## Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.

Classify the following polynomials based on number of terms.

(i) x^{3} – x^{2}

(ii) 5x

(iii) 4x^{4} + 2x^{3} + 1

(iv) 4.x^{3}

(v) x + 2

(vi) 3x^{2}

(vii) y^{4} + 1

(viii) y^{20} + y^{18} + y^{2}

(ix) 6

(x) 2u^{3} + u^{2} + 3

(xi) u^{23} – u^{4}

(xii) y

Solution:

5x, 3x^{2}, 4x^{3}, y and 6 are monomials because they have only one term.

x^{3} – x^{2}, x + 2, y^{4} + 1 and u^{23} – u^{4} are binomials as they contain only two terms.

4x^{4} + 2x^{3} + 1 , y^{20} + y^{18} + y^{2} and 2u^{3} + u^{2} + 3 are trinomials as they contain only three terms.

Question 2.

Classify the following polynomials based on their degree.

Solution:

p(x) = 3, p(x) = -7, p(x) = \(\frac{3}{2}\) are constant polynomials

p(x) = x + 3, p(x) = 4x, p(x) = \(\sqrt{3}\)x + 1 are linear polynomials, since the highest degree of the variable x is one.

p(x) = 5x^{2} – 3x + 2, p(y) = \(\frac{5}{2}\) y^{2} + 1, p(x) = 3x^{2} are quadratic polynomials, since the highest degree of the variable is two.

p(x) = 2x^{3} – x^{2} + 4x + 1, p(x) = x^{3} + 1, p(y) = y^{3} + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.

Find the product of given polynomials p(x) = 3x^{3}+ 2x – x^{2} + 8 and q (x) = 7x + 2.

Solution:

(7x +2) (3x^{3} + 2x – x^{2} + 8) = 7x(3x^{3} + 2x – x^{2} + 8) + 2x

(3x^{3} + 2x – x^{2} + 8) = 21x^{4} + 14x^{2} – 7x^{3} + 56x + 6x^{3} + 4x – 2x^{2}+ 16 = 21x^{4} – x^{3} + 12x^{4} + 60x + 16

Question 4.

Let P(x) = 4x^{4} – 3x + 2x^{3} + 5 and q(x) = x^{2} + 2x + 4 find p (x) – q(x).

Solution:

Exercise 3.2

Question 1.

If p(x) = 5x^{3} – 3x^{2} + 7x – 9, find

(i) p(-1)

(ii) p(2).

Solution:

Given that p(x) = 5x^{3} – 3x^{2} + 7x – 9

(i) p(-1) = 5(-1)^{3} – 3(-1)^{2} + 7(-1) – 9 = -5 – 3 – 7 – 9

= -24

(ii) p(2) = 5(2)^{3} – 3(2)^{3} + 7(2) – 9 = 40 – 12 + 14 – 9

∴ p(2) = 33

Question 2.

Find the zeros of the following polynomials.

(i) p(x) = 2x – 3

(ii) p(x) = x – 2

Solution:

Exercise 3.3

Question 1.

Find the remainder using remainder theorem, when

(i) 4x^{3} – 5x^{2} + 6x – 2 is divided by x – 1.

(ii) x^{3} – 7x^{2} – x + 6 is divided by x + 2.

Solution:

(i) Let p(x) = 4x^{3} – 5x^{2} + 6x – 2. The zero of (x – 1) is 1.

When p(x) is divided by (x – 1) the remainder is p( 1). Now,

p(1) = 4(1)^{3} – 5(1)^{2} + 6(1) – 2 = 4 – 5 + 6 – 2 = 3

∴ The remainder is 3.

(ii) Let p(x) = x^{3} – 7x^{2} – x + 6. The zero of x + 2 is -2.

When p(x) is divided by x + 2, the remainder is p(-2). Now,

p(-2) = (-2)^{3} – 7(-2)^{2} – (-2) +6 = – 8 – 7(4) + 2 + 6

= – 8 – 28 + 2 + 6 = -28.

∴ The remainder is -28.

Question 2.

Find the value of a if 2x^{3} – 6x^{2} + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.

Solution:

Let p(x) = 2x^{3} – 6x^{2} + 5ax – 9

When p(x) is divided by (x – 2) the remainder is p(2).

Given that p(2) = 13

⇒ 2(2)^{3} – 6(2)^{2} + 5a(2) – 9 = 13

⇒ 2(8) – 6(4) + 10a – 9 = 13

⇒ 16 – 24 + 10a – 9 = 13

⇒ 10a – 17 = 13

⇒ 10a = 30

⇒ ∴ a = 3

Question 3.

If the polynomials f(x) = ax^{3}+ 4x^{2} + 3x – 4 and g (x) = x^{3} – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.

Solution:

Let f(x) = ax^{3} + 4x^{2} + 3x – 4 and g(x) = x^{3} – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).

Now f(3) = a(3)^{3}4(3)^{2} + 3(3) – 4 = 27a + 36 + 9 – 4

f (3) = 27a + 41 …(1)

When g(x) is divided by (x – 3), the remainder is g(3).

Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)

Since the remainders are same, (1) = (2)

Given that, f(3) = g(3)

That is 27a + 41 = 15 +a

27a – a = 15 – 41

26a = -26

Sustituting a = -1, in f(3), we get

f(3) = 27(-1) + 41 = -27 + 41

f(3) = 14

∴ The remainder is 14.

Question 4.

Show that x + 4 is a factor of x^{3} + 6x^{2} – 7x – 60.

Solution:

Question 5.

In (5x + 4) a factor of 5x^{3} + 14x^{2} – 32x – 32

Solution:

Question 6.

Find the value of k, if (x – 3) is a factor of polynomial x^{3} – 9x^{2} + 26x + k.

Solution:

Let p (x) = x^{3} – 9x^{2} + 26x + k

By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0

p(3) = 0

3^{3} – 9(3)^{2} + 26 (3) + k = 0

27 – 81 + 78 + k = 0

k = -24

Question 7.

Show that (x – 3) is a factor of x^{3} + 9x^{2} – x – 105.

Solution:

Let p(x) = x^{3} + 9x^{2} – x – 105

By factor theorem, x – 3 is a factor of p(x), if p(3) = 0

p (3) = 3^{3} + 9(3)^{3} – 3 – 105

= 27 + 81 – 3 – 105

= 108 – 108

p(3) = 0

Therefore, x – 3 is a factor of x^{3} + 9x^{2} – x – 105.

Question 8.

Show that (x + 2) is a factor of x^{3} – 4x^{2} – 2x + 20.

Solution:

Let p(x) = x^{3} – 4x^{2} – 2x + 20

By factor theorem,

(x + 2) is factor of p(x), if p(-2) = 0

p(-2) = (-2)^{3} – 4(-2)^{2} – 2(-2) + 20 = -8 – 4(4) + 4 + 20

P(- 2) = 0

Therefore, (x + 2) is a factor of x^{3} – 4x^{2} – 2x + 20

Exercise 3.4

Question 1.

Expand the following using identities :

(i) (7x + 2y)^{2}

(ii) (4m – 3m)^{2}

(iii) (4a + 3b) (4a – 3b)

(iv) (k + 2)(k – 3)

Solution:

(i) (7x + 2y)^{2} = (7x)^{2} + 2 (7x) (2y) + (2y)^{2} = 49x^{2} + 28xy + 4y^{2}

(ii) (4m – 3m)^{2} = (4m)^{2} – 2(4m) (3m) + (3m)^{2} = 16m^{2} – 24mn + 9n^{2}

(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a^{2} – b^{2} ]

Put [a = 4a, b = 3b]

(4a + 3b) (4a – 3b) = (4a)^{2} – (3b)^{2} = 16a^{2} – 9b^{2}

(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x^{2} + (a – b)x – ab]

Put [x = k, a = 2, b = 3]

(k + 2) (k – 3) = k^{2} + (2 – 3)x – 2 × 3 = k^{2} – x – 6

Question 2.

Expand : (a + b – c)^{2}

Solution:

Replacing ‘c’ by ‘-c’ in the expansion of

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

(a + b + (-c))^{2} = a^{2} + b^{2} + (-c)^{2} + 2ab + 2b(-c) + 2(-c)a

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

Question 3.

Expand : (x + 2y + 3z)^{2}

Solution:

We know that,

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

Substituting, a = x, b = 2y and c = 3z

(x + 2y + 3 z)^{2} = x^{2} + (2y)^{2} + (3z)^{2} + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)

= x^{2} + 4y^{2} + 9z^{2} + 4xy + 12y^{2} + 6zx

Question 4.

Find the area of square whose side length is m + n – q.

Solution:

Area of square = side × side = (m + n – q)^{2}

We know that,

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

[m + n + (-q)]^{2} = m^{2} + n^{2} + (-q)^{2} + 2mn + 2n(-q) + 2(-q)m

= m^{2} + n^{2} + q^{2} + 2mn – 2nq – 2qm

Therefore, Area of square = m^{2} + n^{2} + q^{2} + 2mn – 2nq – 2qm = [m^{2} + n^{2} + q^{2} + 2mn – 2nq – 2qm] sq. units.

EXERCISE 3.5

Question 1.

Factorise the following

(i) 25m^{-2} – 16n^{2}

(ii) x^{4} – 9x^{2}

Solution:

(i) 25m^{2} – 16n^{2} = (5m)^{2} – (4n)^{2}

= (5m – 4n) (5m + 4n) [∵ a^{2} – b^{2} = (a – b)(a + b)]

(ii) x^{4} – 9x^{2} = x^{2}(x^{2} – 9) = x^{2}(x^{2} – 3^{2}) = x^{2}(x – 3)(x + 3)

Question 2.

Factorise the following.

(i) 64m^{3} + 27n^{3}

(ii) 729x^{3} – 343y^{3}

Solution:

(i) 64m^{3} + 27n^{3} = (4m)^{3} + (3n)^{3}

= (4m + 3n) ((4m)^{2} – (4m) (3n) + (3n)^{2}) [∵ a^{2} + b^{2} = (a + b) (a^{2} – ab + b^{2})]

= (4m + 3n) (16m^{2} – 12mn + 9n^{2})

(ii) 729x^{3} – 343y^{3} = (9x)^{3} – (7y)^{3} = (9x – 7y) ((9x)^{2} + (9x) (7y) + (7y)^{2}

= (9x – 7y) (81x^{2} + 63xy + 49y^{2})

[∵ a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})]

Question 3.

Factorise 81x^{2} + 90x + 25

Solution:

81x^{2} + 90x + 25 = (9x)^{2} + 2 (9x) (5) + 5^{2}

= (9x + 5)^{2} [ ∵ a^{2} + 2ab + b^{2} = (a + b)^{2}]

Question 4.

Find a^{3} + b^{2} if a + b = 6, ab = 5.

Solution:

Given, a + b = 6, ab = 5

a^{2} + b^{2} = (a + b)^{2} – 3ab(a + b) = (6)^{3} – 3(5)(6) = 126

Question 5.

Factorise (a – b)^{2} + 7(a – b) + 10

Solution:

Let a – b = p, we get p^{2} + 7p + 10,

p^{2} + 7p + 10 = p^{2} + 5p + 2p + 10

= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)

Put p = a – b we get,

(a – b)^{2} + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.

Factorise 6x^{2} + 17x + 12

Solution:

Exercise 3.6

Question 1.

Factorise x^{2} + 4x – 96

Solution:

Question 2.

Factorise x^{2} + 7xy – 12y^{2}

Solution:

Question 3.

Find the quotient and remainder when 5x^{3} – 9x^{2} + 10x + 2 is divided by x + 2 using synthetic division.

Solution:

Hence the quotient is 5x^{2} – 19x + 48 and remainder is -94.

Question 4.

Find the quotient and remainder when -7 + 3x – 2x^{2} + 5x^{3} is divided by -1 + 4x using synthetic division.

Solution:

Question 5.

If the quotient on dividing 5x^{4} + 4x^{3} + 2x + 1 by x + 3 is 5x^{3} + 9x^{2} + bx – 97 then find the values of a, b and also remainder.

Solution:

Quotient 5x^{3} + 11x^{2} + 33x – 97 is compared with given quotient 5x^{3} + ax^{2} + bx – 97

co-efficient of x^{2} is – 11 = a and co-efficient of x is 33 = b.

Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.

Find the quotient and the remainder when 10 – 4x + 3x^{2} is divided by x – 2.

Solution:

Let us first write the terms of each polynomial in descending order (or ascending

order). Thus the given problem becomes (3x^{2} – 4x + 10) ÷ (x – 2)

∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x^{2} – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form

Dividend = (Divisor × Quotient) + Remainder

Question 2.

If 8x^{3} – 14x^{2} – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.

Solution:

Exercise 3.8

Question 1.

Factorise 2x^{3} – x^{2} – 12x – 9 into linear factors.

Solution:

Let p(x) = 2x^{3} – x^{2} – 12x – 9

Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0

Hence x – 1 is not a factor

Sum of co-efficients of even powers with constant = -1 – 9 = -10

Sum of co-efficients of odd powers = 2 – 12 = -10

Hence x + 1 is a factor of x.

Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x^{2} – 3x – 9)

Now 2x^{2} – 3x – 9 = 2x^{2} – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)

= (x – 3)(2x + 3)

Hence 2x^{3} – x^{2} – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.

Factorize x^{3} + 3x^{2} – 13x – 15.

Solution:

Let p(x) = x^{3} + 3x^{2} – 13x – 15

Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0

Hence x – 1 is not a factor

Sum of co-efficients of even powers with constant = 3 – 15 = – 12

Sum of coefficients of odd powers = 1 – 13 = – 12

Hence x + 1 is a factor of p (x)

Now use synthetic division to find the other factors.

Hence p(x) = (x + 1) (x^{2} + 2x – 15)

Now x^{3} + 3x^{2} – 13x – 15 = (x + 1)(x^{2} + 2x – 15)

Now x^{2} + 2x – 15 = x^{2} + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)

Hence x^{3} + 3x^{2} – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.

Factorize x^{3} + 13x^{2} + 32x + 20 into linear factors.

Solution:

Let, p(x) = x^{3} + 13x^{2} + 32x + 20

Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0

Hence, (x – 1) is not a factor.

Sum of coefficients of even powers with constant = 13 + 20 = 33

Sum of coefficients of odd powers = 1 + 32 = 33

Hence, (x + 1) is a factor of p(x)

Now we use synthetic division to find the other factors

Exercise 3.9

Question 1.

Find GCD of 25x^{3}y^{2} z, 45x^{2}y^{4}z^{3}b

Solution:

Question 2.

Find the GCD of (y^{3} – 1) and (y – 1).

Solution:

y^{3} – 1 = (y – 1)(y^{3} + y + 1)

y – 1 = y – 1

Therefore, GCD = y – 1

Question 3.

Find the GCD of 3x^{2} – 48 and x^{2} – 7x + 12.

Solution:

3x^{2} – 48 = 3(x^{2} – 16) = 3(x^{2} – 4^{3}) = 3(x + 4)(x – 4)

x^{2} – 7x + 12 = x^{2} – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)

Therefore, GCD = x – 4

Question 4.

Find the GCD of a^{x} , a^{x + y}, a^{x + y + z}.

Solution:

Exercise 3.10

Question 5.

Solve graphically. x – y = 3 ; 2x – y = 11

Solution:

Exercise 3.11

Question 1.

Solve using the method of substitution. 5x – y = 5, 3x + y = 11

Solution:

Exercise 3.12

Question 2.

Solve by the method of elimination. 2x + y = 10; 5x – y = 11

Solution:

Exercise 3.13

Question 3.

Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.

Solution:

Exercise 3.14

Question 1.

The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Solution:

Let the age of the father be x

Let the sum of the age of his sons be y

At present x = 2y ⇒ x – 2y = 0 …. (1)

After 20 years x + 20 = y + 40

x – y = 40 – 20

Exercise 3.15

Multiple Choice Questions :

Question 1.

The polynomial 3x – 2 is a

(1) linear polynomial

(2) quadratic polynomial

(3) cubic polynomial

(4) constant polynomial

Hint: A polynomial is linear if its degree is 1

Solution:

(1) linear polynomial

Question 2.

The polynomial 4x^{2} + 2x – 2 is a

(1) linear polynomial

(2) quadratic polynomial

(3) cubic polynomial

(4) constant polynomial

Hint: A polynomial is quadratic of its highest power of x is 2

Solution:

(2) quadratic polynomial]

Question 3.

The zero of the polynomial 2x – 5 is

Hint:

Solution:

(1) \(\frac{5}{2}\)

Question 4.

The root of the polynomial equation 3x – 1 = 0 is

Hint:

Solution:

(2) x = \(\frac{1}{3}\)

Question 5.

Zero of (7 + 4x) is ___

Solution:

(2) \(\frac{-7}{4}\)

Question 6.

Which of the following has as a factor?

(1) x^{2} + 2x

(2) (x – 1)^{2}

(3) (x + 1)^{2}

(4) (x^{2} – 2^{2})

Solution:

(1) x^{2} + 2x

Question 7.

If x – 2 is a factor of q (x), then the remainder is _____

(1) q (-2)

(2) x – 2

(3) 0

(4) -2

Solution:

(3) 0

Question 8.

(a – b) (a^{2} + ab + b^{2}) =

Solution:

(1) a^{3} + b^{3} + c^{3} – 3abc

(2) a^{2} – b^{2}

(3) a^{3} + b^{3}

(4) a^{3} – b^{3}

Solution:

(4) a^{3} – b^{3}

Question 9.

The polynomial whose factors are (x + 2) (x + 3) is

(1) x^{2} + 5x + 6

(2) x^{2} – 4

(3) x^{2} – 9

(4) x^{2} + 6x + 5

Solution:

(1) x^{2} + 5x + 6

Question 10.

(-a – b – c)^{2} is equal to

(1) (a – b + c)^{2}

(2) (a + b – c)^{2}

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

(4) (a + b + c)^{2}

Solution:

(4) (a + b + c)^{2}

Text Book Activities

Activity – 1

Write the Variable, Coefficient and Constant in the given algebraic expression,

Solution:

Activity – 2

Write the following polynomials in standard form.

Activity – 3

Question 1.

Find the value of k for the given system of linear equations satisfying the condition below:

(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution

(ii) kx + 3y = 3; 12x + ky = 6 has no solution

(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution

Solution:

Question 2.

Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b

Solution:

Activity – 4

For the given linear equations, find another linear equation satisfying each of the given condition

Solution:

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