 8 – 8 = 0
   The sum of the coefficients of the odd powers of x
   = (-3) + (-5)
   = (-8)
   The sum of the coefficients of the even powers of x
   = 2 + 4 + 2
   = 8 ≠ (-8)
   Hence, (x – 1) is a factor of p(x), but (x + 1) is not a
   actor of p(x).
4. p(x) = x³ + 13x² + 32x + 20
   The sum of all the coefficients
   = 1 + 13 + 32 + 20
   = 66 ≠ 0
   The sum of the coefficients of the odd powers of x
   = 1 + 32
   = 33
   The sum of the coefficients of the even powers of x
   = 13 + 20
   = 33
   Thus, (x + 1) is a factor of p(x), but (x – 1) is not a
   actor of p(x).

Solution 6:

p(x) = ax⁴ – 7x³ – 3x² – 2x – 8 and x – 4 is a factor of p(x).
∴ p(4) = 0
∴ a(4)⁴ – 7(4)³ – 3(4)² – 2(4) – 8 = 0
∴ 256a – 448 – 48 – 8 – 8 = 0
∴ 256a – 512 = 0
∴ 256a = 512
∴ a = 2

Exercise 3.5

Solution 1(1):

(x – 7)(x – 12)
= [x + (-7)][x + (-12)]
= x² + [(-7) + (-12)]x + (-7)(-12)
= x² – 19x + 84

Solution 1(2):

(5 – 4x)(7 – 4x)
= (-4x + 5)(-4x + 7)
= (-4x)² + (5 + 7)(-4x) + (5)(7)
= 16x² – 48x + 35

Solution 1(3):

Solution 1(4):

Solution 2:

1. 97 × 103
   = (100 – 3)(100 + 3)
   = (100)² – (3)²
   = 10000 – 9
   = 9991
2. 57 × 63
   = (60 – 3)(60 + 3)
   = (60)² – (3) ²
   = 3600 – 9
   = 3591
3. 34 × 26
   = (30 + 4)(30 – 4)
   = (30)² – (4)²
   = 900 – 16
   = 884

Solution 3(1):

16x² – 40xy + 25y²
= (4x)² – 2(4x)(5y) + (5y)²
= (4x – 5y)²

Solution 3(2):

Solution 3(3):

9a² + 25b² + 49c² – 30ab + 70bc – 42ac
= (3a)² + (-5b)² + (-7c)² + 2(3a)(-5b) + 2(-5b)(-7c) + 2(-7c)(3a)
= (3a – 5b – 7c)²
OR
9a² + 25b² + 49c² – 30ab + 70bc – 42ac
= (-3a)² + (5b)² + (7c)² + 2(-3a)(5b) + 2(5b)(7c) + 2 (7c)(-3a)
= (-3a + 5b + 7c)²

Solution 3(4):

16a⁴ – 625b⁴
= (4a²)² – (25b²)²
= (4a² + 25b²)(4a² – 25b²)
= (4a² + 25b²)(2a + 5b)(2a – 5b)

Solution 3(5):

Solution 3(6):

125a³ + 600a²b + 960ab² + 512b³
= 125a³ + 512b³ + 600a²b + 960ab²
= (5a)³ + (8b)³ + 120ab(5a + 8b)
= (5a)³ + (8b)³ + 3(5a)(8b)(5a + 8b)
= (5a + 8b)³

Solution 3(7):

64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 36ab(4a – 3b)
= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³

Solution 4:

1. 105 × 102
   = (100 + 5)(100 + 2)
   = (100)² + (5 + 2)100 + (5)(2)
   = 10000 + 700 + 10
   = 10710
2. (92)²
   = (90 + 2)²
   = (90)² + 2(90)(2) + (2)²
   = 8100 + 360 + 4
   = 8464
3. (8)³ – (4)³
   = (8 – 4)³ + 3(8)(4)(8 – 4)
   = (4)³ + 3(32)(4)
   = 64 + 384
   = 448

Solution 5:

(-28)³ + (15)³ + (13)³
Let a = -28, b = 15 and c = 13
Then, a + b + c = (-28) + 15 + 13 = 0
Now, if a + b + c = 0 then a³ + b³ + c³ = 3abc
∴ (-28)³ + (15)³ + (13)³
= 3(-28)(15)(13)
= (-84)(195)
= -16380