NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials.

## NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

**Exercise 2.1**

Question 1.

Which one of the following is a polynomial?

Solution:

(C)

It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number.

It is not a polynomial, because exponent of x is 1/2 which is not a whole number.

It is a polynomial, because each exponent of x is a whole number.

It is not a polynomial because it is a rational function.

Question 2.

√2 is a polynomial of degree

(a) 2

(b) 0

(c) 1

(d)½

Solution:

(b) √2 = -√2x°. Hence, √2 is a polynomial of degree 0, because exponent of x is 0.

Question 3.

Degree of the polynomial 4x^{4} + 0x^{3} + 0x^{5} + 5x + 7 is

(a) 4

(b) 5

(c) 3

(d) 7

Solution:

(a) 4

4x^{4} + 0x^{3} + 0x^{5} + 5x + 7 = 4x^{4} + 5x + 7

As we know that the degree of a polynomial is equal to the highest power of variable x.

Here, the highest power of x is 4. Therefore, the degree of the given polynomial is 4.

Question 4.

Degree of the zero polynomial is

(A) 0

(B) 1

(C) Any natural number

(D) Not defined

Solution:

(D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x^{5} etc.

Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial.

Question 5.

If p (x) = x^{2} – 2√2x + 1, then p (2√2) is equal to

(a) 0

(b) 1

(c) 4√2

(d) 8 √2 +1

Solution:

(b) Given, p(x) = x^{2} – 2√2x + 1 …(i)

On putting x = 2√2 in Eq. (i), we get

P(2√2) = (2√2)^{2}– (2√2)(2√2) + 1 = 8 – 8 + 1 = 1

Question 6.

The value of the polynomial 5x – 4x^{2} + 3, when x = – 1 is

(a)-6

(b) 6

(c) 2

(d) -2

Solution:

(a) Let p (x) = 5x – 4x^{2} + 3 …(i)

On putting x = -1 in Eq. (i), we get

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

Question 7.

If p (x) = x + 3, then p(x) + p(- x) is equal to

(a) 3

(b) 2x

(c) 0

(d) 6

Solution:

(d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3

Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6

Question 8.

Zero of the zero polynomial is

(a) 0

(b) 1

(c) any real number

(d) not defined

Solution:

(c) Zero of the zero polynomial is any real number.

e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k

Hence, zero of the zero polynomial be any real number.

Question 9.

Zero of the polynomial p(x) = 2x + 5 is

(a) -2/5

(b) -5/2

(c) 2/5

(d) 5/2

Solution:

(b) Given, p(x) = 2x + 5

For zero of the polynomial, put p(x) = 0

∴ 2x + 5 = 0

⇒ -5/2

Hence, zero of the polynomial p(x) is -5/2.

Question 10.

One of the zeroes of the polynomial 2x^{2} + 7x – 4 is

(a) 2

(b) ½

(c) -1

(d) -2

Solution:

(b) Let p (x) = 2x^{2} + 7x – 4

= 2x^{2} + 8x – x – 4 [by splitting middle term]

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

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

For zeroes of p(x), put p(x) = 0

⇒ (2x -1) (x + 4) = 0

⇒ 2x – 1 = 0 and x + 4 = 0

⇒ x = ½ and x = -4

Hence, one of the zeroes of the polynomial p(x) is ½.

Question 11.

If x^{51} + 51 is divided by x + 1, then the remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Solution:

(d) Let p(x) = x^{51} + 51 . …(i)

When we divide p(x) by x+1, we get the remainder p(-1)

On putting x = -1 in Eq. (i), we get p(-1) = (-1)^{51} + 51

= -1 + 51 = 50

Hence, the remainder is 50.

Question 12.

If x + 1 is a factor of the polynomial 2x^{2} + kx, then the value of k is

(a) -3

(b) 4

(c) 2

(d)-2

Solution:

(c) Let p(x) = 2x^{2} + kx

Since, (x + 1) is a factor of p(x), then

p(-1)=0

2(-1)2 + k(-1) = 0

⇒ 2 – k = 0

⇒ k = 2

Hence, the value of k is 2.

Question 13.

x + 1 is a factor of the polynomial

(a) x^{3} + x^{2} – x + 1

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

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

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

Solution:

(b) Let assume (x + 1) is a factor of x^{3} + x^{2} + x + 1.

So, x = -1 is zero of x^{3} + x^{2} + x + 1

(-1)^{3} + (-1)^{2} + (-1) + 1 = 0

⇒ -1 + 1 – 1 + 1 = 0

⇒ 0 = 0 Hence, our assumption is true.

Question 14.

One of the factors of (25x^{2} – 1) + (1 + 5x)^{2} is

(a) 5 + x

(b) 5 – x

(c) 5x -1

(d) 10x

Solution:

(d) Now, (25x^{2} -1) + (1 + 5x)^{2}

= 25x^{2} -1 + 1 + 25x^{2} + 10x [using identity, (a + b)^{2} = a^{2} + b^{2} + 2ab]

= 50x^{2} + 10x = 10x (5x + 1)

Hence, one of the factor of given polynomial is 10x.

Question 15.

The value of 249^{2} – 248^{2} is

(a) 1^{2}

(b) 477

(c) 487

(d) 497

Solution:

(d) Now, 249^{2} – 248^{2} = (249 + 248) (249 – 248) [using identity, a^{2} – b^{2} = (a – b)(a + b)]

= 497 × 1 = 497.

Question 16.

The factorization of 4x^{2} + 8x+ 3 is

(a) (x + 1) (x + 3)

(b) (2x + 1) (2x + 3)

(c) (2x + 2) (2x + 5)

(d) (2x – 1) (2x – 3)

Solution:

(b) Now, 4x^{2} + 8x + 3= 4x^{2} + 6x + 2x + 3 [by splitting middle term]

= 2x(2x + 3) + 1 (2x + 3)

= (2x + 3) (2x + 1)

Question 17.

Which of the following is a factor of (x+ y)^{3} – (x^{3} + y^{3})?

(a) x^{2} + y^{2} + 2 xy

(b) x^{2} + y^{2} – xy

(c) xy^{2}

(d) 3xy

Solution:

(d) Now, (x+ y)3 – (x^{3} + y^{3}) = (x + y) – (x + y)(x^{2}– xy + y^{2})

[using identity, a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})] = (x + y)[(x + y)^{2} -(x^{2} – xy + y^{2})]

= (x+ y)(x^{2}+ y^{2}+ 2xy – x^{2}+ xy – y^{2})

[using identity, (a + b)^{2} = a^{2} + b^{2} + 2 ab)]

= (x + y) (3xy)

Hence, one of the factor of given polynomial is 3xy.

Question 18.

The coefficient of x in the expansion of (x + 3)^{3} is

(a) 1

(b) 9

(c) 18

(d) 27

Solution:

(d) Now, (x + 3)^{3} = x^{3} + 3^{3} + 3x (3)(x + 3)

[using identity, (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)]

= x^{3} + 27 + 9x (x + 3)

= x^{3} + 27 + 9x^{2 }+ 27x Hence, the coefficient of x in (x + 3)^{3} is 27.

Question 19.

(a) 1

(b) -1

(c) 0

(d) 1/2

Solution:

Question 20.

Solution:

Question 21.

If a + b + c =0, then a^{3 }+ b^{3} + c^{3} is equal to

(a) 0

(b) abc

(c) 3abc

(d) 2abc

Solution:

(d) Now, a^{3 }+ b^{3} + c^{3 }= (a + b + c) (a^{2} + b^{2} + c^{2} – ab – be – ca) + 3abc

[using identity, a^{3 }+ b^{3} + c^{3} – 3 abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – be – ca)]

= 0 + 3abc [∴ a + b + c = 0, given]

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

Exercise 2.2

Question 1.

Which of the following expressions are polynomials? Justify your answer:

Solution:

(i) Polynomial

Because 8 = 8x°, then exponent of the variable x is 0, which is a whole number.

(ii) Polynomial

Each exponent of the variable x is a whole number.

(iii) Not polynomial

Because exponent of the variable x is 1/2, which is not a whole number.

(iv) Polynomial

Because each exponent of the variable x is a whole number.

(v) Not polynomial

Because one of the exponents of the variable x is -1, which is not a whole number.

(vi) Not polynomial

Because given expression is a rational expression, thus, not a polynomial.

(vii) Polynomial

Because each exponent of the variable a is a whole number.

(viii) Not polynomial

Question 2.

Write whether the following statements are True or False. Justify your answer.

(i) A binomial can have atmost two terms

(ii) Every polynomial is a binomial

(iii) A binomial may have degree 5

(iv) Zero of a polynomial is always 0

(v) A polynomial cannot have more than one zero

(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

Solution:

(i) False

Because a binomial has exactly two terms.

(ii) False

Because every polynomial is not a binomial.

e.g., (a)x² + 4x + 3 [polynomial but not a binomial]

(b) x² + 5 [polynomial and also a binomial]

(iii) True

Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Therefore, a binomial may have degree 5.

(iv) False

Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number.

(v) False

Because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial. e.g. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2.

(vi) False

Because the sum of any two polynomials of same degree has not always same degree.

e.g., Let f(x) = x^{5} + 2 and g(x) = -x^{5} + 2x^{2}

∴ Sum of two polynomials, f(x) + g(x) = x^{5 }+ 2 + (-x^{5} + 2x^{2}) = 2x^{2 }+ 2, which is not a polynomial of degree 5.

**Exercise 2.3**

Question 1.

Classify the following polynomials as polynomials in one variable, two variables etc.

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

(ii) y^{3} – 5y

(iii) xy + yz + zx

(iv) x^{2} – Zxy + y^{2} + 1

Solution:

(i) Polynomial x^{2 }+ x + 1 is a one variable polynomial, because it contains only one variable i.e., x.

(ii) Polynomial y^{3} – 5y is a one variable polynomial, because it contains only one variable i.e., y.

(iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z.

(iv) Polynomial x^{2} – 2xy + y^{2} + 1 is a two variables polynomial, because it contains two variables x and y.

Question 2.

Determine the degree of each of the following polynomials.

(i) 2x – 1

(ii) -10

(iii) x^{3 }– 9x + 3x^{5}

(iv) y^{3}(1 – y^{4})

Solution:

(i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1.

(ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0.

(iii) Degree of polynomial x3 – 9x + 3x^{5} is 5, because the maximum exponent of x is 5.

(iv) Degree of polynomial y^{3}(1 – y^{4}) or y^{3} – y^{7} is 7, because the maximum exponent of y is 7.

Question 3.

(i) the degree of the polynomial

(ii) the coefficient of x^{3
}(iii) the coefficient of x^{6
}(iv) the constant term

Solution:

(i) The degree of the polynomial is the highest power of the variable x i.e., 6.

(ii) The coefficient of x^{3} in given polynomial is 1/5

(iii) The coefficient of x^{6} in given polynomial is -1.

(iv) The constant term in given polynomial is 1/5

Question 4.

Write the coefficient of x² in each of the following

Solution:

(ii) Coefficient of x^{2} in 3x – 5 is 0.

(iii) We have, (x – 1) (3x – 4) = 3x^{2} – 7x + 4

∴ Coefficient of x² in 3x² – 7x + 4 is 3.

(iv) We have, (2x – 5) (2x² – 3x + 1)

= 2x (2x² – 3x + 1) – 5(2x² – 3x + 1)

= 4x³ – 6x² + 2x – 10x² + 15x – 5

= 4x³ – 16x² + 17x – 5

Coefficient of x² in 4x³ – 16x² + 17x – 5 is -16.

Question 5.

Classify the following as constant, linear, quadratic and cubic polynomials:

(i) 2 – x² + x³

(ii) 3x³

(iii) 5t – √7

(iv) 4 – 5y²

(v) 3

(vi) 2 + x

(vii) y³ – y

(viii) 1 + x + x²

(ix) t²

(x) √2x – 1

Solution:

(i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3.

(ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3.

(iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1.

(iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2.

(v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0.

(vi) Polynomial 2 + x is a linear polynomial, because its degree is 1.

(vii) Polynomial y³ – y is a cubic polynomial, because its degree is 3.

(viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2.

(ix) Polynomial t² is a quadratic polynomial, because its degree is 2.

(x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1.

Question 6.

Give an example of a polynomial, which is

(i) monomial of degree 1.

(ii) binomial of degree 20.

(iii) trinomial of degree 2.

Solution:

(i) The example of monomial of degree 1 is 3x.

(ii) The example of binomial of degree 20 is 3x^{20 }+ x^{10}

(iii) The example of trinomial of degree 2 is x^{2} – 4x + 3

Question 7.

Find the value of the polynomial 3x^{3} – 4x^{2} + 7x – 5, when x = 3 and also when x = -3.

Solution:

Let p(x) =3x^{3} – 4x^{2} + 7x – 5

At x= 3, p(3) = 3(3)^{3} – 4(3)^{2} + 7(3) – 5

= 81 – 36 + 21 – 5

∴ P( 3) = 61

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

= -81 – 36 – 21 – 5 = -143

∴ p(-3) = -143

Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Question 8.

If p (x) = x^{2} – 4x + 3, then evaluate p(2) – p (-1) + p ( ½).

Solution:

Question 9.

Find p(0), p( 1) and p(-2) for the following polynomials

(i) p(x) = 10x – 4x^{2} – 3

(ii) p(y) = (y + 2)(y – 2)

Solution:

(i) Given, polynomial is

p(x) = 10x – 4x^{2} – 3

On putting x = 0, 1 and – 2, respectively in Eq. (i),

we get p(0) = 10(0) – 4(0)^{2} – 3 = 0 – 0 – 3 = -3

p(1) = 10 (1) – 4 (1 )^{2} – 3

= 10 – 4 – 3= 10 – 7 = 3

and p(-2) =10 (-2) -4 (-2)^{2} – 3

= -20- 4 × 4 – 3 = -20 – 16 – 3 = -39

Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and – 39.

(ii) Given, polynomial isp(y) = (y+2)(y-2)

On putting y =0,1 and -2, respectively in Eq. (i), we get p(0) = (0+2)(0-2) = -4

p(1) = (1 + 2)(1-2)

= 3 x (-1) = -3

and p(-2) = (-2 + 2)(-2 -2)

= 0 (-4) = 0

Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0.

Question 10.

Verify whether the following are true or false.

(i) -3 is a zero of at – 3

(ii) -1/3 is a zero of 3x + 1

(iii) -4/5 is a zero of 4 – 5y

(iv) 0 and 2 are the zeroes of t^{2} – 2t

(v) -3 is a zero of y^{2} + y – 6

Solution:

(i) False

Put x – 3 = 0 ⇒ x = 3

Hence, zero of x – 3 is 3.

(ii) True

Put 3x + 1 = 0 ⇒ x = -1/3

Hence, zero of 3x + 1 is -1/5.

(iii) False

Put 4 – 5y = 0 ⇒ y = 4/5

Hence, zero of 4 – 5y is 4/5.

(iv) True

Put t² – 2t = 0 ⇒ t(t – 2) = 0

⇒ t = 0 and t – 2 = 0

⇒ t = 0 and t = 2

Hence, the zeroes of t² – 2t are 0 and 2.

(v) True

Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0

⇒ y(y + 3) – 2(y + 3) = 0

= (y-2)(y + 3) = 0

⇒ y – 2 = 0 and y + 3 = 0

⇒ y = 2 and y = -3

Hence, the zeroes of y² + y – 6 are 2 and – 3.

Question 11.

Find the zeroes of the polynomial in each of the following,

(i) p(x)= x – 4

(ii) g(x)= 3 – 6x

(iii) q(x) = 2x – 7

(iv) h(y) = 2y

Solution:

(i) Given, polynomial is

p(x) = x – 4

For zero of polynomial, put p(x) = 0

∴ x – 4 = 0 ⇒ x = 4

Hence, zero of polynomial is 4.

(ii) Given, polynomial is

g(x) = 3 – 6x

For zero of polynomial, put g(x) = 0

∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2.

Hence, zero of polynomial is X

(iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0

∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2

Hence, zero of polynomial q(x) is 7/2

(iv) Given polynomial h(y) = 2 y

For zero of polynomial, put h(y) = 0

∴ 2y = 0 ⇒ y = 0

Hence, the zero of polynomial h(y) is 0.

Question 12.

Find the zeroes of the polynomial p(x) = (x – 2)^{2} – (x + 2)^{2}.

Solution:

Given, polynomial is p(x) = (x – 2)^{2} – (x + 2)^{2}

For zeroes of polynomial, put p(x) = 0

∴ (x – 2)^{2} – (x + 2)^{2 }= 0

⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0

[using identity, a^{2 }– b^{2} = (a – b)(a + b)]

⇒ (2x)(-4) = 0

⇒ -8x = 0

⇒ x = 0

Therefore zero of the polynomial is p(x) is 0.

Question 13.

By actual division, find the quotient and the remainder when the first

polynomial is divided by the second polynomial x^{4} + 1 and x – 1.

Solution:

Using long division method

Hence, quotient = x³ + x² + x + 1 and remainder = 2

Question 14.

By remainder theorem, find the remainder when p(x) is divided by g(x)

(i) p(x) = x^{3 }– 2x^{2 }– 4x – 1, g(x) = x + 1

(ii) p(x) = x^{3} – 3x^{2} + 4x + 50, g(x) = x – 3

(iii) p(x) = x^{3} – 12x^{2} + 14x – 3, g(x) = 2x – 1 – 1

(iv) p(x) = x^{3 }– 6x^{2 }+ 2x – 4, g(x) = 1 – (3/2) x

Solution:

(i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1

Here, zero of g(x) is -1.

When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1)

∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1

= -1 – 2 + 4 – 1 = 0

Therefore, remainder is 0.

(ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3

Here, zero of g(x) is 3.

When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3)

∴ p(3) = (3)³ – 3(3)² + 4(3) + 50

= 27 – 27 + 12 + 50 = 62

Therefore, remainder is 62.

(iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1

Here, zero of g(x) is 1/2

Question 15.

Check whether p(x) is a multiple of g(x) or not

(i) p(x) = x^{3} – 5x^{2} + 4x – 3, g(x) = x – 2.

(ii) p(x) = 2x^{3} – 11x^{2} – 4x + 5, g(x) = 2x + 1

Solution:

(i) We have, g(x) = x – 2

Here, zero of g(x) is 2.

Now, p(x) = x^{3} – 5x^{2} + 4x – 3

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

= 8 – 20 + 8 – 3 = – 7

Since, remainder ≠ 0, then p(x) is not a multiple of g(x).

Since, remainder ≠ 0, then p(x) is not a multiple of g(x).

Question 16.

Show that,

(i) x + 3 is a factor of 69 + 11x – x^{2} + x^{3}

(ii) 2x – 3 is a factor of x + 2x^{3} -9x^{2} +12

Solution:

(i) Let p(x) = x^{3} – x^{2} + 11x + 69

x + 3 is a factor of p(x) if p(-3) = 0

Now, p(-3) = (-3)^{3} – (-3)^{2} + 11(-3) + 69

= – 27 – 9 – 33 + 69 = 0

Therefore, (x + 3) is a factor of p(x).

Question 17.

Determine which of the following polynomial has x – 2 a factor

(i) 3x^{2} + 6x – 24

(ii) 4x^{2}+ x – 2

Solution:

(i) Let p(x) = 3x^{2} + 6x – 24 … (1)

Substituting x = 2 in (1), we get

p(2) = 3(2)^{2} + 6(2) – 24

= 12 + 12 – 24 = 0

Hence, x – 2 is a factor of p(x).

(ii)Let p(x) = 4x^{2} + x – 2 … (2)

Substituting x = 2 in (2), we get

p(2) = 4(2)^{2} + 2 – 2 = 16 ≠ 0

Hence, x – 2 is not a factor of p(x).

Question 18.

Show that p-1 is a factor of p^{10} -1 and also of p^{11} -1.

Solution:

Let g (p) = p^{10} -1 …(1)

and h(p) = p^{11} -1 …(2)

On putting p = 1 in Eq. (i), we get

g(1) = 1^{10 }-1 = 1 – 1 = 0

Hence, p-1 is a factor of g(p).

Again, putting p = 1 in Eq. (2), we get

h (1) = (1)^{11} -1 = 1 – 1 = 0

Hence, p -1 is a factor of h(p).

Question 19.

For what value of m is x^{3} -2mx^{2} +16 divisible by x + 2?

Solution:

Let p(x) = x^{3} – 2mx^{2} + 16

Since, p(x) is divisible by (x+2), then remainder = 0

P(-2) = 0

⇒ (-2)^{3} – 2m(-2)^{2} + 16 = 0

⇒ -8 – 8m + 16 = 0

⇒ 8m = 8

m = 1

Hence, the value of m is 1.

Question 20.

If x + 2a is a factor of a^{5} – 4a^{2}x^{3} + 2x + 2a + 3, then find the value of a.

Solution:

Let p(x) =a^{5} – 4a^{2}x^{3} + 2x + 2a + 3

Since, x + 2a is a factor of p(x), then put p(-2a) = 0

∴ (-2a)^{5 }– 4a^{2} (-2a)^{3} + 2(-2a) + 2a + 3 = 0

⇒ -32a^{5} + 32a^{5} – 4a + 2a + 3 = 0

⇒ -2a + 3 = 0

2a = 3

a = 3/2.

Hence, the value of a is 3/2.

Question 21.

Find the value of m, so that 2x -1 be a factor of

8x^{4} + 4x^{3} – 16x^{2} + 10x + m

Solution:

Let p(x) = 8x^{4} + 4x^{3} – 16x^{2} + 10x + m

Since 2x – 1 is afactor of p(x) then p(1/2) = 0

Question 22.

If x + 1 is a factor of ax^{3} + x^{2} – 2x + 4o – 9, find the value of a.

Solution:

Let p(x) = ax^{3} + x^{2} – 2x + 4a – 9

Since, x + 1 is a factor of p(x), then p(-1) = 0

a(- 1)^{3}+ (- 1)^{2} – 2(-1) + 4a – 9 = 0

⇒ -a + 1 + 2 + 4a – 9 = 0

⇒ 3a = 6

⇒ a = 2

Question 23.

Factorise:

(i) x^{2} + 9x + 18

(ii) 6x^{2} + 7x – 3

(iii) 2x^{2}– 7x – 15

(iv) 84 – 2r – 2r^{2}^{
}Solution:

(i) We have, x^{2} + 9x +18 = x^{2} + 6x + 3x +18

= x(x + 6) + 3(x + 6) = (x + 3) (x + 6)

We have, 6x^{2} + 7x – 3 = 6x^{2} + 9x – 2x – 3

= 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3)

We have, 2x^{2} – 7x – 15 = 2x^{2} – 10x + 3x -15

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

We have, 84 – 2r – 2r^{2} = – 2 (r^{2} + r – 42)

= -2(r^{2} + 7r – 6r – 42)

= -2[r(r + 7) -6(r + 7)]

= 2(6 – r)(r + 7) or 2(6 – r) (7 + r)

Question 24.

Factorise:

(i) 2x^{3} – 3x^{2} – 17x + 30

(ii) x^{3}-6x^{2}+11x-6

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

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

Solution:

(i) We have, 2X^{3} – 3x^{2} – 17x + 30

= 2x^{3} – 4x^{2} + x^{2} – 2x – 15x + 30

= 2x^{2}(x – 2) + x(x – 2) – 15(x – 2)

= (x – 2) (2x^{2} + x – 15)

= (x – 2) (2x^{2} + 6x – 5x – 15)

= (x – 2) [2x(x + 3) – 5(x + 3)]

= (x – 2)(x + 3)(2x – 5)

(ii)We have, x^{3} – 6x^{2} + 11x – 6

= x^{3} – x^{2} – 5x^{2} + 5x + 6x – 6

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

= (x -1) (x^{2} – 5x + 6)

= (x – 1) (x^{2} – 3x – 2x + 6)

= (x – 1) [x(x – 3) – 2(x – 3)]

= (x-1)(x-2)(x-3)

(iii) We have, x^{3} + x^{2} – 4x – 4

= x^{2}(x + 1) – 4(x + 1)

= (x + 1)(x^{2} – 4)

= (x + 1) (x – 2) (x + 2)[∴ a^{2}– b^{2} = (a – b) (a + b)]

(iv) We have, 3x^{3} – x^{2} – 3x + 1 = 3x^{3} – 3x^{2} + 2x^{2}– 2x – x + 1

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

= (x – 1)(3x^{2} + 2x – 1)

= (x – 1) (3x^{2} + 3x – x – 1)

= (x – 1) [3x(x + 1) – 1(x + 1)]

= (x – 1) (x + 1)(3x -1)

Question 25.

Using suitable identity, evaluate the following:

(i) 103^{3}

(ii) 101 × 102

(iii) 999^{2
}Solution:

(i) We have, 103^{3} = (100 + 3)^{3
}= (100)^{3} + (3)^{3} + 3(100)(3)(100 + 3)

[∴ (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)]

= 1000000 + 27 + 900(103)

= 1000027 + 92700 = 1092727

(ii) We have, 101 × 102 = (100 + 1) (100 + 2)

= (100)^{2} + (1 + 2)100 + (1)(2)

[∴ (x + a)(x + b) = x^{2} + (a + b)x + ab]

= 10000 + 300 + 2 = 10302

(iii) We have, (999)^{2} = (1000 -1)^{2
}= (1000)^{2} + (1)^{2} – 2(1000)(1)

[∴ (a – b)^{2} = a^{2} + b^{2} – 2ab]

= 1000000 + 1 – 2000 = 998001

Question 26.

Solution:

Question 27.

Factorise the following:

(i) 9x^{2 }– 12x + 3

(ii) 9x^{2 }– 12x + 4

Solution:

(i) We have, 9x^{2} – 12x + 3 = 3(3x^{2} – 4x + 1)

= 3(3x^{2} – 3x – x + 1)

= 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1)

(ii) We have, 9x^{2} – 12x + 4

= (3x)^{2} – 2 × 3x × 2 + (2)^{2
}= (3x – 2)^{2} [∴ a^{2} – 2ab + b^{2} = (a – b)²]

= (3x – 2) (3x – 2)

Question 28.

Expand the following:

(i) (4o – b + 2c)^{2}

(ii) (3o-5b-c)^{2
}(ii) (-x + 2y – 3z)^{2
}Solution:

We know that

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx

(i) We have, (4a – b + 2c)^{2} = (4a)^{2} + (-b)^{2 }+ (2c)^{2} + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)

= 16a^{2} + b^{2} + 4c^{2} – 8ab – 4bc + 16ac

(ii)We have, (3a – 5b – cf = (3a)^{2} + (-5b² + (- c²) + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a)

= 9a^{2} + 25b^{2} + c^{2 }– 30ab + 10bc – 6ac

(iii) We have, (- x + 2y – 3z)^{2} = (- x)^{2} + (2y)^{2 }+ (-3z)^{2} + 2(-x)(2y) + 2(2y)(- 3z) + 2(- 3z)(- x)

= x^{2} + 4y^{2} + 9z^{2} – 4xy – 12yz + 6xz

Question 29.

Factorise the following:

(i) 9x^{2} + 4y^{2}+16z^{2}+12xy-16yz-24xz

(ii) 25x^{2} + 16y^{2} + 4Z^{2} – 40xy +16yz – 20xz

(iii) 16x^{2} + 4)^ + 9z^{2}-^ 6xy – 12yz + 24xz

Solution:

(i) We have,

9x^{2} + 4y^{2} + 16z^{2} + 12xy – 16yz – 24xz

= (3x)^{2} + (2y)^{2} + (-4z)^{2} + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x)

= (3x + 2y – 4z)^{2
}[∴ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}]

= (3x + 2y- 4z) (3x + 2y – 4z)

(ii)We have, 25X^{2} + 16y^{2} + 4z^{2} – 40xy + 16yz – 20xz

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

= (- 5x + 4y + 2z)^{2
}[∴ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = (a + b + c)^{2}]

= (- 5x + 4y + 2z)(- 5x + 4y + 2z)

(iii) We have, 16x^{2} + 4y^{2} + 9z^{2} – 16xy – 12yz + 24xz

= (4x)^{2} + (- 2y)^{2} + (3z)^{2} + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x)

= (4x – 2y + 3z)^{2}

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

= (4x – 2y + 3z)(4x -2y + 3z)

Question 30.

lf a+b+c=9 and ab + bc + ca = 26, find a^{2} + b^{2} + c^{2}.

Solution:

We have, a + b + c = 9

⇒ (a + b + c)^{2} = (9)^{2} [Squaring on both sides]

⇒ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 81

⇒ a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca) = 81

⇒ a^{2} + b^{2} + c^{2} + 2(26) = 81 [∴ ab + bc + ca = 26]

⇒ a^{2} + b^{2} + c^{2} = 81 – 52 = 29

Question 31.

Expand the following

37

Solution:

(i) We have, (3a – 2b)^{3
}= (3a)^{3} – (2b)^{3} – 3(3a)(2b)(3a – 2b)

[∴ (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)]

= 27a^{3 }– 8b^{3 }– 18ab(3a – 2b)

= 27a^{3 }– 8b^{3 }– 54a^{2}b + 36ab^{2
}= 27a^{3 }– 54a^{2}b + 36ab^{2 }– 8b^{3
}

Question 32.

Factorise the following:

Solution:

Question 33.

Find the following products:

Solution:

(ii) We have, (x^{2} – 1) (x^{4} + x^{2 }+ 1)

= x^{2} (x^{4} + x^{2} + 1) – 1(x^{4} + x^{2} + 1)

= x^{6} + x^{4} + x^{2} – x^{4} – x^{2} – 1 = x^{6 }– 1 ^{ }

Question 34.

Factorise:

(i) 1 + 64x^{3}

(ii) a^{3 }– 2√2b^{3
}Solution:

(i) We have, 1 + 64x^{3} = (1)^{3} + (4x)^{3
}= (1 + 4x)[(1)^{2} – (1)(4x) + (4x)^{2}]

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

= (1 + 4x)(1 – 4x + 16x^{2})

(ii) We have, a^{3} – 2√2b^{3} = (a)^{3} – (√2b)^{3
}= (a – √2b)[a^{2} + a( √2b) + (√2b)^{2}]

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

= (a – √2b)(a^{2} + √2ab + 2b^{2})

Question 35.

Fi nd (2x – y + 3z) (4x^{2} + y^{2} + 9z^{2 }+ 2xy + 3yz – 6xz).

Solution:

We have, (2x – y + 3z) (4x^{2} + y^{2} + 9z^{2} + 2xy + 3yz – 6xz)

= 2x(4x^{2} + y^{2} + 9z^{2} + 2xy + 3yz – 6xz) – y(4x^{2} + y^{2 }+ 9z^{2} + 2xy + 3yz – 6xz) + 3z(4x^{2} + y^{2} + 9z^{2} + 2xy + 3yz – 6xz)

= 8x³ + 2xy^{2} + 18xz^{2} + 4x^{2}y + 6xyz – 12x^{2}z – 4x^{2}y – y^{3} – 9yz^{2} – 2xy^{2} – 3y^{2}z + 6xyz + 12x^{2}z + 3y^{2}z + 27z^{3} + 6xyz + 9yz^{2} – 18xz^{2
}= 8X^{3} – y^{3} + 27z^{3} + 18xyz

Question 36.

Factorise

(i) a^{3} -8b^{3} -64c^{3} -2Aabc

(ii) 2√2a^{3} +8b^{3} -27c^{3} +18√2abc

Solution:

Question 37.

Without actually calculating the cubes, find the value

Solution:

(ii) We have, (0.2)^{3} – (0.3)^{3} + (0.1)^{3
}= (0.2)^{3} + (- 0.3)^{3} + (0.1)^{3}^{
}Since, 0.2 – 0.3 + 0.1 = 0,

∴ (0.2)³ + (-0.3)³ + (0.1)^{3} = 3(0.2) (-0.3) (0.1)

[ ∴ If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc] = -0.018

Question 38.

Without finding the cubes, factorise (x- 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3}.

Solution:

we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0

Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x).

If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc

Question 39.

Find the value of

(i) x^{3} + y^{3} – 12xy + 64,when x + y = -4.

(ii) x^{3} – 8y^{3} – 36xy-216,when x = 2y + 6.

Solution:

(i) Since, x + y + 4 = 0, then

x^{3} + y^{3} + (4)^{3} = 3xy(4)

[∴ If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc]

⇒ x^{3} + y^{3} + 64 = 12xy

⇒ x^{3} + y^{3 }– 12xy + 64 = 0

(ii) Since, x – 2y – 6 = 0, then

x³ + (-2y)^{3} + (-6)^{3} = 3x(-2y)(-6)

[∴ If a + b + c = 0, then a^{3} + b^{3} + c^{3} 3abc]

⇒ x^{3} – 8y^{3} – 216 = 36xy

⇒ x^{3} – 8y^{3} – 36xy – 216 = 0

Question 40.

Give possible expression for the length and breadth of the rectangle whose area is given by 4a^{2} + 4a – 3.

Solution:

Given, area of rectangle = (Length) × (Breadth)

= 4a^{2} + 4a – 3

= 4a^{2} + 6a – 2a – 3

= 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3)

Hence, possible length = 2a -1 and breadth = 2a + 3

**Exercise 2.4**

Question 1.

If the polynomials az^{3} + 4z^{2} + 3z – 4 and z^{3 }– 4z + o leave the same remainder when divided by z – 3, find the value of a.

Solution:

Let p_{1}(z) = az^{3} + 4z^{2} + 3z – 4 and p_{2}(z) = z^{3 }– 4z + o

When we divide p_{1}(z) by z – 3, then we get the remainder p,(3).

Now, p_{1}(3) = a(3)3 + 4(3)2 + 3(3) – 4

= 27a+ 36+ 9-4= 27a+ 41

When we divide p_{2}(z) by z-3 then we get the remainder p_{2}(3).

Now, p_{2}(3) = (3)^{3}-4(3)+a

= 27-12 + a = 15+a

According to the question, both the remainders are same.

p_{1}(3) = p_{2}(3)

27a + 41 = 15 + a

27a – a = 15 – 41 .

26a = 26

∴ a = -1

Question 2.

The polynomial p{x) = x^{4} – 2x^{3} + 3x^{2} – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x + 2.

Solution:

We have, p(x) = x^{4} – 2x^{3} + 3x^{2} – ox + 3a – 7

Since p(x) is divided by x + 1, then remainder is p(-1).

p(-1) = (-1)^{4} – 2(-1)^{3} + 3(-1)^{2} – a(-1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1

p(- 1) = 19 (Given)

⇒ 4a – 1 = 19 ⇒ 4a = 20

∴ a = 5

Thus, required polynomial,

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

= x^{4} – 2x^{3} + 3x^{2} – 5x + 8

Now, this is divided by x + 2, then remainder is p(-2).

p(- 2) = (- 2)^{4} – 2(- 2)^{3} + 3(- 2)^{2} – 5(- 2) + 8

= 16 + 16 + 12 + 10 + 8 = 62

Question 3.

If both x – 2 and x -(1/2) are factors of px^{2}+ 5x+r, then show that p = r.

Solution:

Question 4.

Without actual division, prove that

2x^{4} – Sx^{3} + 2x^{2} – x + 2 is divisible by x^{2} – 3x + 2. [Hint: Factorise x^{2} – 3x + 2]

Solution:

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

Now, x^{2}-3x + 2 = x^{2 }– 2x – x + 2

= (x-2)(x- 1)

Hence, zeroes of x^{2} – 3x + 2 are 1 and 2.

⇒ p(x) is divisible by x^{2} – 3x + 2 i.e., divisible by x – 1 and x – 2, if p(1) = 0 and p(2) = 0

Now, p(1) = 2(1)^{4} – 5(1)^{3} + 2(1)^{2} -1 + 2

= 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0

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

= 32 – 40 + 8 = 40 – 40 = 0

Hence, p(x) is divisible by x^{2} – 3x + 2.

Question 5.

Simplify (2x- 5y)^{3} – (2x+ 5y)^{3}.

Solution:

Question 6.

Multiply x^{2} + 4y^{2} + z^{2} + 2xy + xz – 2yz by (-z + x-2y).

Solution:

We have,

(x^{2} + 4y^{2} + z^{2} + 2xy + xz – 2yz)(- z + x – 2y)

= x^{2} (- z + x – 2y) + 4y^{2}(- z + x – 2y) + z^{2}(- z + x – 2y) + 2xy(- z + x – 2y) + xz(- z + x – 2y) – 2yz (- z + x – 2y)

= -x^{2}z + x^{3} – 2 x^{2}y – 4y^{2}z + 4xy^{2} – 81y^{3} – z^{3} + xz^{2 }– 2yz^{2}– 2xyz + 2x^{2}y – 4xy^{2} – xz^{2} + x^{2}z – 2xyz + 2yz^{2} – 2xyz + 4y^{2}z

= x^{3} – 8y^{3} – z^{3} – 6xyz

Question 7.

Solution:

Question 8.

If a + b + c = 5 and ab + bc + ca = 10, then prove that a^{3} + b^{3} + c^{3 }– 3abc = -25.

Solution:

We have, a + b + c = 5,ab + bc + ca = 10

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

then (5)^{2 }= a^{2} + b^{2}+ c^{2} + 2(10)

⇒ a^{2 }+ b^{2 }+ c^{2 }= 25 – 20

⇒ a^{2} + b^{2} + c^{2 }= 5 … (i)

L.H.S. = a^{3} + b^{3} + c^{3} – 3abc

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

= (5) [5 – (ab + be + ca)] [From (i)]

= 5(5 -10) = 5(-5) = – 25 = R.H.S.

Question 9.

Prove that (a +b +c)^{3} -a^{3} -b^{3 }– c^{3} =3(a +b)(b +c)(c +a).

Solution:

L.H.S. = [(a + b + c)^{3} – a^{3}] – (b^{3}+ c^{3})

= (a + b + c – a)[(a + b + c)^{2} + a^{2} + (a + b + c)a] – [(b + c) (b^{2} + c^{2}– be)]

x^{3} – y^{3} = (x – y)(x^{2} + y^{2} + xy) and

x^{3} + y^{3} = (x + y)(x^{2} + y^{2} – xy)

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

= (b + c)[3a^{2}+3ab + 3ac + 3bc]

= (b + c)[3(a^{2}+ ab + ac + bc)]

= 3 (b + c)[a(a + b) + c(a + b)]

= 3(a + b)(b + c)(c + a) = R.H.S.

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