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

GSEB SolutionsMathsScience

**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 (x^{4} – 3x^{3} + 5x^{2} + 8x + 5) by the number of chocolates received by each friend (x^{2} – 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 = x^{2} – 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

= x^{4} – 3x^{3} + 5x^{2} + 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 = x^{2} – 1

= (3)^{2} – 1

= 9 – 1

= 8

Thus, each friend received 8 chocolates.

The number of friends

= x^{2} – 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.(2x^{3} + x^{2} – 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 x^{3} – 6x^{2} + 4x + 16, split the term other than the first and the last to obtain the other factor.

x^{3} – 6x^{2} + 4x + 16

= __x ^{3} – 4x^{2}__ –

__2x__–

^{2}+ 8x__4x + 16__

= x

^{2}(x – 4) – 2x(x – 4) – 4(x – 4)

= (x – 4)(x

^{2}– 2x – 4)

Thus, the other factor is (x

^{2}– 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 a^{3} + b^{3} + c^{3} = 3abc

Let a = (-7), b = 12 and c = (-5)

∴ a + b + c = (-7) + 12 + (-5) = -7 + 12 + -5 = 0

Now, a^{3} + b^{3} + c^{3} = 3abc

∴ (-7)^{3} + (12)^{3} + (-5)^{3
}= 3(-7)(12)(-5)

= (-21)(-60)

= 1260

**Solution 9:**

4x^{2} + 9y^{2} + 25z^{2} + 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
}∴ a^{2} + b^{2} + c^{2} + 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

a^{3} + b^{3} + c^{3} – 3abc

= (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

= (a + b + c)[(a^{2} + b^{2} + c^{2}) – (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 x^{3} + 2x^{2} – 6x + 9 is divided by (x – 2), then 13 is the remainder

**Solution 12(3) :**

d. 5

The degree of the polynomial x^{5} + 3x^{3} – 7x^{2} + 9x + 11 is 5.

**Solution 12(4) :**

d. -18

(x – 2) is a factor of p(x) = 3x^{4} – 2x^{3 }+ 7x^{2} – 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)

y^{2} + 10y +21

= y^{2} + 3y + 7y + 21

= y(y + 3) + 7(y + 3)

= (y + 3)(y + 7)

Factors of y^{2} + 10y +21 are (y + 3)(y + 7).

**Solution 12(8) :**

c. 26

a^{3} – b^{3} = (a – b)^{3} + 3ab(a – b) = (2)^{3} + 3(3)(2) = 26

**Solution 12(9) :**

d. 0

a^{3} + b^{3} + c^{3} – 3abc

= a^{3} + a^{3} + a^{3} – 3a(a)(a)

= 3a^{3} – 3a^{3
}= 0

**Solution 12(10) :**

c. x^{2} + 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

25x^{2} – 49y^{2 }= (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. 64x^{3} – 343y^{3} – 336x^{2}y + 588xy^{2
}(4x – 7y)^{3
}=(4x)^{3} – (7y)^{3} – 3(4x)(7y)(4x – 7y)

= 64x^{3} – 343y^{3} – 84xy(4x – 7y)

= 64x^{3} – 343y^{3} – 336x^{2}y + 588xy^{2}

**Exercise 3.1:**

**Solution 1:**

- p(x) = 3x
^{7}– 6x^{5}+ 4x^{3}– x^{2}– 5

The degree of the polynomial is 7. - p(x) = x
^{100}– (x^{10})^{20}+ 3x^{50}+ x^{25}+ x^{5}– 7

∴p(x) = x^{100}– x^{200}+ 3x^{50}+ x^{25}+ x^{5}– 7

The degree of polynomial is 200. - p(x) = 7x – 3x
^{2}+ 4x^{3}– x^{4 }Thus, the degree of the polynomial is 4. - p(x) = 3.14x
^{2}+ 1.57x + 1

The degree of the polynomial is 2.

**Solution 2:**

- p(x) = 4x
^{3}+ 3x^{2}+ 2x + 1

In the polynomial p(x), the coefficient of x^{3 }is 4. - p(x) = x
^{2}+ 2x + 1

In the polynomial p(x) the term with x^{3 }is absent.

Hence, the coefficient of x^{3 }in the polynomial p(x) is 0.

**Solution 3:**

- (x) = x
^{2}+ 27

The degree of the polynomial p(x) is 2.

Hence the given polynomial is a quadratic polynomial. - p(x) = 2010x + 2009

The degree of the polynomial p(x) is 1.

Hence the given polynomial is a linear polynomial. - p(x) = 4x
^{2}+ 7x^{3}+ 3

The degree of the polynomial p(x) is 3.

Hence the given polynomial is a cubic polynomial. - p(x) = (x – 1)(x + 1) = x
^{2}– 1

The degree of the polynomial p(x) is 2.

Hence, the given polynomial is a quadratic polynomial.

**Solution 4:**

- p(x) = x
^{7}+ 10x^{5}+ 4x^{3}+ 3x + 1

The index in each term of the given algebraic expression is a whole number.

Thus, p(x) is a polynomial.

**Solution 5:**

8x^{10} is a monomial of degree 10.

4x^{20} + 9x^{15} is a binomial of degree 20.

**Exercise 3.2:**

**Solution 1:**

**Solution 2:**

p(x) = x^{4} + 2x^{3} – x + 5

At x = 2,

p(2) = (2)^{4} + 2(2)^{3} – 2 + 5 = 16 + 16 – 2 + 5

∴p(2) = 35

p(x) = 3x^{3} – 5x^{2} + 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) = 5x^{3} + 11x^{2} + 10

At x = -2,

p(-2)

= 5(-2)^{3} + 11(-2)^{2} + 10

= 5(-8) + 11(4) + 10

= -40 + 44 + 10

∴p(-2) = 14

- p(x) = x
^{4}+ 2x^{3}– x + 5

At x = 2,

p(2) = (2)^{4}+ 2(2)^{3}– 2 + 5 = 16 + 16 – 2 + 5

∴p(2) = 35 - p(x) = 3x
^{3}– 5x^{2}+ 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) = 5x
^{3}+ 11x^{2}+ 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:**

- p(x) = x
^{7 }∴ p(0) = 0^{7}= 0

∴ p(1) = (1)^{7}= 1

∴ p(2) = (2)^{7}= 128 - 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 - p(x) = x
^{2}– 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) = x^{4} – 4x^{3} + 3x – 1 is divided by d (x) = x + 2, the remainder is given by p(-2).

Remainder(R) = p(-2)

p(x) = x^{4} – 4x^{3} + 3x – 1

∴ p(-2)

x^{4} – 4x^{3} + 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) = 12y^{3} – 39y^{2} + 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) = ax^{5} – 23x^{3 }+ 47x + 1 is divided by (x – 2), the remainder is given by p(2).

∴ Remainder = p(2)

p(x) = ax^{5} – 23x^{3} + 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).

- 2x
^{3}– 3x^{2}+ 3x – 2

The sum of all the coefficients

= 2 + (-3) + 3 + (-2)

= 0

Thus, (x – 1) is a factor of 2x^{3}– 3x^{2}+ 3x – 2. - 4x
^{3}+ x^{4 }– x + 1

The sum of all the coefficients

= 4 + 1 + (-1) + 1

≠ 0

Thus, (x – 1) is not a factor of 4x^{3}+ x^{4 }– x + 1. - 5x
^{4 }– 4x^{3 }– 2x + 1

The sum of all the coefficients

= 5 + (-4) + (-2) + 1

= 0

Thus, (x – 1) is a factor of 5x^{4 }– 4x^{3 }– 2x + 1. - 3x
^{3}+ x^{2 }+ x + 11

The sum of all the coefficients

= 3 + 1 + 1 + 11

= 16

≠ 0

Thus, (x – 1) is not a factor of 3x^{3}+ x^{2 }+ x + 11.

**Solution 2:**

- p(x) = 21x
^{2}+ 26x + 8

d(x) = 3x + 2 is a factor of p(x).

21x^{2}+ 26 + 8

= 21x^{2}+ 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). - p(x) = x
^{3}+ 10x^{2}+ 23x + 14

d(x) = x + 1 is a factor of p(x).

x^{3}+ 10x^{2}+ 23x + 14

= x^{3}+ x^{2}+ 9x^{2}+ 9x + 14x + 14

(Splitting the terms to get x + 1 as a factor)

= x^{2}(x + 1) + 9x(x + 1) + 14(x + 1)

= (x + 1)(x^{2}+ 9x +14)

Thus, (x^{2}+ 9x +14) is other factor of p(x). - p(x) = x
^{3}– 9x^{2}+ 20x – 12

d(x) = x – 6 is a factor of p(x).

x^{3}– 9x^{2}+ 20x – 12

= x^{3 }– 6x^{2}– 3x^{2}+ 18x + 2x – 12

= (x – 6)(x^{2}+ 3x + 2)

Thus, (x^{2}+ 3x + 2) is other factor of p(x).

**Solution 3:**

If p(x) = ax^{3} + 3x^{2} + 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:**

- 3x
^{2}+ 7x + 4

= 3x^{2}+ 3x + 4x + 4

=3x(x + 1) + 4(x + 1)

= (x + 1)(3x + 4) - 15x
^{2}+ 16x + 4

= 15x^{2}+ 10x + 6x + 4

= 5x(3x + 2) + 2(3x + 2)

= (3x + 2)(5x + 2) - -21x
^{2}+ 16x + 5

= -(21x^{2 }– 16x – 5)

= -(21x^{2 }– 21x + 5x – 5)

= -[21x(x – 1) + 5(x – 1)]

= -(x – 1)(21x + 5)

**Solution 5:**

- p(x) = 3x
^{3}– 7x^{2}+ 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). - p(x) = 21x
^{3}+ 16x^{2}+ 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). - p(x) = 2x
^{4 }– 3x^{3 }+ 4x^{2 }– 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). - p(x) = x
^{3}+ 13x^{2}+ 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) = ax^{4} – 7x^{3} – 3x^{2} – 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)]

= x^{2} + [(-7) + (-12)]x + (-7)(-12)

= x^{2} – 19x + 84

**Solution 1(2):**

(5 – 4x)(7 – 4x)

= (-4x + 5)(-4x + 7)

= (-4x)^{2} + (5 + 7)(-4x) + (5)(7)

= 16x^{2} – 48x + 35

**Solution 1(3):**

**Solution 1(4):**

**Solution 2:**

- 97 × 103

= (100 – 3)(100 + 3)

= (100)^{2}– (3)^{2 }= 10000 – 9

= 9991 - 57 × 63

= (60 – 3)(60 + 3)

= (60)^{2 }– (3)^{ 2 }= 3600 – 9

= 3591 - 34 × 26

= (30 + 4)(30 – 4)

= (30)^{2 }– (4)^{2 }= 900 – 16

= 884

**Solution 3(1):**

16x^{2} – 40xy + 25y^{2
}= (4x)^{2} – 2(4x)(5y) + (5y)^{2
}= (4x – 5y)^{2}

**Solution 3(2):**

**Solution 3(3):**

9a^{2} + 25b^{2} + 49c^{2} – 30ab + 70bc – 42ac

= (3a)^{2} + (-5b)^{2} + (-7c)^{2} + 2(3a)(-5b) + 2(-5b)(-7c) + 2(-7c)(3a)

= (3a – 5b – 7c)^{2
}OR

9a^{2} + 25b^{2} + 49c^{2} – 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):**

16a^{4} – 625b^{4
}= (4a^{2})^{2} – (25b^{2})^{2
}= (4a^{2} + 25b^{2})(4a^{2 }– 25b^{2})

= (4a^{2} + 25b^{2})(2a + 5b)(2a – 5b)

**Solution 3(5):**

**Solution 3(6):**

125a^{3} + 600a^{2}b + 960ab^{2} + 512b^{3
}= 125a^{3} + 512b^{3} + 600a^{2}b + 960ab^{2
}= (5a)^{3} + (8b)^{3} + 120ab(5a + 8b)

= (5a)^{3} + (8b)^{3} + 3(5a)(8b)(5a + 8b)

= (5a + 8b)^{3}

**Solution 3(7):**

64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2
}= (4a)^{3 }– (3b)^{3 }– 36ab(4a – 3b)

= (4a)^{3 }– (3b)^{3} – 3(4a)(3b)(4a – 3b)

= (4a – 3b)^{3}

**Solution 4:**

- 105 × 102

= (100 + 5)(100 + 2)

= (100)^{2}+ (5 + 2)100 + (5)(2)

= 10000 + 700 + 10

= 10710 - (92)
^{2 }= (90 + 2)^{2 }= (90)^{2}+ 2(90)(2) + (2)^{2 }= 8100 + 360 + 4

= 8464 - (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 a^{3} + b^{3} + c^{3} = 3abc

∴ (-28)^{3} + (15)^{3} + (13)^{3
}= 3(-28)(15)(13)

= (-84)(195)

= -16380