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Answers of Maths NCERT class 9 Chapter 2 – Polynomials

Class 9 Maths Chapter 2 Exercise 2.1

Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7

Solution:-

The equation 4x² – 3x + 7 can be written as 4x² – 3x¹ + 7x⁰

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 4x² – 3x + 7 is a polynomial in one variable.

(ii) y² + √2

Solution:-

The equation y² + can be written as y² + y⁰

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 y² + is a polynomial in one variable.

(iii) 3√t + t√2

Solution:-

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

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

(iv) y + 2/y

Solution:-

The equation y + can 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 + is not a polynomial in one variable.

(v) x¹⁰ + y³ + t⁵⁰

Solution:-

Here, in the equation x¹⁰ + y³ + t⁵⁰

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

x¹⁰ + y³ + t⁵⁰. Hence, it is not a polynomial in one variable.

Q2. Write the coefficients of x² in each of the following:
(i) 2 + x² + x

Solution:-

The equation 2 + x² + x can be written as 2 + (1) x² + x

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

Here, the number that multiplies the variable x² is 1

, the coefficients of x² in 2 + x² + x is 1.

(ii) 2 – x² + x³

Solution:-

The equation 2 – x² + x³ can be written as 2 + (–1) x² + x³

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

Here, the number that multiplies the variable x² is -1

, the coefficients of x² in 2 – x² + x³ is -1.

(iii) Π/2 x² +x

Solution:-

The equation Π/2x² +x can be written as (Π/2 ) x² + x

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

Here, the number that multiplies the variable x² is

, the coefficients of x² in Π/2x² +x is Π/2.

(iv)√2x-1

Solution:-

The equationx√2x-1 can be written as 0x² +√2x-1 [Since 0x² is 0]

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

Here, the number that multiplies the variable x² is 0

, the coefficients of x² in√2x-1 is 0.

Q3. 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., 3x³⁵+5

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

Eg., 4x¹⁰⁰

Q4. Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x

Solution:-

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

Here, 5x³ + 4x² + 7x= 5x³ + 4x² + 7x¹

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

, the degree of 5x³ + 4x² + 7x is 3 as 3 is the highest power of x in the equation.

(ii) 4 – y²

Solution:-

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

Here, in 4 – y²,

The power of the variable y is: 2

, the degree of 4 – y² 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== 3x⁰

The power of the variable here is: 0

, the degree of 3 is 0.

Q5. 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 a cubic polynomial.

(i) x² + x

Solution:-

The highest power of x² + x is 2

, the degree is 2

Hence, x² + x is a quadratic polynomial

(ii) x – x³

Solution:-

The highest power of x – x³ is 3

, the degree is 3

Hence, x – x³ is a cubic polynomial

(iii) y + y² + 4

Solution:-

The highest power of y + y² + 4 is 2

, the degree is 2

Hence, y + y² + 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) r²

Solution:-

The highest power of r² is 2

, the degree is 2

Hence, r² is a quadratic polynomial

(vii) 7x³

Solution:-

The highest power of 7x³ is 3

, the degree is 3

Hence, 7x³ is a cubic polynomial

Exercise 2.2 Page: 34

Q1. Find the value of the polynomial (x)=5x−4x²+3 
(i) x= 0
(ii) x = – 1
(iii) x = 2

Solution:-

Let f(x)= 5x−4x²+3

(i) When x=0

f(0)=5(0)+4(0)²+3

=3

(ii) When x= -1

f(x)=5x−4x²+3

f(−1)=5(−1) −4(−1)²+3

=−5–4+3

=−6

(iii) When x=2

f(x)=5x−4x²+3

f(2)=5(2) −4(2)²+3

=10–16+3

=−3

Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y²−y+1

Solution:-

p(y)=y²–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+2t²−t³

Solution:-

p(t)= 2+t+2t²−t³

∴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)=x³

Solution:-

p(x)=x³

∴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

Q3. 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

∴ 4/5is not a zero of p(x).

(iii) p(x)=x²−1, x=1, −1

Solution:-

For, x=1, −1;

p(x)=x²−1

∴p(1)=1²−1=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)=x², x=0

Solution:-

For, x=0 p(x)= x²

p(0)=0²=0

∴0 is a zero of p(x).

(vi) p(x)=lx+m, x=−m/t

Solution:-

For, x=−m/t; p(x)=lx+m

∴p(−m/t)=l(−m/t)+m=−m+m=0

∴−m/tis a zero of p(x).

(vii) p(x)=3x²−1,x=−1/√3,2/√3,

Solution:-

For, x=−1/√3,2/√3,; p(x)=3x²−1

∴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).

Q4. Find the zero of the polynomial 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).

Class 9 Maths Chapter 2 Exercise 2.3 Page: 40

Q1. Find the remainder when x³+3x²+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

=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

=−π³+3π²−3π+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

Q2. Find the remainder when x³−ax²+6x−a is divided by x-a.

Solution:-

Let p(x)=x³−ax²+6x−a

x−a=0

∴x=a

Remainder:

p(a)= (a)³ −a(a²)+6(a)−a

=a³−a³+6a−a = 5a

Q3. Check whether 7+3x is a factor of 3x³+7x.

Solution:-

7+3x=0

⇒3x=−7 only if 7+3x divides 3x³+7x leaving no remainder.

⇒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 3x³+7x

Exercise 2.4 Page: 43

Q1. Determine which of the following polynomials has (x + 1) a factor:

(i) x³+x²+x+1

Solution:-

Let p(x)= x³+x²+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 x³+x²+x+1

(ii) x⁴ + x³ + x² + x + 1

Solution:-

Let p(x)= x⁴ + x³ + x² + 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 x⁴ + x³ + x² + x + 1

(iii) x⁴ + 3x³ + 3x² + x + 1

Solution:-

Let p(x)= x⁴ + 3x³ + 3x² + 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 x⁴ + 3x³ + 3x² + x + 1

(iv) x³ – x² – (2 +√2 )x +√2

Solution:-

Let p(x)= x³ – x² – (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

∴By factor theorem, x+1 is not a factor of x³ – x² – (2 +√2 )x +√2

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x)=2x³+x²–2x–1, g(x) = x + 1

Solution:-

p(x)= 2x³+x²–2x–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)=x³+3x²+3x+1, g(x) = x + 2

Solution:-

p(x)=x3+3×2+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)=x³–4x²+x+6, g(x) = x – 3

Solution:-

p(x)= x³–4x²+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).

Q3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x)=x²+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)=2x²+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)=kx²–√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)=kx²–3x+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= \(\frac{3}{2}\)

Q4. Factorize:

(i) 12x²–7x+1

Solution:-

Using the splitting the middle term method,

We have to find a number whose sum=-7 and product=112=12

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

12x²–7x+1=12x²-4x-3x+1

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

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

(ii) 2x²+7x+3

Solution:-

Using the splitting the middle term method,

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

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

2x²+7x+3 =2x²+6x+1x+3

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

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

(iii) 6x²+5x-6

Solution:-

Using the splitting the middle term method,

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

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

6x²+5x-6=6x²+ 9x – 4x – 6

=3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

(iv) 3x² – x – 4

Solution:-

Using the splitting the middle term method,

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

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

3x² – x – 4 =3x²–x–4

=3x²–4x+3x–4

=x(3x–4)+1(3x–4)

=(3x–4)(x+1)

Q5. Factorize:
(i) x³–2x²–x+2

Solution:-

Let p(x)=x³–2x²–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)= x³–2x²–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)(x²–3x+2) =(x+1)(x²–x–2x+2)

=(x+1)(x(x−1)−2(x−1))

=(x+1)(x−1)(x-2)

(ii) x³–3x²–9x–5

Solution:-

Let p(x) = x³–3x²–9x–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) = x³–3x²–9x–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)(x²+2x+1) =(x−5)(x²+x+x+1)

=(x−5)(x(x+1)+1(x+1))

=(x−5)(x+1)(x+1)

(iii) x³+13x²+32x+20

Solution:-

Let p(x) = x³+13x²+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) = x³+13x²+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)(x²+12x+20) =(x+1)(x²+2x+10x+20)

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

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

(iv) 2y³+y²–2y–1

Solution:-

Let p(y) = 2y³+y²–2y–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) = 2y³+y²–2y–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)(2y²+3y+1) =(y−1)(2y²+2y+y+1)

=(y−1)(2y(y+1)+1(y+1))

=(y−1)(2y+1)(y+1)

Exercise 2.5 Page: 48

Q1. 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) =x²+(4+10)x+(4×10)

=x²+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) =x²+(8+(−10))x+(8×(−10))

=x²+(8−10)x–80

=x²−2x−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)

=9x²+3x(4–5)–20

=9x²–3x–20

(iv) (y²+3/2)(y²–3/2 )

Solution:-

Using the identity, (x + y) (x – y) = x ² – y ²

[Here, x=y² and y=3/2]

We get,

(y²+3/2)(y²–3/2) = (y²)²–(3/2)²

=y⁴–(9/4)

Q2. Evaluate the following products without multiplying directly:
(i) 103 × 107

Solution:-

103×107=(100+3)×(100+7)

Using identity, [(x+a)(x+b)=x2+(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)=x²+(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)= a²-b²]

Here, a=100

b=4

We get, 104×96=(100+4)×(100–4)

=(100)²–(4)²

=10000–16

=9984

Q3. Factorize the following using appropriate identities:
(i) 9x²+6xy+y²

Solution:-

9x²+6xy+y²=(3x)²+(2×3x×y)+y²

Using identity, x² + 2xy + y²= (x + y)²

Here, x=3x

y=y

9x²+6xy+y²=(3x)²+(2×3x×y)+y²

=(3x+y)²

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

(ii) 4y²−4y+1

Solution:-

4y²−4y+1=(2y)²–(2×2y×1)+12

Using identity, x² – 2xy + y²= (x – y)²

Here, x=2y

y=1

4y²−4y+1=(2y)²–(2×2y×1)+1²

=(2y–1)²

=(2y–1)(2y–1)

(iii) x²–y^2/100

Solution:-

x²–y^{2/100 = x2–(y/10)2}

Using identity, x² – y²= (x – y) (x y)

Here,

x=x

y=y/10

x² – y²/100= x²–(y/10)²

=(x–y/10)(x+y/10)

Q4. 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) (a–b+1)²

Solutions:

(i) (x+2y+4z)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Here, x=x

y=2y

z=4z

(x+2y+4z)² =x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)

=x²+4y²+16z²+4xy+16yz+8xz

(ii) (2x−y+z)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Here, x=2x

y=−y

z=z

(2x−y+z)² =(2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)

=4x²+y²+z²–4xy–2yz+4xz

(iii) (−2x+3y+2z)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 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)

=4x²+9y²+4z²–12xy+12yz–8xz

(iv) (3a – 7b – c)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 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)

=9a² + 49b² + c²– 42ab+14bc–6ca

(v) (–2x + 5y – 3z)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 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)

=4x² + 25y² + 9z²– 20xy–30yz+12zx

(vi) (1/4a – 1/2b+1)²

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Here, x= 1/4a

y= –1/2b

z= 1

(1/4a –1/2b +1)² =(1/4a)²+(–1/2b)²+(1)²+(2×1/4a × –1/2b)+(2× –1/2b ×1)+(2×1×1/4a)

=1/16a²+1/4b²+1²–2/8ab– 2/2b +2/4a

= 1/16a²+1/4b²+1–1/4ab – b +1/2a

Q5. Factorize:

(i) 4x²+9y²+16z²+12xy–24yz–16xz
(ii) 2x²+y²+8z²–2xy+4yz–8xz

Solutions:

(i) 4x²+9y²+16z²+12xy–24yz–16xz

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

We can say that, x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²

4x²+9y²+16z²+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) 2x²+y²+8z²–2√2xy+4√2yz–8xz

Solution:-

Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

We can say that, x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²

2x²+y²+8z²–2√2xy+4√2yz–8xz =(-√2x)²+(y)²+(2√2z)²+(2×−√2x×y)+(2×y×2√2z)+(2×2√2√2z×−√2x)

=(−√2x+y+2√2z)²

=(−√2x+y+2√2z)(−√2x+y+2√2z)

Q6. Write the following cubes in expanded form:
(i) (2x+1)³
(ii) (2a−3b)³
(iii) (x+1)³
(iv) (x−y)³

Solutions:

(i) (2x+1)³

Solution:-

Using identity, (x + y)³ = x³ + y³ + 3xy (x + y)

(2x+1)³=(2x)³+1³+(3×2x×1)(2x+1)

=8x³+1+6x(2x+1)

=8x³+12x²+6x+1

(ii) (2a−3b)³

Solution:-

Using identity, (x – y)³ = x³ – y³ – 3xy(x – y)

(2a−3b)³=(2a)³−(3b)³–(3×2a×3b)(2a–3b)

=8a³–27b³–18ab(2a–3b)

=8a³–27b³–36a²b+54ab²

(iii) (3/2x+1)³

Solution:-

Using identity, (x + y)³ = x³ + y³ + 3xy (x + y)

(3/2x+1)³ =(3/2x)³+1³+(3×3/2x×1)(3/2x+1)

=27/8x³+1+9/2x(3/2x+1)

=27/8x³+1+27/4x²+9/2x

=27/8x³+27/4x²+9/2x+1

(iv) (x−2/3y)³

Solution:-

Using identity, (x – y)³ = x³ – y³ – 3xy(x – y)

(x−2/3y)³ =(x)³–(2/3y)³–(3×x×2/3y)(x–2/3y)

=(x)³–8/27y³–2xy(x– 2/3y)

=(x)³–8/27y³–2x²y+4/3xy²

Q7. 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 = x³ – y³ – 3xy(x – y)

(99)³ = (100–1)³

=(100)³–1³–(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 = x³ + y³ + 3xy (x + y)

(100+2)³ =(100)³+2³+(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 = x³ – y³ – 3xy(x – y)

(998)³ =(1000–2)³

=(1000)³–2³–(3×1000×2)(1000–2)

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000

= 994011992

Q8. Factorise each of the following:
(i) 8a³+b³+12a²b+6ab²
(ii) 8a³–b³–12a²b+6ab²
(iii) 27 – 125a³ – 135a + 225a² 
(iv) 64a³–27b³–144a²b+108ab²
(v) 27p³ –1/216 −(9/2)p²+(1/4)p

Solutions:

(i) 8a³+b³+12a²b+6ab²

Solution:-

The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²

8a³+b³+12a²b+6ab² =(2a)³+b³+3(2a)²b+3(2a)(b)²

=(2a+b)³

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

Here, the identity, (x + y)³ = x³ + y³ + 3xy (x + y) is used.

(ii) 8a³–b³–12a²b+6ab²

Solution:-

The expression, 8a³–b³−12a²b+6ab² can be written as (2a)³–b³–3(2a)²b+3(2a)(b)²

8a³–b³−12a²b+6ab² =(2a)³–b³–3(2a)²b+3(2a)(b)²

=(2a–b)³

=(2a–b)(2a–b)(2a–b)

Here, the identity, (x – y)³ = x³ – y³ – 3xy(x – y) is used.

(iii) 27 – 125a³ – 135a + 225a²

Solution:-

The expression, 27 – 125a³ – 135a + 225a² can be written as 3³–(5a)³–3(3)²(5a)+3(3)(5a)²

27–125a³–135a+225a² = 3³–(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 = x³ – y³ – 3xy(x – y) is used.

(iv) 64a3–27b3–144a²b+108ab²

Solution:-

The expression, 64a³–27b³–144a²b+108ab² can be written as (4a)³–(3b)³–3(4a)²(3b)+3(4a)(3b)²

64a³–27b³–144a²b+108ab² = (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 = x³ – y³ – 3xy(x – y) is used.

(v) 27p³ – 1/216 − 9/2p²+ 1/4p

Solution:-

The expression, 27p³ –1/216 − 9/2p²+ 1/4p can be written as (3p)³–(1/6)³–3(3p)²(1/6)+3(3p)(1/6)²

27p³ – 1/216 − 9/2p²+ 1/4p = (3p)³–(1/6)³–3(3p)²(1/6)+3(3p)(1/6)²

= (3p–(1/6))³

= (3p–(1/6))(3p–(1/6))(3p–(1/6))

Q9. Verify:
(i) x³+y³=(x+y)(x²–xy+y²)
(ii) x³–y³=(x–y)(x²+xy+y²)

Solutions:

(i) x³+y³=(x+y)(x²–xy+y²)

We know that, (x+y)³ =x³+y³+3xy(x+y)

⇒x³+y³=(x+y)³–3xy(x+y)

⇒x³+y³=(x+y)[(x+y)²–3xy]

Taking(x+y) common ⇒x³+y³=(x+y)[(x²+y²+2xy)–3xy]

⇒x³+y³=(x+y)(x²+y²–xy)

(ii) x³–y³=(x–y)(x²+xy+y²)

We know that,(x–y)³ =x³–y³–3xy(x–y)

⇒x³−y³=(x–y)³+3xy(x–y)

⇒x³−y³=(x–y)[(x–y)²+3xy]

Taking(x+y) common⇒x³−y³=(x–y)[(x²+y²–2xy)+3xy]

⇒x³+y³=(x–y)(x²+y²+xy)

Q10. Factorize each of the following:
(i) 27y³+125z³
(ii) 64m³–343n³

Solutions:

(i) 27y³+125z³

The expression, 27y³+125z³ can be written as (3y)³+(5z)³

27y³+125z³ =(3y)³+(5z)³

We know that, x³+y³=(x+y)(x²–xy+y²)

27y³+125z³ =(3y)³+(5z)³

=(3y+5z)[(3y)²–(3y)(5z)+(5z)²]

=(3y+5z)(9y²–15yz+25z²)

(ii) 64m³–343n³

The expression, 64m³–343n³ can be written as (4m)³–(7n)³

64m³–343n³ = (4m)³–(7n)³

We know that, x³–y³=(x–y)(x²+xy+y²)

64m³–343n³ =(4m)³–(7n)³

=(4m-7n)[(4m)²+(4m)(7n)+(7n)²]

=(4m-7n)(16m²+28mn+49n²)

Q11. Factorise : 27x³+y³+z³–9xyz

Solution:-

The expression 27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)

27x³+y³+z³–9xyz =(3x)³+y³+z³–3(3x)(y)(z)

We know that, x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

27x³+y³+z³–9xyz =(3x)³+y³+z³–3(3x)(y)(z)

=(3x+y+z)(3x)²+y²+z²–3xy–yz–3xz

=(3x+y+z)(9x²+y²+z²–3xy–yz–3xz)

Q12. Verify that:

x³+y³+z³–3xyz=(x+y+z)[(x–y)²+(y–z)²+(z–x)²]

Solution:-

We know that,

x³+y³+z³−3xyz=(x+y+z)(x²+y²+z²–xy–yz–xz)

⇒x³+y³+z³–3xyz =×(x+y+z)[2(x²+y²+z²–xy–yz–xz)]

= (x+y+z)(2x²+2y²+2z²–2xy–2yz–2xz)

= (x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]

= (x+y+z)[(x–y)²+(y–z)²+(z–x)²]

Q13. If x + y + z = 0, show that x³+y³+z³=3xyz.

Solution:-

We know that,

x³+y³+z³=3xyz = (x + y + z)(x² + y² + z² – xy – yz – xz)

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

then, x³+y³+z³=3xyz =(0)(x²+y²+z²–xy–yz–xz)

⇒x³+y³+z³–3xyz =0

⇒ x³+y³+z³ =3xyz

Hence Proved

Q14. 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

(i) (−12)³+(7)³+(5)³

Solution:-

(−12)³+(7)³+(5)³

Let a= −12

b= 7

c= 5

We know that if x + y + z = 0, then x³+y³+z³=3xyz.

Here, −12+7+5=0

(−12)³+(7)³+(5)³ = 3xyz

=

=

(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 x³+y³+z³=3xyz.

Here, x + y + z = 28 – 15 – 13 = 0

(28)³+(−15)³+(−13)³= 3xyz

= 0+3(28)(−15)(−13)

=16380

Q15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : 25a²–35a+12
(ii) Area : 35y²+13y–12

Solution:-

(i) Area : 25a²–35a+12

Using the splitting the middle term method,

We have to find a number whose sum= -35 and product=2512=300

We get -15 and -20 as the numbers [-15+-20=-35 and -3-4=300]

25a²–35a+12 =25a²–15a−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 : 35y²+13y–12

Using the splitting the middle term method,

We have to find a number whose sum= 13 and product=3512=420

We get -15 and 28 as the numbers [-15+28=-35 and -15=420]

35y²+13y–12 =35y²–15y+28y–12

=5y(7y–3)+4(7y–3)

=(5y+4)(7y–3)

Possible expression for length = (5y + 4)

Possible expression for breadth = (7y – 3)

Q16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 
(i) Volume : 3x²–12x
(ii) Volume : 12ky²+8ky–20k

Solution:-

(i) Volume : 3x²–12x

3x²–12x 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 : 12ky²+8ky –20k

12ky²+8ky –20k can be written as 4k(3y²+2y–5) by taking 4k out of both the terms.

12ky²+8ky–20k =4k(3y²+2y–5)

[Here, 3y²+2y–5 can be written as 3y²+5y–3y–5 using splitting the middle term method.]

=4k(3y²+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)