# NCERT Solutions Class 9 Maths Chapter 2 – Polynomials (Updated Pattern)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The chapter’s introduction includes whole numbers, integers, and rational numbers.

The chapter starts with the introduction of Polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3

Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.

Zeroes of a Polynomial – A zero of a polynomial need not be zero and can have more than one zero.

Real Numbers and their Decimal Expansions – Here, you study the decimal expansions of real numbers and see whether they can help in distinguishing between rational and irrational numbers.

Next, it discusses the following topics:

Representing Real Numbers on the Number Line – The solutions for 2 problems in Exercise 2.4.

Operations on Real Numbers – Here, you explore operations like addition, subtraction, multiplication, and division on irrational numbers.

Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

## Exercise 2.1

**1. Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer.**

**(i) 4x ^{2}–3x+7**

Solution:

The equation 4x^{2}–3x+7 can be written as 4x^{2}–3x^{1}+7x^{0}

Since *x* is the only variable in the given equation and the powers of x (i.e. 2, 1 and 0) are whole numbers, we can say that the expression 4x^{2}–3x+7 is a polynomial in one variable.

**(ii) y ^{2}+√2**

Solution:

The equation y^{2}+**√2** can be written as y^{2}+**√**2y^{0}

Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y^{2}+**√**2 is a polynomial in one variable.

**(iii) 3√t+t√2**

Solution:

The equation 3√t+t√2 can be written as 3t^{1/2}+√2t

Though *t* is the only variable in the given equation, the power of *t* (i.e., 1/2) is not a whole number. Hence, we can say that the expression 3√t+t√2 is **not **a polynomial in one variable.

**(iv) y+2/y**

Solution:

The equation y+2/y can be written as y+2y^{-1}

Though *y *is the only variable in the given equation, the power of *y* (i.e., -1) is not a whole number. Hence, we can say that the expression y+2/y is **not **a polynomial in one variable.

**(v) x ^{10}+y^{3}+t^{50}**

Solution:

Here, in the equation x^{10}+y^{3}+t^{50}

Though the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression

x^{10}+y^{3}+t^{50}. Hence, it is **not **a polynomial in one variable.

**NCERT Class 9 Maths Chapter 1 – Number Systems (Updated Pattern)**

**2. Write the coefficients of x ^{2} in each of the following:**

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

Solution:

The equation 2+x^{2}+x can be written as 2+(1)x^{2}+x

We know that the coefficient is the number which multiplies the variable.

Here, the number that multiplies the variable x^{2} is 1

Hence, the coefficient of x^{2 }in 2+x^{2}+x is 1.

**(ii) 2–x ^{2}+x^{3}**

Solution:

The equation 2–x^{2}+x^{3 }can be written as 2+(–1)x^{2}+x^{3}

We know that the coefficient is the number (along with its sign, i.e. – or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2} is -1

Hence, the coefficient of x^{2 }in 2–x^{2}+x^{3 }is -1.

**(iii) (π/2)x ^{2}+x**

Solution:

The equation (π/2)x^{2 }+x can be written as (π/2)x^{2} + x

We know that the coefficient is the number (along with its sign, i.e. – or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2} is π/2.

Hence, the coefficient of x^{2 }in (π/2)x^{2 }+x is π/2.

**(iii)√2x-1**

Solution:

The equation √2x-1 can be written as 0x^{2}+√2x-1 [Since 0x^{2} is 0]

We know that the coefficient is the number (along with its sign, i.e. – or +) which multiplies the variable.

Here, the number that multiplies the variable x^{2}is 0

Hence, the coefficient of x^{2 }in √2x-1 is 0.

**3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.**

Solution:

Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35.

For example, 3x^{35}+5

Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100.

For example, 4x^{100}

**4. Write the degree of each of the following polynomials:**

**(i) 5x ^{3}+4x^{2}+7x**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 5x^{3}+4x^{2}+7x = 5x^{3}+4x^{2}+7x^{1}

The powers of the variable x are: 3, 2, 1

The degree of 5x^{3}+4x^{2}+7x is 3, as 3 is the highest power of x in the equation.

**(ii) 4–y ^{2}**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 4–y^{2},

The power of the variable y is 2

The degree of 4–y^{2} is 2, as 2 is the highest power of y in the equation.

**(iii) 5t–√7**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, in 5t**–√7 **

The power of the variable t is: 1

The degree of 5t**–√7 **is 1, as 1 is the highest power of y in the equation.

**(iv) 3**

Solution:

The highest power of the variable in a polynomial is the degree of the polynomial.

Here, 3 = 3×1 = 3× x^{0}

The power of the variable here is: 0

Hence, the degree of 3 is 0.

**5. Classify the following as linear, quadratic and cubic polynomials:**

Solution:

We know that,

Linear polynomial: A polynomial of degree one is called a linear polynomial.

Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.

Cubic polynomial: A polynomial of degree three is called a cubic polynomial.

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

Solution:

The highest power of x^{2}+x is 2

The degree is 2

Hence, x^{2}+x is a quadratic polynomial

**(ii) x–x ^{3}**

Solution:

The highest power of x–x^{3 }is 3

The degree is 3

Hence, x–x^{3} is a cubic polynomial

**(iii) y+y ^{2}+4**

Solution:

The highest power of y+y^{2}+4 is 2

The degree is 2

Hence, y+y^{2}+4 is a quadratic polynomial

**(iv) 1+x**

Solution:

The highest power of 1+x is 1

The degree is 1

Hence, 1+x is a linear polynomial.

**(v) 3t**

Solution:

The highest power of 3t is 1

The degree is 1

Hence, 3t is a linear polynomial.

**(vi) r ^{2}**

Solution:

The highest power of r^{2 }is 2

The degree is 2

Hence, r^{2}is a quadratic polynomial.

**(vii) 7x ^{3}**

Solution:

The highest power of 7x^{3 }is 3

The degree is 3

Hence, 7x^{3} is a cubic polynomial.

## Exercise 2.2

**1. Find the value of the polynomial (x)=5x−4x ^{2}+3. **

**(i) x = 0**

**(ii) x = – 1**

**(iii) x = 2**

Solution:

Let f(x) = 5x−4x^{2}+3

(i) When x = 0

f(0) = 5(0)-4(0)^{2}+3

= 3

(ii) When x = -1

f(x) = 5x−4x^{2}+3

f(−1) = 5(−1)−4(−1)^{2}+3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x^{2}+3

f(2) = 5(2)−4(2)^{2}+3

= 10–16+3

= −3

**2. Find p(0), p(1) and p(2) for each of the following polynomials:**

**(i) p(y)=y ^{2}−y+1**

Solution:

p(y) = y^{2}–y+1

∴ p(0) = (0)^{2}−(0)+1 = 1

p(1) = (1)^{2}–(1)+1 = 1

p(2) = (2)^{2}–(2)+1 = 3

**(ii) p(t)=2+t+2t ^{2}−t^{3}**

Solution:

p(t) = 2+t+2t^{2}−t^{3}

∴ p(0) = 2+0+2(0)^{2}–(0)^{3 }= 2

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

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

**(iii) p(x)=x ^{3}**

Solution:

p(x) = x^{3}

∴ p(0) = (0)^{3 }= 0

p(1) = (1)^{3 }= 1

p(2) = (2)^{3 }= 8

**(iv) P(x) = (x−1)(x+1)**

Solution:

p(x) = (x–1)(x+1)

∴ p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

**3. Verify whether the following are zeroes of the polynomial indicated against them.**

**(i) p(x)=3x+1, x = −1/3**

Solution:

For, x = -1/3, p(x) = 3x+1

∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

**(ii) p(x) = 5x–π, x = 4/5**

Solution:

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- π = 4-π

∴ 4/5 is not a zero of p(x).

**(iii) p(x) = x ^{2}−1, x = 1, −1**

Solution:

For, x = 1, −1;

p(x) = x^{2}−1

∴ p(1)=1^{2}−1=1−1 = 0

p(−1)=(-1)^{2}−1 = 1−1 = 0

∴ 1, −1 are zeros of p(x).

**(iv) p(x) = (x+1)(x–2), x =−1, 2**

Solution:

For, x = −1,2;

p(x) = (x+1)(x–2)

∴ p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴ −1, 2 are zeros of p(x).

**(v) p(x) = x ^{2}, x = 0**

Solution:

For, x = 0 p(x) = x^{2}

p(0) = 0^{2 }= 0

∴ 0 is a zero of p(x).

**(vi) p(x) = lx+m, x = −m/l**

Solution:

For, x = -m/*l *; p(x) = *l*x+m

∴ p(-m/*l)*= *l*(-m/*l*)+m = −m+m = 0

∴ -m/*l* is a zero of p(x).

**(vii) p(x) = 3x ^{2}−1, x = -1/√3 , 2/√3**

Solution:

For, x = -1/√3 , 2/√3 ; p(x) = 3x^{2}−1

∴ p(-1/√3) = 3(-1/√3)^{2}-1 = 3(1/3)-1 = 1-1 = 0

∴ p(2/√3 ) = 3(2/√3)^{2}-1 = 3(4/3)-1 = 4−1 = 3 ≠ 0

∴ -1/√3 is a zero of p(x), but 2/√3 is not a zero of p(x).

**(viii) p(x) =2x+1, x = 1/2**

Solution:

For, x = 1/2 p(x) = 2x+1

∴ p(1/2) = 2(1/2)+1 = 1+1 = 2≠0

∴ 1/2 is not a zero of p(x).

**4. Find the zero of the polynomials in each of the following cases:**

**(i) p(x) = x+5 **

Solution:

p(x) = x+5

⇒ x+5 = 0

⇒ x = −5

∴ -5 is a zero polynomial of the polynomial p(x).

**(ii) p(x) = x–5**

Solution:

p(x) = x−5

⇒ x−5 = 0

⇒ x = 5

∴ 5 is a zero polynomial of the polynomial p(x).

**(iii) p(x) = 2x+5**

Solution:

p(x) = 2x+5

⇒ 2x+5 = 0

⇒ 2x = −5

⇒ x = -5/2

∴x = -5/2 is a zero polynomial of the polynomial p(x).

**(iv) p(x) = 3x–2 **

Solution:

p(x) = 3x–2

⇒ 3x−2 = 0

⇒ 3x = 2

⇒x = 2/3

∴ x = 2/3 is a zero polynomial of the polynomial p(x).

**(v) p(x) = 3x **

Solution:

p(x) = 3x

⇒ 3x = 0

⇒ x = 0

∴ 0 is a zero polynomial of the polynomial p(x).

**(vi) p(x) = ax, a≠0**

Solution:

p(x) = ax

⇒ ax = 0

⇒ x = 0

∴ x = 0 is a zero polynomial of the polynomial p(x).

**(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.**

Solution:

p(x) = cx + d

⇒ cx+d =0

⇒ x = -d/c

∴ x = -d/c is a zero polynomial of the polynomial p(x).

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## Exercise 2.3

**1. Determine which of the following polynomials has (x + 1) a factor:**

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

Solution:

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

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(−1) = (−1)^{3}+(−1)^{2}+(−1)+1

= −1+1−1+1

= 0

∴ By factor theorem, x+1 is a factor of x^{3}+x^{2}+x+1

**(ii) x ^{4}+x^{3}+x^{2}+x+1**

Solution:

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

The zero of x+1 is -1. [x+1= 0 means x = -1]

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

= 1−1+1−1+1

= 1 ≠ 0

∴ By factor theorem, x+1 is not a factor of x^{4} + x^{3} + x^{2} + x + 1

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

Solution:

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

The zero of x+1 is -1.

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

=1−3+3−1+1

=1 ≠ 0

∴ By factor theorem, x+1 is not a factor of x^{4}+3x^{3}+3x^{2}+x+1

**(iv) x ^{3 }– x^{2}– (2+√2)x +√2**

Solution:

Let p(x) = x^{3}–x^{2}–(2+√2)x +√2

The zero of x+1 is -1.

p(−1) = (-1)^{3}–(-1)^{2}–(2+√2)(-1) + √2 = −1−1+2+√2+√2

= 2√2 ≠ 0

∴ By factor theorem, x+1 is not a factor of x^{3}–x^{2}–(2+√2)x +√2

**2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:**

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

Solution:

p(x) = 2x^{3}+x^{2}–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴ Zero of g(x) is -1.

Now,

p(−1) = 2(−1)^{3}+(−1)^{2}–2(−1)–1

= −2+1+2−1

= 0

∴ By factor theorem, g(x) is a factor of p(x).

**(ii) p(x)=x ^{3}+3x^{2}+3x+1, g(x) = x+2**

Solution:

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

g(x) = 0

⇒ x+2 = 0

⇒ x = −2

∴ Zero of g(x) is -2.

Now,

p(−2) = (−2)^{3}+3(−2)^{2}+3(−2)+1

= −8+12−6+1

= −1 ≠ 0

∴ By factor theorem, g(x) is not a factor of p(x).

**(iii) p(x)=x ^{3}–4x^{2}+x+6, g(x) = x–3**

Solution:

p(x) = x^{3}–4x^{2}+x+6, g(x) = x -3

g(x) = 0

⇒ x−3 = 0

⇒ x = 3

∴ Zero of g(x) is 3.

Now,

p(3) = (3)^{3}−4(3)^{2}+(3)+6

= 27−36+3+6

= 0

∴ By factor theorem, g(x) is a factor of p(x).

**3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:**

**(i) p(x) = x ^{2}+x+k**

Solution:

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

By Factor Theorem

⇒ (1)^{2}+(1)+k = 0

⇒ 1+1+k = 0

⇒ 2+k = 0

⇒ k = −2

**(ii) p(x) = 2x ^{2}+kx+**√2

Solution:

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

⇒ 2(1)^{2}+k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

**(iii) p(x) = kx ^{2}–**√

**2x+1**

Solution:

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

By Factor Theorem

⇒ k(1)^{2}-√2(1)+1=0

⇒ k = √2-1

**(iv) p(x)=kx ^{2}–3x+k**

Solution:

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

By Factor Theorem

⇒ k(1)^{2}–3(1)+k = 0

⇒ k−3+k = 0

⇒ 2k−3 = 0

⇒ k= 3/2

**4. Factorise:**

**(i) 12x ^{2}–7x+1**

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1×12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]

12x^{2}–7x+1= 12x^{2}-4x-3x+1

= 4x(3x-1)-1(3x-1)

= (4x-1)(3x-1)

**(ii) 2x ^{2}+7x+3**

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2×3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]

2x^{2}+7x+3 = 2x^{2}+6x+1x+3

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

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

**(iii) 6x ^{2}+5x-6 **

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6×-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]

6x^{2}+5x-6 = 6x^{2}+9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

**(iv) 3x ^{2}–x–4 **

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3×-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]

3x^{2}–x–4 = 3x^{2}–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

**5. Factorise:**

**(i) x ^{3}–2x^{2}–x+2**

Solution:

Let p(x) = x^{3}–2x^{2}–x+2

Factors of 2 are ±1 and ± 2

Now,

p(x) = x^{3}–2x^{2}–x+2

p(−1) = (−1)^{3}–2(−1)^{2}–(−1)+2

= −1−2+1+2

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

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

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

**(ii) x ^{3}–3x^{2}–9x–5**

Solution:

Let p(x) = x^{3}–3x^{2}–9x–5

Factors of 5 are ±1 and ±5

By the trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x^{3}–3x^{2}–9x–5

p(5) = (5)^{3}–3(5)^{2}–9(5)–5

= 125−75−45−5

= 0

Therefore, (x-5) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x^{2}+2x+1) = (x−5)(x^{2}+x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

**(iii) x ^{3}+13x^{2}+32x+20**

Solution:

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

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By the trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

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

p(-1) = (−1)^{3}+13(−1)^{2}+32(−1)+20

= −1+13−32+20

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder

(x+1)(x^{2}+12x+20) = (x+1)(x^{2}+2x+10x+20)

= (x+1)x(x+2)+10(x+2)

= (x+1)(x+2)(x+10)

**(iv) 2y ^{3}+y^{2}–2y–1**

Solution:

Let p(y) = 2y^{3}+y^{2}–2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

By the trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y^{3}+y^{2}–2y–1

p(1) = 2(1)^{3}+(1)^{2}–2(1)–1

= 2+1−2

= 0

Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y^{2}+3y+1) = (y−1)(2y^{2}+2y+y+1)

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

## Exercise 2.4

**1. Use suitable identities to find the following products:**

**(i) (x+4)(x +10) **

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, a = 4 and b = 10]

We get,

(x+4)(x+10) = x^{2}+(4+10)x+(4×10)

= x^{2}+14x+40

**(ii) (x+8)(x –10) **

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, a = 8 and b = −10]

We get,

(x+8)(x−10) = x^{2}+(8+(−10))x+(8×(−10))

= x^{2}+(8−10)x–80

= x^{2}−2x−80

**(iii) (3x+4)(3x–5)**

Solution:

Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab[Here, x = 3x, a = 4 and b = −5]

We get,

(3x+4)(3x−5) = (3x)^{2}+[4+(−5)]3x+4×(−5)

= 9x^{2}+3x(4–5)–20

= 9x^{2}–3x–20

**(iv) (y ^{2}+3/2)(y^{2}-3/2)**

Solution:

Using the identity, (x+y)(x–y) = x^{2}–y^{ 2}[Here, x = y^{2}and y = 3/2]

We get,

(y^{2}+3/2)(y^{2}–3/2) = (y^{2})^{2}–(3/2)^{2}

= y^{4}–9/4

**2. Evaluate the following products without multiplying directly:**

**(i) 103×107**

Solution:

103×107= (100+3)×(100+7)

Using identity, [(x+a)(x+b) = x^{2}+(a+b)x+ab

Here, x = 100

a = 3

b = 7

We get, 103×107 = (100+3)×(100+7)

= (100)^{2}+(3+7)100+(3×7)

= 10000+1000+21

= 11021

**(ii) 95×96 **

Solution:

95×96 = (100-5)×(100-4)

Using identity, [(x-a)(x-b) = x^{2}-(a+b)x+ab

Here, x = 100

a = -5

b = -4

We get, 95×96 = (100-5)×(100-4)

= (100)^{2}+100(-5+(-4))+(-5×-4)

= 10000-900+20

= 9120

**(iii) 104×96**

Solution:

104×96 = (100+4)×(100–4)

Using identity, [(a+b)(a-b)= a^{2}-b^{2}]

Here, a = 100

b = 4

We get, 104×96 = (100+4)×(100–4)

= (100)^{2}–(4)^{2}

= 10000–16

= 9984

**3. Factorise the following using appropriate identities:**

**(i) 9x ^{2}+6xy+y^{2}**

Solution:

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2×3x×y)+y^{2}

Using identity, x^{2}+2xy+y^{2 }= (x+y)^{2}

Here, x = 3x

y = y

9x^{2}+6xy+y^{2 }= (3x)^{2}+(2×3x×y)+y^{2}

= (3x+y)^{2}

= (3x+y)(3x+y)

**(ii) 4y ^{2}−4y+1**

Solution:

4y^{2}−4y+1 = (2y)^{2}–(2×2y×1)+1

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

Here, x = 2y

y = 1

4y^{2}−4y+1 = (2y)^{2}–(2×2y×1)+1^{2}

= (2y–1)^{2}

= (2y–1)(2y–1)

**(iii) x ^{2}–y^{2}/100**

Solution:

x^{2}–y^{2}/100 = x^{2}–(y/10)^{2}

Using identity, x^{2}-y^{2 }= (x-y)(x+y)

Here, x = x

y = y/10

x^{2}–y^{2}/100 = x^{2}–(y/10)^{2}

= (x–y/10)(x+y/10)

**4. Expand each of the following using suitable identities:**

**(i) (x+2y+4z) ^{2}**

**(ii) (2x−y+z) ^{2}**

**(iii) (−2x+3y+2z) ^{2}**

**(iv) (3a –7b–c) ^{2}**

**(v) (–2x+5y–3z) ^{2}**

**(vi) ((1/4)a-(1/2)b +1) ^{2}**

Solution:

**(i) (x+2y+4z) ^{2}**

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

Here, x = x

y = 2y

z = 4z

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

= x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz

**(ii) (2x−y+z) ^{2} **

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

Here, x = 2x

y = −y

z = z

(2x−y+z)^{2 }= (2x)^{2}+(−y)^{2}+z^{2}+(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x^{2}+y^{2}+z^{2}–4xy–2yz+4xz

**(iii) (−2x+3y+2z) ^{2}**

Solution:

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

Here, x = −2x

y = 3y

z = 2z

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

= 4x^{2}+9y^{2}+4z^{2}–12xy+12yz–8xz

**(iv) (3a –7b–c) ^{2}**

Solution:

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

Here, x = 3a

y = – 7b

z = – c

(3a –7b– c)^{2 }= (3a)^{2}+(– 7b)^{2}+(– c)^{2}+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

= 9a^{2} + 49b^{2 }+ c^{2}– 42ab+14bc–6ca

**(v) (–2x+5y–3z) ^{2}**

Solution:

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

Here, x = –2x

y = 5y

z = – 3z

(–2x+5y–3z)^{2 }= (–2x)^{2}+(5y)^{2}+(–3z)^{2}+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)

= 4x^{2}+25y^{2 }+9z^{2}– 20xy–30yz+12zx

**(vi) ((1/4)a-(1/2)b+1) ^{2}**

Solution:

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

Here, x = (1/4)a

y = (-1/2)b

z = 1

**5. Factorise:**

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

**(ii) 2x ^{2}+y^{2}+8z^{2}–2√2xy+4√2yz–8xz**

Solution:

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

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

We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}

4x^{2}+9y^{2}+16z^{2}+12xy–24yz–16xz = (2x)^{2}+(3y)^{2}+(−4z)^{2}+(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

= (2x+3y–4z)^{2}

= (2x+3y–4z)(2x+3y–4z)

**(ii) 2x ^{2}+y^{2}+8z^{2}–2√2xy+4√2yz–8xz**

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

We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}

2x^{2}+y^{2}+8z^{2}–2√2xy+4√2yz–8xz

= (-√2x)^{2}+(y)^{2}+(2√2z)^{2}+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)

= (−√2x+y+2√2z)^{2}

= (−√2x+y+2√2z)(−√2x+y+2√2z)

**6. Write the following cubes in expanded form:**

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

**(ii) (2a−3b) ^{3}**

**(iii) ((3/2)x+1) ^{3}**

**(iv) (x−(2/3)y) ^{3}**

Solution:

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

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

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

= 8x^{3}+1+6x(2x+1)

= 8x^{3}+12x^{2}+6x+1

**(ii) (2a−3b) ^{3}**

Using identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y)

(2a−3b)^{3 }= (2a)^{3}−(3b)^{3}–(3×2a×3b)(2a–3b)

= 8a^{3}–27b^{3}–18ab(2a–3b)

= 8a^{3}–27b^{3}–36a^{2}b+54ab^{2}

**(iii) ((3/2)x+1) ^{3}**

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

((3/2)x+1)^{3}=((3/2)x)^{3}+1^{3}+(3×(3/2)x×1)((3/2)x +1)

**(iv) (x−(2/3)y) ^{3}**

Using identity, (x –y)^{3} = x^{3}–y^{3}–3xy(x–y)

**7. Evaluate the following using suitable identities: **

**(i) (99) ^{3}**

**(ii) (102) ^{3}**

**(iii) (998) ^{3}**

Solutions:

**(i) (99) ^{3}**

Solution:

We can write 99 as 100–1

Using identity, (x –y)^{3} = x^{3}–y^{3}–3xy(x–y)

(99)^{3 }= (100–1)^{3}

= (100)^{3}–1^{3}–(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

= 970299

**(ii) (102) ^{3}**

Solution:

We can write 102 as 100+2

Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

(100+2)^{3 }=(100)^{3}+2^{3}+(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

**(iii) (998) ^{3}**

Solution:

We can write 99 as 1000–2

Using identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y)

(998)^{3 }=(1000–2)^{3}

=(1000)^{3}–2^{3}–(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

**8. Factorise each of the following:**

**(i) 8a ^{3}+b^{3}+12a^{2}b+6ab^{2}**

**(ii) 8a ^{3}–b^{3}–12a^{2}b+6ab^{2}**

**(iii) 27–125a ^{3}–135a +225a^{2} **

**(iv) 64a ^{3}–27b^{3}–144a^{2}b+108ab^{2}**

**(v) 27p ^{3}–(1/216)−(9/2) p^{2}+(1/4)p**

Solutions:

**(i) 8a ^{3}+b^{3}+12a^{2}b+6ab^{2}**

Solution:

The expression, 8a^{3}+b^{3}+12a^{2}b+6ab^{2} can be written as (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}

8a^{3}+b^{3}+12a^{2}b+6ab^{2 }= (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}

= (2a+b)^{3}

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y)^{3} = x^{3}+y^{3}+3xy(x+y) is used.

**(ii) 8a ^{3}–b^{3}–12a^{2}b+6ab^{2}**

Solution:

The expression, 8a^{3}–b^{3}−12a^{2}b+6ab^{2} can be written as (2a)^{3}–b^{3}–3(2a)^{2}b+3(2a)(b)^{2}

8a^{3}–b^{3}−12a^{2}b+6ab^{2 }= (2a)^{3}–b^{3}–3(2a)^{2}b+3(2a)(b)^{2}

= (2a–b)^{3}

= (2a–b)(2a–b)(2a–b)

Here, the identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y) is used.

**(iii) 27–125a ^{3}–135a+225a^{2} **

Solution:

The expression, 27–125a^{3}–135a +225a^{2} can be written as 3^{3}–(5a)^{3}–3(3)^{2}(5a)+3(3)(5a)^{2}

27–125a^{3}–135a+225a^{2 }=

3^{3}–(5a)^{3}–3(3)^{2}(5a)+3(3)(5a)^{2}

= (3–5a)^{3}

= (3–5a)(3–5a)(3–5a)

Here, the identity, (x–y)^{3} = x^{3}–y^{3}-3xy(x–y) is used.

**(iv) 64a ^{3}–27b^{3}–144a^{2}b+108ab^{2}**

Solution:

The expression, 64a^{3}–27b^{3}–144a^{2}b+108ab^{2}can be written as (4a)^{3}–(3b)^{3}–3(4a)^{2}(3b)+3(4a)(3b)^{2}

64a^{3}–27b^{3}–144a^{2}b+108ab^{2}=

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

=(4a–3b)^{3}

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y) is used.

**(v) 27p ^{3}– (1/216)−(9/2) p^{2}+(1/4)p**

Solution:

The expression, 27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p can be written as

(3p)^{3}–(1/6)^{3}−(9/2) p^{2}+(1/4)p = (3p)^{3}–(1/6)^{3}−3(3p)(1/6)(3p – 1/6)

Using (x – y)^{3} = x^{3} – y^{3} – 3xy (x – y)

27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p = (3p)^{3}–(1/6)^{3}−3(3p)(1/6)(3p – 1/6)

Taking x = 3p and y = 1/6

= (3p–1/6)^{3}

= (3p–1/6)(3p–1/6)(3p–1/6)

**9. Verify:**

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

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

Solutions:

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

We know that, (x+y)^{3} = x^{3}+y^{3}+3xy(x+y)

⇒ x^{3}+y^{3 }= (x+y)^{3}–3xy(x+y)

⇒ x^{3}+y^{3 }= (x+y)[(x+y)^{2}–3xy]

Taking (x+y) common ⇒ x^{3}+y^{3 }= (x+y)[(x^{2}+y^{2}+2xy)–3xy]

⇒ x^{3}+y^{3 }= (x+y)(x^{2}+y^{2}–xy)

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

We know that, (x–y)^{3} = x^{3}–y^{3}–3xy(x–y)

⇒ x^{3}−y^{3 }= (x–y)^{3}+3xy(x–y)

⇒ x^{3}−y^{3 }= (x–y)[(x–y)^{2}+3xy]

Taking (x+y) common ⇒ x^{3}−y^{3 }= (x–y)[(x^{2}+y^{2}–2xy)+3xy]

⇒ x^{3}+y^{3 }= (x–y)(x^{2}+y^{2}+xy)

**10. Factorise each of the following:**

**(i) 27y ^{3}+125z^{3}**

**(ii) 64m ^{3}–343n^{3}**

Solutions:

**(i) 27y ^{3}+125z^{3}**

The expression, 27y^{3}+125z^{3 }can be written as (3y)^{3}+(5z)^{3}

27y^{3}+125z^{3 }= (3y)^{3}+(5z)^{3}

We know that, x^{3}+y^{3 }= (x+y)(x^{2}–xy+y^{2})

27y^{3}+125z^{3 }= (3y)^{3}+(5z)^{3}

= (3y+5z)[(3y)^{2}–(3y)(5z)+(5z)^{2}]

= (3y+5z)(9y^{2}–15yz+25z^{2})

**(ii) 64m ^{3}–343n^{3}**

The expression, 64m^{3}–343n^{3}can be written as (4m)^{3}–(7n)^{3}

64m^{3}–343n^{3 }=

(4m)^{3}–(7n)^{3}

We know that, x^{3}–y^{3 }= (x–y)(x^{2}+xy+y^{2})

64m^{3}–343n^{3 }= (4m)^{3}–(7n)^{3}

= (4m-7n)[(4m)^{2}+(4m)(7n)+(7n)^{2}]

= (4m-7n)(16m^{2}+28mn+49n^{2})

**11. Factorise: 27x ^{3}+y^{3}+z^{3}–9xyz. **

Solution:

The expression 27x^{3}+y^{3}+z^{3}–9xyz can be written as (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)

27x^{3}+y^{3}+z^{3}–9xyz = (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)

We know that, x^{3}+y^{3}+z^{3}–3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}–xy –yz–zx)

27x^{3}+y^{3}+z^{3}–9xyz = (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)

= (3x+y+z)[(3x)^{2}+y^{2}+z^{2}–3xy–yz–3xz]

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

**12. Verify that:**

**x ^{3}+y^{3}+z^{3}–3xyz = (1/2) (x+y+z)[(x–y)^{2}+(y–z)^{2}+(z–x)^{2}]**

Solution:

We know that,

x^{3}+y^{3}+z^{3}−3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}–xy–yz–xz)

⇒ x^{3}+y^{3}+z^{3}–3xyz = (1/2)(x+y+z)[2(x^{2}+y^{2}+z^{2}–xy–yz–xz)]

= (1/2)(x+y+z)(2x^{2}+2y^{2}+2z^{2}–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x^{2}+y^{2}−2xy)+(y^{2}+z^{2}–2yz)+(x^{2}+z^{2}–2xz)]

= (1/2)(x+y+z)[(x–y)^{2}+(y–z)^{2}+(z–x)^{2}]

**13. If x+y+z = 0, show that x ^{3}+y^{3}+z^{3 }= 3xyz.**

Solution:

We know that,

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

Now, according to the question, let (x+y+z) = 0,

Then, x^{3}+y^{3}+z^{3 }-3xyz = (0)(x^{2}+y^{2}+z^{2}–xy–yz–xz)

⇒ x^{3}+y^{3}+z^{3}–3xyz = 0

⇒ x^{3}+y^{3}+z^{3 }= 3xyz

Hence Proved

**14. Without actually calculating the cubes, find the value of each of the following:**

**(i) (−12) ^{3}+(7)^{3}+(5)^{3}**

**(ii) (28) ^{3}+(−15)^{3}+(−13)^{3}**

Solution:

**(i) (−12) ^{3}+(7)^{3}+(5)^{3}**

Let a = −12

b = 7

c = 5

We know that if x+y+z = 0, then x^{3}+y^{3}+z^{3}=3xyz.

Here, −12+7+5=0

(−12)^{3}+(7)^{3}+(5)^{3 }= 3xyz

= 3×-12×7×5

= -1260

**(ii) (28) ^{3}+(−15)^{3}+(−13)^{3}**

Solution:

(28)^{3}+(−15)^{3}+(−13)^{3}

Let a = 28

b = −15

c = −13

We know that if x+y+z = 0, then x^{3}+y^{3}+z^{3 }= 3xyz.

Here, x+y+z = 28–15–13 = 0

(28)^{3}+(−15)^{3}+(−13)^{3 }= 3xyz

= 0+3(28)(−15)(−13)

= 16380

**15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: **

**(i) Area: 25a ^{2}–35a+12**

**(ii) Area: 35y ^{2}+13y–12**

Solution:

(i) Area: 25a^{2}–35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25×12 = 300

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 300]

25a^{2}–35a+12 = 25a^{2}–15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length = 5a–4

Possible expression for breadth = 5a –3

(ii) Area: 35y^{2}+13y–12

Using the splitting the middle term method,

We have to find a number whose sum = 13 and product = 35×-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]

35y^{2}+13y–12 = 35y^{2}–15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length = (5y+4)

Possible expression for breadth = (7y–3)

**By these five methods, you can find whether your children are suffering from weakness**

**16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? **

**(i) Volume: 3x ^{2}–12x**

**(ii) Volume: 12ky ^{2}+8ky–20k**

Solution:

(i) Volume: 3x^{2}–12x

3x^{2}–12x can be written as 3x(x–4) by taking 3x out of both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x–4)

(ii) Volume:

12ky^{2}+8ky–20k

12ky^{2}+8ky–20k can be written as 4k(3y^{2}+2y–5) by taking 4k out of both the terms.

12ky^{2}+8ky–20k = 4k(3y^{2}+2y–5)[Here, 3y^{2}+2y–5 can be written as 3y^{2}+5y–3y–5 using splitting the middle term method.]

= 4k(3y^{2}+5y–3y–5)

= 4k[y(3y+5)–1(3y+5)]

= 4k(3y+5)(y–1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

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