## Ex 2.1

**Question 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**

**(ii) y**

^{2}+ √2**(iii) 3 √t + t√2**

**(iv) y+**

**(v) x**

^{10}+ y^{3}+t^{50}**(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 powers 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 an be written as y+2y^{-1}

Though, *y *is the only variable in the given equation, the powers 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.

Question 2.

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

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

(ii) 2 – x^{2} + x^{3}

(iii) x^{2} + x

(iv) √2 x – 1

Solution:

(i) The given polynomial is 2 + x^{2} + x.

The coefficient of x^{2} is 1.

(ii) The given polynomial is 2 – x^{2} + x^{3}.

The coefficient of x^{2} is -1.

(iii) The given polynomial is + x.

The coefficient of x^{2} is .

(iv) The given polynomial is √2 x – 1.

The coefficient of x^{2} 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

Eg., 3x^{35}+5

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

Eg., 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 y 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

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}+4is 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.

## Ex 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)- = 4-

∴ 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, a0**

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).

## Ex 2.3

**1. Find the remainder when x ^{3}+3x^{2}+3x+1 is divided by**

**(i) x+1**

Solution:

x+1= 0

⇒x = −1

∴Remainder:

p(−1) = (−1)^{3}+3(−1)^{2}+3(−1)+1

= −1+3−3+1

= 0

**(ii) x−1/2**

Solution:

x-1/2 = 0

⇒ x = 1/2

∴Remainder:

p(1/2) = (1/2)^{3}+3(1/2)^{2}+3(1/2)+1

= (1/8)+(3/4)+(3/2)+1

= 27/8

**(iii) x**

Solution:

x = 0

∴Remainder:

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

= 1

**(iv) x+π**

Solution:

x+π = 0

⇒ x = −π

∴Remainder:

p(0) = (−π)^{3 }+3(−π)^{2}+3(−π)+1

= −π^{3}+3π^{2}−3π+1

**(v) 5+2x**

Solution:

5+2x=0

⇒ 2x = −5

⇒ x = -5/2

∴Remainder:

(-5/2)^{3}+3(-5/2)^{2}+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1

= -27/8

**2. Find the remainder when x ^{3}−ax^{2}+6x−a is divided by x-a.**

Solution:

Let p(x) = x^{3}−ax^{2}+6x−a

x−a = 0

∴x = a

Remainder:

p(a) = (a)^{3}−a(a^{2})+6(a)−a

= a^{3}−a^{3}+6a−a = 5a

**3. Check whether 7+3x is a factor of 3x ^{3}+7x.**

Solution:

7+3x = 0

⇒ 3x = −7

⇒ x = -7/3

∴Remainder:

3(-7/3)^{3}+7(-7/3) = -(343/9)+(-49/3)

= (-343-(49)3)/9

= (-343-147)/9

= -490/9 ≠ 0

∴7+3x is not a factor of 3x^{3}+7x

## Ex 2.4

**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. Factorize:**

**(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}–x–4

= 3x^{2}–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

**5. Factorize:**

**(i) x ^{3}–2x^{2}–x+2**

Solution:

Let p(x) = x^{3}–2x^{2}–x+2

Factors of 2 are ±1 and ± 2

By trial method, we find that

p(1) = 0

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

Now,

p(x) = x^{3}–2x^{2}–x+2

p(−1) = (−1)^{3}–2(−1)^{2}–(−1)+2

= −1−1+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 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 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−5)x(x+2)+10(x+2)

= (x−5)(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 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)

### Ex 2.5

**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. Factorize 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)+12

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}**

**((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. Factorize:**

**(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) 64a3–27b3–144a ^{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) 7p ^{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}–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p =

(3p)^{3}–(1/6)^{3}–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

= (3p–16)^{3}

= (3p–16)(3p–16)(3p–16)

**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. Factorize 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 expression27x^{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)

**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)