NCERT Exemplar Solutions: Polynomials | Mathematics (Maths) Class 9 PDF Download

Exercise 2.1

Write the correct answer in each of the following:
Q.1. Which one of the following is a polynomial?
(a)
(b)
(c)
(d)

Correct Answer is option (c)

The power of x must be a non-negative integer in a polynomial.
Hence, option C is correct.
As we can write option c as

Q.2. √2 is a polynomial of degree
(a)  2
(b) 0
(c) 1
(d) 1/2

Correct Answer is option (b)

We can write it as √2  = √2  × x⁰ 
Thus, degree is 0.

Q.3. Degree of the polynomial 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is
(a) 4
(b) 5
(c) 3
(d) 7

Correct Answer is option (a)
Degree of the polynomial 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is 4
We know that, the degree of given polynomial 4x⁴ + 0 x x³ + 0 x x⁵ + 5x + 7 will be the highest power of variable that is 4.

Q.4. Degree of the zero polynomial is
(a) 0
(b) 1
(c) Any natural number
(d) Not defined

Correct Answer is option (d)
The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.

Degree of zero polynomial is thus not-defined.

Q.5. If p (x ) = x² –2√2 x+ 1 , then p(2√2) is equal to
(a) 0
(b) 1
(c) 4√2
(d)

Correct Answer is option (b)
p(x) = x² −2√2 x + 1
p(2√2 ) = (2√2)² − 2√2  × 2√2  +1
p(2√2) = 8 − 8 + 1
p(2√2) = 1
Hence, option B.

Q.6. The value of the polynomial 5x – 4x² + 3, when x = –1 is
(a) - 6
(b) 6
(c) 2
(d) - 2

Correct Answer is option (a)
Given that: a polynomial as
P(x) = 5x − 4x² + 3
To find value of polynomial at x = −1 or P(−1)
So we put x = −1 in polynomial,
P(−1) = 5 × (−1) − 4 × (−1)² + 3
P(−1) = − 5 − 4 + 3
P(−1) = − 6
So value of polynomial at x = − 1 is − 6.

Q.7. If p(x) = x + 3, then p(x) + p(–x) is equal to
(a) 3
(b) 2x
(c) 0
(d) 6

Correct Answer is option (d)
Given, p(x) = x + 3
Therefore, p(−x) = −x + 3
Now, p(x) + p(−x) = (x+3) + (−x+3) = 6
Option D is correct.

Q.8. Zero of the zero polynomial is
(a) 0
(b) 1
(c) Any real number
(d) Not defined

Correct Answer is option (c)
Zero of the 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.

Q.9. Zero of the polynomial p(x) = 2x + 5 is
(a) -(2/5)
(b) -(5/2
(c) 2/5
(d) 5/2

Correct Answer is option (b)
Zero of a polynomial is the value of the variable for which the polynomial becomes 0.
Now, p(x) = 2x + 5.
For, p(x) = 0,
2x + 5 = 0.
or,  x = −5/2.
Therefore, option D is correct.

Q.10. One of the zeroes of the polynomial 2x² + 7x –4 is
(a) 2
(b) 1/2
(c) -(1/2)
(d) –2

Correct Answer is option (b)
To find the zero of 2x 2 +7x−4 is same as solving the equation 2x² + 7x − 4 = 0
⇒ 2x² + 8x − x − 4 = 0
⇒ 2x(x+4) − 1(x+4) = 0
⇒ (2x−1)(x+4) = 0
⇒ (2x−1) = 0, (x + 4) = 0
⇒ x = 1/2 , −4
Therefore, option B is correct.

Q.11. If x⁵¹ +  51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50

Correct Answer is option (d)
By remainder theorem,

Remainder =p(−1), since the divisor is x+1 

p(x) = x ⁵¹ + 51
p(−1) = (−1)⁵¹ + 51 = −1 + 51
Hence, the remainder is 50

Q.12. If x + 1 is a factor of the polynomial 2x² + kx, then the value of k is
(a) –3
(b) 4
(c) 2
(d) –2

Correct Answer is option (c)
(x + 1) is a factor of p (x)
x + 1 = 0
⇒ x = − 1
Therefore, p (− 1) = 0
∴ 2 (− 1)² + k(− 1) = 0
⇒ 2 − k = 0 ⇒ k = 2
Hence, the value of k is 2.

Q.13. x + 1 is a factor of the polynomial
(a) x³ + x² – x + 1
(b) x³ + x² + x + 1
(c) x⁴ + x³ + x² + 1
(d) x⁴ + 3 x³ + 3x² + x + 1

Correct Answer is option (b)

We know that if x + a is a factor of f(x) then, f(-a) = 0.
Let f(x) = x⁴ + x³ + x² + 1
Now, f(-1) = (-1)⁴ + (-1)³ + (-1)² + 1
= 1 - 1 + 1 + 1
= 2 ≠ 0
So, f(x) is not a factor of x + 1.

Q.14. One of the factors of (25x² – 1) + (1 + 5x)² is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x

Correct Answer is option (d)
(25x² – 1) + (1 + 5x)²
= 25x² − 1 + 1 + 25x² + 10 x [Using identity, (a + b)² = a² + b² + 2ab]
= 50x² + 10 x = 10 x (5x + 1)
Hence, one of the factor of given polynomial is 10x.

Q.15. The value of 249² – 248² is
(a) 1²
(b) 477
(c) 487
(d) 497

Correct Answer is option (d)

We know, a² −b² = (a + b) (a − b)
So, 249² −248² = (249 + 248) (249 − 248) = 497(1) = 497

Q.16. The factorisation of 4x² + 8x + 3 is
(a) (x + 1) (x + 3)
(b) (2x + 1) (2x + 3)
(c) (2x + 2) (2x + 5)
(d) (2x –1) (2x –3)

Correct Answer is option (b)

4x² + 8x + 3
⇒ 4x² + 6x + 2x + 3
⇒ 2x(2x+3) + 1(2x+3)
⇒ (2x+3) (2x+1)

Q.17. Which of the following is a factor of (x + y)³ – (x³ + y³)?
(a)  x² + y² + 2 xy
(b)  x² + y² – xy
(c) xy²
(d) 3xy

Correct Answer is option (d)

(x + y)³ −(x³ + y³)
= x³ + y³ + 3xy(x+y) − x³ − y³
= 3xy(x+y)
Hence, 3xy and (x+y) are the factors of the equation

Q.18. The coefficient of x in the expansion of (x + 3)³ is
(a) 1
(b) 9
(c) 18
(d) 27

Correct Answer is option (d)
Now, (x + 3)³ = x³ + 3³ + 3x (3)(x + 3)
[using identity, (a + b)³ = a³ + b³+ 3ab (a + b)]
= x³ + 27 + 9x (x + 3)
= x³ + 27 + 9x² + 27x
Hence, the coefficient of x in (x + 3)³ is 27.

Q.19. If (x, y ≠ 0) , the value of x³ – y³ is
(a) 1
(b) –1
(c) 0
(d) 1/2

Correct Answer is option (c)

Consider the equation:  
Simplify the above expression as follows:
x² + y² = -xy
Now, x³ - y³ = (x-y) (x²+y² + xy)
= (x-y)(-xy + xy)  ...... [Substitute: x² + y² = –xy]
= (x-y) × 0
= 0

Q.20. If 49x² – b = then the value of b is
(a) 0
(b) 1/√2
(c) 1/4
(d) 1/2

Correct Answer is option (c)

Given (49x² – b) =
......[Using identity, (a + b)(a – b) = a² – b²]

⇒ (√b)² = (1/2)²  ...... [Multiplying both sides by –1]
∴ b = 1/4

Q.21. If α + b + c = 0, then α³ + b³ + c³ is equal to
(a) 0
(b) αbc
(c) 3αbc
(d) 2αbc

Correct Answer is option (c)

Since, α³ + b³ + c³ −3αbc = (a + b + c)(a² + b² + c² −bc − ca − ab)
Given, a + b + c = 0
∴ a³ + b³ + c³ − 3abc = 0
∴ a³ + b³ + c³ = 3abc
Option b is correct.

Exercise 2.2

Q.1. Which of the following expressions are polynomials? Justify your answer: 
(i) 8

Polynomial. Because, the exponent of the variable of 8 or 8x⁰ is 0 which is a whole number.

(ii)

Polynomial. Because, the exponent of the variable of  is a whole number.

(iii) 1 - √5x

Not a polynomial. Because, the exponent of the variable of 1 − √5x or 1 − √5x^1/2  is 1/2 which is not a whole number.

(iv)

Polynomial. Because, the exponents of the variable of = 1/5x² + 5x + 7 are whole numbers.

(v)

Not a polynomial. Because, the exponent of the variable of = x − 6 + 8x⁻¹ is -1, which is not a whole number.

(vi) 1/x+1

Not a polynomial as the polynomial is expressed as a₀ + a₁x + a₂xⁿ , where a₀ , a₁ , a₂ ⋯ , a_n are constants. Now, the given function is in the form f(x) = p(x)/q(x) is a rational expression but not a polynomial.

(vii)

Polynomial. Because, the exponents of the variable of + 4 a − 7 are whole numbers.

(viii) 1/2x

Not a polynomial. Because, the exponent of the variable of 1/2x or 1/2^x-1 is -1, which is not a whole number.

Q.2. Write whether the following statements are True or False. Justify your answer. 
(i) A binomial can have at most two terms

False
because a binomial has exactly two terms. 

(ii) Every polynomial is a binomial

False
x³ + x + 1 is a polynomial but not a binomial.

(iii) A binomial may have degree 5

True
because a binomial is a polynomial whose degree is a whole number ≥ 1, so, degree can be 5 also.

(iv) Zero of a polynomial is always 0

False

because zero of a polynomial can be any real number.

(v) A polynomial cannot have more than one zero

False
A polynomial can have any number of zeroes. It depends upon the degree of the polynomial.

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

False
x⁵ + 1 and – x⁵ + 2 x + 3 are two polynomials of degree 5 but the degree of the sum of the two polynomials is 1.

Exercise 2.3

Q.1. Classify the following polynomials as polynomials in one variable, two variables etc. 
(i) x² + x + 1 

Here, the polynomial contains only one variable, i.e., x. Hence, the given polynomial is a polynomial in one variable.

(ii) y³ – 5 y 

Here, the polynomial contains only one variable, i.e., y. Hence, the given polynomial is a polynomial in one variable.

(iii) xy + yz + zx 

Here, the polynomial contains three variables, i.e., x, y and z. Hence, the given polynomial is a polynomial in three variable.

(iv) x² – 2 xy + y² + 1

Here, the polynomial contains two variables, i.e., x and y. Hence, the given polynomial is a polynomial in two variable.

Q.2. Determine the degree of each of the following polynomials:
(i) 2x – 1

Degree of a polynomial in one variable = highest power of the variable in algebraic expression
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial 2x – 1 = 1

(ii) –10

There is no variable in the given term.
Let us assume that the variable in the given expression is x.
– 10 = –10x⁰
Power of x = 0
Highest power of the variable x in the given expression = 0 Hence, degree of the polynomial – 10 = 0

(iii) x³ – 9 x + 3 x⁵

Powers of x = 3, 1 and 5 respectively.
Highest power of the variable x in the given expression = 5 Hence, degree of the polynomial x³ – 9x + 3x⁵ = 5

(iv) y³ (1 – y⁴)

The equation can be written as, y³(1 – y⁴) = y³ – y⁷
Powers of y = 3 and 7 respectively.
Highest power of the variable y in the given expression = 7 Hence, degree of the polynomial y³(1 – y⁴) = 7

Q.3. For the polynomial
write
(i) the degree of the polynomial 

Powers of x = 3, 1, 2 and 6 respectively.

Highest power of the variable x in the given expression = 6 Hence, degree of the polynomial = 6

(ii) the coefficient of x³ 

The given equation can be written as,

Hence, the coefficient of x3 in the given polynomial is 1/5.

(iii) the coefficient of x⁶ 

The coefficient of x⁶ in the given a polynomial is – 1

(iv) the constant term

Since the given equation can be written as,

The constant term in the given polynomial is 1/5 as it has no variable x associated with it.

Q.4. Write the coefficient of x² in each of the following :

(i) (π/6)x + x² – 1

(π/6) x + x²−1 = (π/6) x + (1) x²−1
The coefficient of x² in the polynomial (π/6) x + x²−1 = 1.

(ii) 3x – 5

3x – 5 = 0x² + 3x – 5
The coefficient of x² in the polynomial 3x – 5 = 0, zero.

(iii) (x –1) (3x – 4)

(x – 1)(3x – 4) = 3x² – 4x – 3x + 4
= 3x² – 7x + 4
The coefficient of x² in the polynomial 3x² – 7x + 4 = 3.

(iv) (2x – 5) (2x² – 3x + 1)

(2x – 5) (2x² – 3x + 1)
= 4x³ – 6x² + 2x – 10x² + 15x– 5
= 4x³ – 16x² + 17x – 5
The coefficient of x² in the polynomial (2x – 5) (2x² – 3x + 1) = – 16

Q.5. Classify the following as a constant, linear, quadratic and cubic polynomials:
(i) 2 – x² + x³

Powers of x = 2, and 3 respectively.
Highest power of the variable x in the given expression = 3 Hence, degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.

(ii) 3x³

Power of x = 3.
Highest power of the variable x in the given expression = 3
Hence, the degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.

(iii) 5t – √7

Power of t = 1.
Highest power of the variable t in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.

(iv) 4 – 5y²

Power of y = 2.
Highest power of the variable y in the given expression = 2 Hence, degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(v) 3

There is no variable in the given expression.
Let us assume that x is the variable in the given expression. 3 can be written as 3x0.
i.e., 3 = x⁰ Power of x = 0.
Highest power of the variable x in the given expression = 0 Hence, degree of the polynomial = 0
Since it is a polynomial of the degree 0, it is a constant polynomial.

(vi) 2 + x

Power of x = 1.
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.

(vii) y³ – y

Powers of y = 3 and 1, respectively.
Highest power of the variable x in the given expression = 3 Hence, degree of the polynomial = 3
Since it is a polynomial of the degree 3, it is a cubic polynomial.

(viii) 1 + x + x²

Powers of x = 1 and 2, respectively.
Highest power of the variable x in the given expression = 2 Hence, degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(ix) t²

Power of t = 2.
Highest power of the variable t in the given expression = 2
Hence, the degree of the polynomial = 2
Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(x) √2x – 1

Power of x = 1.
Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial = 1
Since it is a polynomial of the degree 1, it is a linear polynomial.

Q.6. Give an example of a polynomial, which is:
(i) monomial of degree 1

Monomial = an algebraic expression that contains one term
An example of a polynomial, which is a monomial of degree 1 = 2t

(ii) binomial of degree 20

Binomial = an algebraic expression that contains two terms
An example of a polynomial, which is a binomial of degree 20 = x²⁰ + 5

(iii) trinomial of degree 2

Trinomial = an algebraic expression that contains three terms
An example of a polynomial, which is a trinomial of degree 2 = y² + 3y + 11

Q.7. Find the value of the polynomial 3x³ – 4x² + 7 x – 5, when x = 3 and also when x = –3.

Given that,

p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5 According to the question, When x = 3,
p(x) = p(3)
p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5 Substituting x = 3,
p(3)= 3(3)³ – 4(3)² + 7(3) – 5
p(3) = 3(3)³ – 4(3)² + 7(3) – 5
= 3(27) – 4(9) + 21 – 5
= 81 – 36 + 21 – 5
= 102 – 41
= 61
When x = – 3, p(x) = p(– 3)
p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5 Substituting x = – 3,
p(– 3)= 3(– 3)³ – 4(– 3)² + 7(– 3) – 5
p(– 3) = 3(–3)³ – 4(–3)² + 7(–3) – 5
= 3(–27) – 4(9) – 21 – 5
= –81 – 36 – 21 – 5 = –143

Q.8. If p(x) = x² – 4 x + 3, evaluate : p(2) – p(–1) + p(1/2)

Given that,
p(𝑥) = 𝑥² – 4𝑥 + 3
According to the question, When x = 2,
p(x) = p(2)
p(𝑥) = 𝑥² – 4𝑥 + 3
Substituting x = 2,
p(2) = (2)² – 4(2) + 3
= 4 – 8 + 3
= – 4 + 3
= – 1
When x = – 1, p(x) = p(– 1) p(𝑥) = 𝑥² – 4𝑥 + 3
Substituting x = – 1,
p(– 1) = (– 1)² – 4(– 1) + 3
= 1 + 4 + 3
= 8
When x = 1/2 , p(x) = p(1/2)
p(𝑥) = 𝑥² – 4𝑥 + 3
Substituting x = 1/2,
p(1/2) = (1/2)² – 4(1/2) + 3
= 1/4 – 2 + 3
= 1/4 + 1
= 5/4
Now,
p(2)− p(−1) + p(1/2) = – 1 – 8 + (5/4)
= – 9 + (5/4)
= ( – 36 + 5)/4
= – 31/4

Q.9. Find p(0), p(1), p(–2) for the following polynomials:
(i) p(x) = 10x – 4 x² – 3

According to the question, p(x) = 10𝑥−4𝑥2 –3
When x = 0, p(x) = p(0)
Substituting x = 0,
p(0) = 10(0)−4(0)2 –3
= 0 – 0 – 3
= – 3
When x = 1, p(x) = p(1)
Substituting x = 1,
p(1) = 10(1) − 4(1)² –3
= 10 – 4 – 3
= 6 – 3
= 3
When x = – 2, p(x) = p(– 2)
Substituting x = – 2,
p(– 2) = 10(– 2) − 4(– 2)² –3
= – 20 – 16 – 3
= – 36 – 3
= – 39

(ii) p(y) = (y + 2) (y – 2)

According to the question, p(𝑦) = (y + 2) (y – 2)
When y = 0, p(y) = p(0)
Substituting y = 0, p(0) = (0 + 2) (0 – 2)
= (2)(– 2)
= – 4
When y = 1, p(y) = p(1)
Substituting y = 1, p(1)=(1 + 2) (1 – 2)
=(3) (– 1)
= – 3
When y = – 2, p(y) = p(– 2)
Substituting y = – 2,
p(– 2) =(– 2 + 2) (– 2 – 2)
= (0) (– 4)
= 0

Q.10. Verify whether the following are true or false:
(i) –3 is a zero of x – 3

False
Zero of x – 3 is given by, x – 3 = 0
⇒ x=3

(ii) – 1/3 is a zero of 3x + 1

True
Zero of 3x + 1 is given by, 3x + 1 = 0
⇒ 3x = – 1
⇒ x = – 1/3

(iii) – 4/5 is a zero of 4 –5y

False
Zero of 4 – 5y is given by, 4 – 5y =0
⇒ – 5y = – 4
⇒ y = 4/5

(iv) 0 and 2 are the zeroes of t² – 2t

True
Zeros of t2 – 2t is given by, t2 – 2t = t(t – 2) = 0
⇒ t = 0 or 2

(v) –3 is a zero of y² + y – 6

Zero of y² + y – 6 is given by, y² + y – 6 = 0
⇒ y² + 3x – 2x – 6 = 0
⇒ y(y + 3) – 2(x + 3) = 0
⇒ (y – 2) (y + 3) = 0
⇒ y = 2 or – 3

Q.11. Find the zeroes of the polynomial in each of the following :
(i) p(x) = x – 4 

Zero of the polynomial p(x) ⇒ p(x) = 0
P(x) = 0
⇒ x – 4 = 0
⇒ x = 4
Therefore, the zero of the polynomial is 4.

(ii) g(x) = 3 – 6x 

Zero of the polynomial g(x) ⇒ g(x) = 0 g(x) = 0
⇒3 – 6x = 0
⇒ x = 3/6 =1/2
Therefore, the zero of the polynomial is 1/2

(iii) q(x) = 2x –7

Zero of the polynomial q(x) ⇒ q(x) = 0 q(x) = 0
⇒2x – 7 = 0
⇒ x = 7/2
Therefore, the zero of the polynomial is 7/2

(iv) h(y) = 2y

Zero of the polynomial h(y) ⇒ h(y) = 0 h(y) = 0
⇒ 2y = 0
⇒ y = 0
Therefore, the zero of the polynomial is 0

Q.12. Find the zeroes of the polynomial : p(x) = (x – 2)² – ( x + 2)²

We know that,
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)² − (x + 2⁾² = 0
Expanding using the identity, a² – b² = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
⇒ – 8 x = 0
Therefore, the zero of the polynomial = 0

Q.13. By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x⁴ + 1; x –1

Performing the long division method, we get,
Hence, from the above long division method, we get, Quotient = x³ + x² + x + 1
Remainder = 2.

Q.14. By Remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) p(x) = x³ – 2x² – 4 x – 1, g(x) = x + 1

Let g(x) = x + 1
x + 1 = 0
x = –1
p(x) = x³ – 2x² – 4x – 1
p(–1) = (–1)³ – 2(– 1)² – 4(–1) – 1
= –1 – 2 × 1 + 4 – 1
= – 4 – 4 = 0
∴ Remainder = 0.

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

p(x) = x³ – 3x² + 4x + 50, g(x) = x – 3
Let g(x) = x – 3
x – 3 = 0
x = 3
p(3) = 3³ – 3(3²) + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
∴ Remainder = 62.

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

Let us denote the given polynomials as
p(x) = 4x² - 12x³ - 12x² + 14x - 3,
g(x) = 2x - 1
⇒ g(x) =
We have to find the remainder when f(x) is divided by g(x).
By the remainder theorem, when f(x) is divided by g(x) the remainder is

= 1/2 - 3 + 7 - 3
= 3/2
Now we will calculate the remainder by actual division

So the remainder by actual division is 3/2 .

(iv) p(x) = x³ – 6x² + 2x – 4, g(x) = 1 – 3/2 x

Given, p(x) = x³ – 6x² + 2x – 4 and g(x) = 1 - 3/2x
Here, zero of g(x) is 2/3.
When we divide p(x) by g(x) using remainder theorem, we get the remainder p (2/3).

Hence, remainder is -136/27

Q.15. Check whether p(x) is a multiple of g(x) or not:
(i) p(x) = x³ – 5x² + 4 x – 3, g(x) = x – 2

According to the question, g(x) = x – 2,
Then, zero of g(x), g(x) = 0
x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) =(2)³ – 5(2)² + 4(2) – 3
= 8 – 20 + 8 – 3
= – 7 ≠ 0
Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0.

(ii) p(x) = 2x³ – 11x² – 4 x + 5,    g (x) = 2x + 1

According to the question, g(x)= 2𝑥 + 1
Then, zero of g(x), g(x) = 0
2x + 1 = 0
2x = – 1
x = –1/2
Therefore, zero of g(x) = – 1/2
So, substituting the value of x in p(x), we get,
p(–1/2) = 2 × ( – 1/2 )3 – 11 × ( – 1/2 )2 – 4 × ( – 1/2) + 5
= – 1/4 - 11/4 + 7
= 16/4
= 4 ≠ 0
Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0.

Q.16. Show that: (i) x + 3 is a factor of 69 + 11x – x² + x³.

According to the question,
Let p(x) = 69 + 11x − x² + x³ and
g(x) = x + 3 g(x) = x + 3
zero of g(x) ⇒ g(x) = 0 x + 3 = 0
x = – 3
Therefore, zero of g(x) = – 3
So, substituting the value of x in p(x),
we get, p( – 3) = 69 + 11(– 3) –(–3)² + (– 3)³
= 69 – 69
= 0
Since, the remainder = zero, We can say that,
g(x) = x + 3 is factor of p(x) = 69 + 11x − x² + x³

(ii) 2x – 3 is a factor of x + 2x³ – 9x² + 12.

Let p(x) = 2x³ - 9x² + x + 12
We have to show that, 2x - 3 is a factor of p(x).

= (81 – 81)/4
= 0
Hence, (2x – 3) is a factor of p(x).

Q.17. Determine which of the following polynomials has x – 2 a factor:
(i) 3x² + 6x – 24

According to the question,
Let p(x) = 3𝑥² + 6𝑥 − 24 and g(x) = x – 2 g(x) = x – 2
zero of g(x) ⇒ g(x) = 0 x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) = 3(2)² + 6 (2) – 24
= 12 + 12 – 24
= 0
Since, the remainder = zero, We can say that,
g(x) = x – 2 is factor of p(x) = 3𝑥² + 6𝑥 − 24

(ii) 4x² + x – 2

According to the question,
Let p(x) = 4𝑥² + 𝑥 − 2 and g(x) = x – 2 g(x) = x – 2
zero of g(x) ⇒ g(x) = 0 x – 2 = 0
x = 2
Therefore, zero of g(x) = 2
So, substituting the value of x in p(x), we get, p(2) = 4(2)² + 2−2
= 16 ≠ 0
Since, the remainder = zero, We can say that,
g(x) = x – 2 is factor of p(x) = 4𝑥² + 𝑥 − 2

Q.18. Show that p – 1 is a factor of p¹⁰ – 1 and also of p¹¹ – 1.

Let f(p) = p¹⁰ – 1 and g(p) = p¹¹ – 1
Putting p = 1 in f(p), we get
f(1) = 1¹⁰ − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p¹⁰ – 1)
Now, putting p = 1 in g(p), we get
g(1) = 1¹¹ − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p¹¹ – 1)

Q.19. For what value of m is x³ – 2mx² + 16 divisible by x + 2 ?

According to the question,
Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2 g(x) = 0
⟹ x + 2 = 0
⟹ x = – 2
Therefore, zero of g(x) = – 2 We know that,
According to factor theorem,
if p(x) is divisible by g(x), then the remainder p(−2) should be zero. So, substituting the value of x in p(x), we get,
p( – 2) = 0
⟹ ( – 2)³ – 2m( – 2)² + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
⟹ 8m = 8
⟹ m = 1

Q.20. If x + 2α is a factor of x⁵ – 4 α²x³ + 2 x + 2α + 3, find α.

According to the question,
Let p(x) = x⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a g(x) = 0
⟹ x + 2a = 0
⟹ x = – 2a
Therefore, zero of g(x) = – 2a We know that,
According to the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0 So, substituting the value of x in p(x), we get,
p ( – 2a) = (– 2a)⁵ – 4a²(– 2a)3 + 2(– 2a) + 2a + 3 = 0
⟹ – 32a⁵ + 32a⁵ – 2a + 3 = 0
⟹ – 2a = – 3
⟹ a = 3/2

Q.21. Find the value of m so that 2x – 1 be a factor of 8x⁴ + 4x³ – 16x² + 10x + m.

Let p(x) = 8x⁴ + 4x³ – 16x² + 10x + m
Since, 2x - 1 is a factor of p(x), then put p(1/2) = 0

⇒ 1 + 1 + m = 0
∴ m = – 2
Hence, the value of m is – 2.

Q.22. If x + 1 is a factor of ax³ + x² – 2 x + 4 a – 9, find the value of a.

Let p(x) = ax³ + x² – 2x + 4a – 9
Since, x + 1 is a factor of p(x), then put p(– 1) = 0
∴ a(-1)³ + (-1)² - 2(-1) + 4a - 9 = 0
⇒ -a + 1 + 2 + 4a - 9 = 0
⇒ 3a - 6 = 0
⇒ 3a = 6
⇒ a = 6/3 = 2
Hence, the value of a is 2.

Q.23. Factorise:

(i) x2 + 9 x + 18

x² + 9x + 18 = x² + 3x + 6x + 18
= x(x + 3) + 6(x + 3)
= (x + 3)(x + 6)

(ii) 6x² + 7 x – 3

6x² + 7x − 3 = 6x² + 9x − 2x − 3
= 3x(2x + 3) − 1(2x+3)
= (3x − 1)(2x + 3)

(iii) 2x² – 7 x – 15

2x² − 7x − 15 = 2x² − 10x + 3x − 15
= 2x(x − 5) + 3(x − 5)
= (x − 5)(2x + 3)

(iv) 84 – 2r – 2 r²

84 − 2r − 2r² = −(2r² + 2r − 84)
= −(2r² + 14r + 12r − 84)
= −(2r(r + 7) − 12(r + 7))
= −(r + 7)(2r − 12)

Q.24. Factorise: 

(i) 2x³ – 3x² – 17x + 30

2x³ − 3x² − 17x + 30
= 2x³ + 2x² − 12x − 5x² − 5x + 30
= 2x(x² + x − 6) − 5(x² + x - 6)
= (x² + x−6)(2x − 5)
= (x² + 3x − 2x − 6)(2x − 5)
= [x(x + 3) − 2(x + 3)](2x − 5)
= (x + 3)(x − 2)(2x − 5)

(ii) x³ – 6x² + 11x – 6

Let p(x) = x³ - 6x² + 11x - 6
By trial, we find that
p(1) = (1)³ - 6(1)² + 11(1) - 6 = 0
∴ By converse of factor theorem, (x - 1) is a factor of p(x).
Now, x³ - 6x² + 11x - 6
= x² (x - 1)- 5x (x - 1) + 6 (x - 1)
= (x - 1) (x² - 5x + 6)
= (x - 1) {x² - 2x - 3x + 6}
= (x - 1) {x(x - 2)-3 (x - 2)}
= (x - 1)(x - 2)(x - 3)

(iii) x³ + x² – 4x – 4

Let x + 1 = 0
∴ x = -1
On substituting value of x in the expression
∴ f(-1) = (-1)3 + (-1)2 - 4(-1) -4 = 0
Clearly x + 1 is a factor of
f(x) = x³ + x² - 4x - 4
∴ f(x) = (x + 1) (x² - 4)   ...(By actual division)
= (x + 1) (x - 2) (x + 2)

(iv) 3x³ – x² – 3x + 1

Let f(x) = 3x³ - x² - 3x + 1 be the given polynomial.
Now, putting  x = 1, we get
f(1) = 3(1)³ - (1)² - 3(1) + 1
= 3 - 1 - 3 + 1 = 0
Therefore, (x-1) is a factor of polynomial f(x).
Now,
f(x) = 3x²(x-1) + 2x(x-1) - 1(x - 1)
= (x - 1){3x² + 2x - 1}
= (x - 1){3x² + 2x - 1}
= (x - 1)(x + 1)(3x - 1)
Hence (x - 1), (x + 1) and (3x - 1 )are the factors of polynomial f(x).

Q.25. Using suitable identity, evaluate the following:
(i) 103³

= (100 + 3)³
= (100)3 + (3)³ + 3 x 100 x 3 x (100 + 3)
= 1000000 + 27 + 900 (103)
= 1000027 + 92700
= 1092727

(ii) 101 × 102

= (100 + 1) (100 + 2)
= (100) + 100(1 + 2) + 1 x 2
= 10000 + 300 + 2
= 10302

(iii) 999²

= (1000 - 1)²
= (1000)2 + (1)² - 2 x 1000 x 1
= 1000000 + 1 - 2000
= 998001

Q.26. Factorise the following: 

(i) 4x² + 20x + 25

4x² + 20x + 25 = (2x)² + 2 × 2x × 5 + (5)²
= (2x+5)²  ........[Using identity, a² + 2ab + b² = (a + b)²]

(ii) 9y² – 66yz + 121z²

Using (a−b)² = a² + b² −2ab
= (3y)² + (11z)² −2 × 3y × 11
= (3y−11z)²

(iii)

Using a² − b² = (a+b) (a−b)

Q.27. Factorise the following:
(i) 9x² – 12x + 3

⇒ 9x² − 3x − 9x + 3
⇒ 3x(3x − 1) − 3(3x −1)
⇒ (3x − 3)(3x − 1)
⇒ 3(x − 1)(3x − 1)

(ii) 9x² – 12x + 4

9x² – 12x + 4 = (3x)² - 2 × 3x × 2 + (2)²
= (3x-2)²   .....[Using identity, (a – b)² = a² – 2ab + b²]
= (3x-2) (3x-2)

Q.28. Expand the following : 
(i) (4a – b + 2 c)² 

We have,
(4a − b + 2c)²
=(4a)² +(−b)² +(2c)² + 2(4a)(−b) + 2(−b)(2c) + 2(2c)(4a)
[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
=16a² + b² + 4c² −8ab − 4ac + 16ca

(ii) (3a – 5b – c)²

(3a − 5b − c)²
= (3a)² + (−5b)² + (−c)² + 2(3a)² − 5b+2(−5b)(−c) + 2(−c)(3a)
[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
= 9a² + 25b² + c² − 30ab + 10bc − 6ca. 

(iii) (– x + 2y – 3z)²

We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12y² + 6zx

Q.29.  Factorise the following : 

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

= (3x)² + (2y)² + (−4z)² + 2(3x)(2y) + 2(2y)(−4z) + 2(−4z)(3x)
= (3x + 2y−4z)²

= (3x + 2y − 4z)(3x + 2y − 4z) 

(ii) 25x² + 16y² + 4z² – 40xy + 16yz – 20xz

25x² +16y² +4z² −40xy + 16yz − 20xz
= (−5x)² +(4y)² +(2z)² + 2(−5x)(4y) + 2(4y)(2z) + 2(−5x)(2z)
Suitable identities is (x + y + z)³ = x³ + y³ + z³ + 2xy + 2yz + 2xz
Therefore, (−5x)² +(4y)² +(2z)² + 2(−5x)(4y) + 2(4y)(2z) + 2(−5x)(2z)
= (−5x + 4y + 2z)²

(iii) 16x² + 4y² + 9z² – 16xy – 12yz + 24 xz

16x² + 4y² + 9z² −16xy −12yz + 24xz
Using identity,(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
16x² + 4y² + 9z² − 16xy − 12yz + 24xz
=(4x)² + (−2y)² + (3z)² + 2(4x)(−2y) + 2(−2y)(3z) + 2(3z)(4x)
=(4x − 2y + 3z)²

Q.30. If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².

Given, a + b + c = 9 and ab + bc + ca = 26.
We know, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
⇒(a + b + c)² = a² + b² + c² + 2(ab + bc + ca).
Putting the values here, we get,
(9)² = a² + b² + c² + 2(26)
⇒ 81 = a² + b² + c² + 52
⇒ a² + b² + c² = 81 − 52 = 29.
 Hence, option D is correct.

Q.31. Expand the following :

(i) (3a – 2b)³

= (3a)³ – (2b)³ – 3(3a)(2b) [3a – 2b]
= 27a³ – 8b^{3 –} 18ab (3a – 2b)
= 27a³ – 8b³ – 54a²b + 36ab²
= 27a³ – 54a²b + 36ab² – 8b³

(ii) (1/x + y/3)³

......[Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)]

Q.32. Factorise the following 

(i) 1 – 64a³ – 12a + 48a²

= (1)³ −(4a)³ − 3(1)² (4a) + 3(1)(4a)² 
Suitable identities is x³ − y³ - 3x² y + 3xy² = (x−y) 3
∴(1)³ −(4a)³ −3(1)² (4a) + 3(1)(4a)² 
= (1−4a)³

(ii) 

Q.33. Find the following products :
(i)

= (x/2)³ + (2y)³ [(a + b)(a² − ab + b²) = a³ + b³]
= x³/8 + 8y³

(ii) (x² – 1) (x⁴ + x² + 1)

Given (x² - 1) (x⁴ + x² + 1)
We shall use the identity  (a - b) (a² + ab + b²) = a³ - b³
We can rearrange the (x² - 1) (x⁴ + x² + 1) as
(x² − 1) [(x²)² + (x²) (1) + (1)²]
=(x²)³ - (1)³
=(x²) × (x²) × (x²) - (1) × (1) × (1)
= x⁶ - 1³
= x⁶ - 1
Hence the Product value of (x² - 1)(x⁴ + x² + 1) is  x⁶ - 1.

Q.34. Factorise :
(i) 1 + 64x³

We know, (a³ + b³) = a³ + a²b − a²b − ab² + ab² + b³ 
= (a³ + a²b) − (a²b + ab²) + (ab² + b³)
= a²(a + b) − ab(a + b) + b²(a + b)
= (a + b)(a² − ab + b²)
Now, given, 1 + 64x³
=1³ + (4x)³
= (1 + 4x)(1 − 4x(1) + (4x²))
= (1 + 4x)(1 − 4x + 16x²)

(ii) a³ –2√2b³

 (a³ - b³) = a³ - a²b + a²b − ab² + ab² - b³ 

= (a³ - a²b) + (a²b - ab²) + (ab² - b³)
= a²(a - b) + ab(a - b) + b²(a - b)
= (a - b)(a² + ab + b²)
Now, given a³ − 2√2b³
=(a − √2b)(a² + a(√2b) + (√2b)²)
= (a − √2b)(a² + √2ab + 2b²)

Q.35. Find the following product: (2x – y + 3z) (4x² + y² + 9z² + 2xy + 3yz – 6xz)

= (2x − y + 3z)(4x² + y² + 9z² + 2xy + 3yz − 6xz)
= (a + b + c)(a² + b² + c² - ab − bc − ac)
= a³ + b³ + c³−3abc
= (2x)³ + (−y)³ + (3z)³ − 3.(2x)(−y)(3z)
= 8x³ − y³ + 27z³ + 18xyz

Q.36. Factorise :

(i) a³ – 8b³ – 64c³ – 24abc

Here the given expression can be written as,
a³ − 8b³ − 64c³ −24abc = (a)³ + (−2b)³ + (−4c)³ −3(a)(−2b)(−4c)
Comparing with the given identity,
x³ + y³ + z³ −3xyz ≡ (x + y + z)(x² + y² + z² − xy − yz − xz)
We get factor as
= (a − 2b − 4c)[(a)² − (−2b)² + (−4c)² − (a)(−2b) − (−2b)(−4c) − (−4c)(a)]
= (a − 2b − 4c)(a² + 4b² + 16c² + 2ab − 8bc + 4ca)

(ii) 2√2a³ + 8b³ – 27c³ + 18√2 abc.

We know that,
a³  + b³ + c³ −3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Thus, 2√2a³ + 8b³ − 27c³ + 18√2 abc
Here, a = √2 a, b = 2b and c = −3c
= (√2a + 2b − 3c)[(√2a)² + (2b)² + (−3c)² − √2a(2b) − (2b)(−3c)−(−3c)(√2a)]
= (√2a + 2b − 3c)(2a² + 4b² + 9c² − 2√2ab + 6bc + 3√2ac)

Q.37. Without actually calculating the cubes, find the value of :

(i)

Let

We know that a³ + b³ + c3 = 3abc when a + b + c = 0

= - 5/12

(ii) (0.2)³ – (0.3)³ + (0.1)³

Given,
0.2³ − 0.3³ + 0.1³

= (0.1 × 2)³ − (0.1 × 3)³ + (0.1 × 1)³
= 0.1³(2³ − 3³ + 1³)
= 0.1³(−18) = −0.018

Q.38. Without finding the cubes, factorise
(x – 2y)³ + (2y – 3z)³ + (3z – x)³

We know that, a³ + b³ + c³ − 3abc = (a + b + c) (a² + b² + c² − ab − bc − ca)
Also, if a + b + c = 0,
then a³ + b³ + c³ = 3abc .....condition(1)
Here, we see that a + b + c = (x − 2y) + (2y − 3z) + (3z − x) = 0 

∴ Using condition (1), we get
⇒ (x − 2y)³ + (2y − 3z)³ + (3 z − x)³
= 3(x − 2y) (2y − 3z) (3z − x)

Q.39. Find the value of
(i) x³ + y³ – 12xy + 64, when x + y = – 4

We know that, a³ + b³ + c³ = (a + b + c) (a² + b² + c² − ab − bc − ca) + 3abc
If a + b + c = 0, then a³ + b³ + c³ = 3abc
Now, given x³  + y³ −12xy + 64 and
x + y = −4
=> x + y + 4 = 0
Here, a = x, b = y,  c = 4 and a + b + c = x + y + 4 = 0
Therefore x³ + y³ + 64 = 3xy(4)
= 12xyz
Now, x³ + y³ + 64 − 12xyz = 12xyz − 12xyz = 0

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6

x³ − 8y³ − 36xy − 216
= x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6)
= (x−2y−6)(x² + 4y² + 36 + 2xy − 12y + 6x)
= 0 × (x² + 4y² + 36 + 2xy − 12y + 6x) = 0

Q.40. Give possible expressions for the length and breadth of the rectangle whose area is given by 4a² + 4a –3.

Given

Area = 4a² + 4a − 3.

We know that

Area of rectangle = length × breadth 

So, to find the possible expressions for the length and breadth we have to factorise the given expression.
Using the method of splitting the middle term, 

4a + 4a−3

= 4a² + 6a − 2a − 3

= 2a(2a+3)−1(2a+3)

= (2a−1)(2a+3)
∴ length × breadth =(2a−1)(2a+3)
Hence, the possible expressions for the length and breadth of the rectangle are :
length = (2a−1) and breadth =(2a+3) or, length = (2a + 3) and breadth = (2a − 1).

Exercise 2.4

Q1. If the polynomials az³ + 4 z² + 3 z – 4 and z³ – 4 z + a leave the same remainder when divided by z – 3, find the value of a.

Solution: Let p₁(z) = az³ + 4z² + 3z – 4 and p²(z) = z³ – 4z + 0
When we divide p₁(z) by z – 3, then we get the remainder p,(3).
Now, p₁(3) = a(3)³ + 4(3)² + 3(3) – 4
= 27a + 36 + 9 – 4
= 27a + 41
When we divide p₂(z) by z – 3 then
we get the remainder p₂(3).
Now, p2(3) = (3)³ – 4(3) + a
= 27 – 12 + a
= 15 + a
Both the remainders are same.
p₁(3) = p₂(3)
27a + 41 = 15 + a
27a – a = 15 – 41 .
26a = 26a = – 1

Q2. The polynomial p(x) = x⁴ – 2x³ + 3x² – 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: Given, p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7
When we divide p(x) by x + 1, then we get the remainder p(– 1).
Now, p(– 1) = (– 1)⁴ – 2(– 1)³ + 3(– 1)² – a(– 1) + 3a – 7
= 1 + 2 + 3 + a + 3a – 7
= 4a – 1
p(– 1) = 19
⇒ 4a – 1 = 19
⇒ 4a = 20
∴ a = 5
∴ Required polynomial = x⁴ – 2x³ + 3x² – 5x + 3(5) – 7   .....[Put a = 5 on p(x)]
= x⁴ – 2x³ + 3x² – 5x + 15 – 7
= x⁴ – 2x³ + 3x² – 5x + 8
When we divide p(x) by x + 2, then we get the remainder p(– 2)
Now, p(– 2) = (– 2)⁴ – 2(– 2)³ + 3(– 2)² – 5(– 2) + 8
= 16 + 16 + 12 + 10 + 8
= 62
Hence, the value of a 5 and remainder is 62.

Q3. If both x – 2 and x – 1/2 are factors of px² + 5x + r, show that p = r.

Solution: As (x - 2)and  (x - 1/2)are the factors of the polynomial px² + 5x + r
i.e., f(2) = 0 and f(1/2) = 0
Now,
f(2) = p(2)² + 5(2) + r = 0
4p + r = -10  .....(1)
And

p/4 + 5/2 + r = 0
p + 10 + 4x = 0

p + 4x = -10     ........(2)
From equation (1) and (2), we get
4p + r = p + 4r
3p = 3x
p = r

Q4. Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

Solution: Let p(x) = 2x⁴ – 5x³ + 2x² – x + 2 firstly, factorise x² - 3x + 2.
Now, x² - 3x + 2 = x² - 2x - x + 2 [by splitting middle term]
= x(x-2)-1 (x-2)
= (x-1)(x-2)
Hence, 0 of x² - 3x + 2 are land 2.
We have to prove that, 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² - 3x + 2 i.e., to prove that 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
= 2x¹⁶ - 5x⁸ + 2x⁴ + 0
= 32 – 40 + 8
= 40 – 40
= 0
Hence, p(x) is divisible by x² - 3x + 2.

Q5. Simplify (2x – 5y)³ – (2x + 5y)³.

Solution: (2x -5y)³ – (2x + 5y)³ 
= [(2x)³ – (5y)³ – 3(2x)(5y)(2x – 5y)] -[(2x)³ + (5y)³ + 3(2x)(5y)(2x+5y)]
[using identity, (a – b)³ = a³ -b³ – 3ab  and (a + b)³ = a³ +b³ + 3ab]
= (2x)³ – (5y)³ – 30xy(2x – 5y) – (2x)³ – (5y)³ – 30xy (2x + 5y)
= -2 (5y)³ – 30xy(2x – 5y + 2x + 5y)
= -2 x 125y³ – 30xy(4x)
= -250y³ -120x²y

Q6. Multiply x² + 4y² + z² + 2xy + xz – 2yz by (– z + x – 2 y).

Solution: (x–2y–z)(x² + 4y² + z² + 2xy + xz − 2yz)
= (x − 2y − z)[(x)² + (−2y)² + (−z)² − (x)(−2y) − (−2y)(−z) − (x)(−z)]
= (x)³ + (−2y)³  +(−z)³ – 3(x)(−2y)(−z)
[Using the identity, a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)]
= x³ – 8y³ – z³ – 6xyz

Q7. If a, b, c are all non-zero and a + b + c = 0, prove that

Solution: To prove,
We know that, a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
= 0(a² + b² + c² – ab – bc – ca) [∵ a + b + c = 0 , given]
= 0
→ a³ + b³ + c³ = 3 abc
On dividing both sides by abc; we get,

Q8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25.

Solution: Given: a + b + c = 5 and ab + bc + ca = 10
We know that: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c)[a² + b² + c² – (ab + bc + ca)]
= 5{a² + b² + c² – (ab + bc + ca)} = 5(a² + b² + c² – 10)
Given: a + b + c = 5
Now, squaring both sides, get: (a + b + c)² = 5²
a² + b² + c² + 2(ab + bc + ca)
= 25 a² + b² + c² + 2 × 10
= 25 a² + b² + c² = 25 – 20
= 5
Now, a³ + b³ + c³ – 3abc = 5(a² + b² + c² – 10)
= 5 × (5 – 10)
= 5 × (– 5)
= – 25

Hence proved.

Q9. Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b ) ( b + c) (c + a).

Solution: To prove: (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a)
L.H.S = [(a + b + c)³ – a³] – (b³ + c³)
= (a + b + c – a)[(a + b + c)² + a² + a(a + b + c)] – [(b + c)(b² + c² – bc)]  .......[Using identity, a³ + b³ = (a + b)(a² + b² – ab) and a³ – b³ = (a – b)(a² + b² + ab)]
= (b + c)[a² + b² + c² + 2ab + 2bc + 2ca + a² + a² + ab + ac] – (b + c)(b² + c² – bc)
= (b + c)[b² + c² + 3a² + 3ab + 3ac – b² – c² + 3bc]
= (b + c)[3(a² + ab + ac + bc)]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)[(a + c)(a + b)]
= 3(a + b)(b + c)(c + a) = R.H.S
Hence proved.