NCERT solution for class 9 maths polynomials ( Chapter 2)

Get it on Google Play

Solution for Exercise 2.1

1.Which of the following expression are polynomials in one variable and which are not?
State reason for your answer.
i)\(4x^2 - 3x + 7\)
ii)\(y^2 + \sqrt{2}\)
iii)\(3\sqrt{t} + t\sqrt{2}\)
iv)\(y + 2/y\)
v)\(x^{10} + y^3 + t^{50}\)

Open in new tab link

Answer :

i) \(4x^2 - 3x + 7\) is a polynomail in one variable. As it is having only non-negative integral powers of x.

ii)\(y^2 + \sqrt{2}\) is also a polynomial in one variable. As it is having only non-negative integral power of y.

iii)\(3\sqrt{t} + t\sqrt{2}\) is not a polynomial. As the integral power of the variable is not a whole number.

iv)\(y + 2/y\) is not a polynomial. As it having a negative integral power of y.

v) \(x^{10} + y^3 + t^{50}\) is a polynomial with three variables.

2. Write the coefficients of \(x^2\) in each of the following :
i)\(2 + x^2 + x\)
ii)\(2 - x^2 + x^3\)
iii)\(\pi{x^2}/2 + x\)
iv)\(\sqrt{2}x - 1\)

Open in new tab link

Answer :

i) The coefficient of \(x^2\) in \(2 + x^2 + x\) is \(1\).

ii) The coefficient of \(x^2\) in \(2 - x^2 + x^3\) is \(-1\).

iii) The coefficient of \(x^2\) in \(\pi{x^2}/2 + x\) is \(\pi/2\).

iv) The coefficient of \(x^2\) in \(\sqrt{2}x - 1\) is \(0\).

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

Open in new tab link

Answer :

A binomial of degree 35 = \(4y^35 + 3\sqrt{2}\) and a monomial of degree 100 = \(9y^{100}\)

4. Write the degree of each of the following polynomials :
i)\(5x^3 + 4x^2 + 7x\)
ii)\(4 - y^2\)
iii)\(5t - \sqrt{7}\)
iv)\(3\)

Open in new tab link

Answer :

i)\(5x^3 + 4x^2 + 7x\) is a polynomial with highest power of variable 3.
Hence, the degree of given polynomial is 3.

ii)\(4 - y^2\) is a polynomial with highest power of variable 2.
Hence, the degree of given polynomial is 2.

iii)\(5t - \sqrt{7}\) is a polynomial with highest power of variable 1.
Hence, the degree of given polynomial is 1.

iv)\(3\) is a polynomial with highest power of variable 0.
Hence, the degree of given polynomial is 0.

5. Classify the following as linear, quadratic and cubic polynomials :
i)\(x^2 + x\)
ii)\(x - x^3\)
iii)\(y + y^2 + 4\)
iv)\(1 + x\)
v)\(3t\)
vi)\(r^2\)
vii)\(7x^3\)

Open in new tab link

Answer :

i) The degree of polynomial \(x^2 + x\) is 2. Hence, it is a quadratic polynomial.

ii) The degree of polynomial \(x - x^3\) is 3. Hence, it is a cubic polynomial.

iii) The degree of polynomial \(y + y^2 + 4\) is 2. Hence, it is a quadratic polynomial.

iv) The degree of polynomial \(1 + x\) is 1. Hence, it is a linear polynomial.

v) The degree of polynomial \(3t\) is 1. Hence, it is a linear polynomial.

vi) The degree of polynomial \(r^2\) is 2. Hence, it is a quadratic polynomial.

vii) The degree of polynomial \(7x^3\) is 3. Hence, it is a cubic polynomial.

Solution for Exercise 2.2

1.Find the value of the polynomial \(5x - 4x^2 + 3\) at :
i) x = 0
ii) x = -1
iii) x = 2

Open in new tab link

Answer :

Let f(x) = \(5x - 4x^2 + 3\)
i)The value of f(x) at x = 0 is :
= \(5(0) - 4(0)^2 + 3\)
= 3
f(0) = 3.

ii)The value of f(x) at x = -1 is :
= \(5(-1) - 4(-1)^2 + 3\)
= -5 - 4 + 3
= -6
f(-1) = -6.

iii)The value of f(x) at x = 2 is :
= \(5(2) - 4(2)^2 + 3\)
= 10 - 16 + 3
= -3
f(2) = -3.

2.Find p(0), p(1)and p(2) for each of the following polynomials :
i)\(p(y) = y^2 -y + 1\)
ii)\(p(t) = 2 + t + 2t^2 - t^3\)
iii)\(p(x) = x^3\)
iv)\(p(x) = (x - 1)(x + 1)\)

Open in new tab link

Answer :

i)\(p(y) = y^2 - y + 1\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 0^2 - 0 +1\)
Hence, p(0) = 1
Now, let us put x = 1,
Therefore, \(p(1) = 1^2 - 1 +1\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = 2^2 - 2 +1\)
i.e. \(p(2) = 4 - 2 +1\)
Hence, p(2) = 3

ii)\(p(t) = 2 + t + 2t^2 - t^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 2 + 0 + 2{0}^2 - {0}^3\)
Hence, p(0) = 2
Now, let us put x = 1,
Therefore, \(p(1) = 2 + 1 + 2{1}^2 - {1}^3\)
\(p(1) = 3 + 2 - 1\)
Hence, p(1) = 4
Now, let us put x = 2,
Therefore, \(p(2) = 2 + 2 + 2{2}^2 - {2}^3\)
i.e. \(p(2) = 4 + 8 - 8\)
Hence, p(2) = 4

iii)\(p(x) = x^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = {0}^3\)
Hence, p(0) = 0
Now, let us put x = 1,
Therefore, \(p(1) = {1}^3\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = {2}^3\)
Hence, p(2) = 8

iv)\(p(x) = (x - 1)(x + 1)\)
Firstly, let us put x = 0,
Therefore, \(p(0) = (0 - 1)(0 + 1)\)
Hence, p(0) = -1
Now, let us put x = 1,
Therefore, \(p(1) = (1 - 1)(1 + 1)\)
Hence, p(1) = 0
Now, let us put x = 2,
Therefore, \(p(2) = (2 - 1)(2 + 1)\)
i.e. \(p(2) = 1 × 3\)
Hence, p(2) = 3

3.Verify whether the following are zeroes of the polynomial, indicated against them.
i)\(p(x) = 3x + 1\), x = -1/3
ii)\(p(x) = 5x - \pi\), x = 4/5
iii)\(p(x) = x^2 - 1\), x = 1, -1
iv)\(p(x) = (x + 1)(x - 2)\), x = -1, 2
v)\(p(x) = x^2\), x = 0
vi)\(p(x) = lx + m\), x = -m/l
vii)\(p(x) = 3x^2 - 1\), \(x = -1/\sqrt{3}\), \(2/\sqrt{3}\)
viii)\(p(x) = 2x + 1\), x = 1/2

Open in new tab link

Answer :

i)\(p(x) = 3x + 1\)
To check, if x = -1/3 is zero of given polynomial then p(-1/3) should be equal to 0.
So, p(-1/3) = 3(-1/3) + 1 = -1 +1 = 0
Therefore, x = -1/3 is zero of given polynomial.

ii)\(p(x) = 5x - \pi\)
To check, if 4/5 is zero of given polynomial then p(4/5) should be equal to 0.
So, \(p(4/5) = 5(4/5) - \pi = 4 - \pi\)
Therefore, x = 4/5 is not a zero of given polynomial.

iii)\(p(x) = x^2 - 1\)
To check, if x = 1 is zero of given polynomial then p(1) should be equal to 0.
So, p(1) = \(1^2 - 1\) = 1 - 1 = 0
Therefore, x = 1 is a zero of given polynomial.
To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.
So, p(-1) = \({-1}^2 - 1\) = 1 - 1 = 0
Therefore, x = 1 is too, a zero of given polynomial.

iv)\(p(x) = (x + 1)(x - 2)\)
To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.
So, p(-1) = (-1 + 1)(-1 - 2) = 0 × -3 = 0
Therefore, x = -1 is zero of given polynomial.
To check, if x = 2 is zero of given polynomial then p(2) should be equal to 0.
So, p(2) = (2 + 1)(2 - 2) = 3 × 0 = 0
Therefore, x = 2 is too, a zero of given polynomial.

v)\(p(x) = x^2\)
To check, if x = 0 is zero of given polynomial then p(0) should be equal to 0.
So,\( p(0) = (0^3) = 0\)
Therefore, x = 0 is zero of given polynomial.

vi)\(p(x) = lx + m\)
To check, if x = -m/l is zero of given polynomial then p(-m/l) should be equal to 0.
So, p(-m/l) = l(-m/l) + m = -m + m = 0
Therefore, x = -m/l is zero of given polynomial.

vii)\(p(x) = 3x^2 - 1\)
To check, if\( x = -1/\sqrt{3}\) is zero of given polynomial then p(\(-1/\sqrt{3}\)) should be equal to 0.
So, p(\(-1/\sqrt{3}\)) = \(3(-1/\sqrt{3})^2 - 1\) = 1 - 1 = 0
Therefore,\( x = -1/\sqrt{3}\) is zero of given polynomial.
To check, if \(x = 2/\sqrt{3}\) is zero of given polynomial then p(\(2/\sqrt{3}\)) should be equal to 0.
So, p(\(2/\sqrt{3}\)) = \(3(2/\sqrt{3})^2 - 1\) = 3
Therefore, \(x = 2/\sqrt{3}\) is not a zero of given polynomial.

viii)\(p(x) = 2x + 1\)
To check, if x = 1/2 is zero of given polynomial then p(1/2) should be equal to 0.
So, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial.

4. Find the Zero of the polynomial in each of the following cases.
i)\(p(x) = x + 5\)
ii)\(p(x) = x - 5\)
iii)\(p(x) = 2x + 5\)
iv)\(p(x) = 3x - 2\)
v)\(p(x) = 3x\)
vi)\(p(x) = ax, a \ne 0\)
vii)\(p(x) = cx + d ,c \ne 0\), c,d, are real numbers.

Open in new tab link

Answer :

i)We have, \(p(x) = x + 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(x + 5 = 0\)
i.e. x = -5
Hence, -5 is the zero of given ploynomial.

ii)We have, \(p(x) = x - 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(x - 5 = 0\)
i.e. x = 5
Hence, 5 is the zero of given ploynomial.

iii)We have, \(p(x) = 2x + 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(2x + 5 = 0\)
i.e. x = -5/2
Hence, -5/2 is the zero of given ploynomial.

iv)We have, \(p(x) = 3x - 2\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(3x - 2 = 0\)
i.e. x = 2/3
Hence, 2/3 is the zero of given ploynomial.

v)We have, \(p(x) = 3x\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(3x = 0\)
i.e. x = 0
Hence, 0 is the zero of given ploynomial.

vi)We have, \(p(x) = ax\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(ax = 0\)
i.e.\( x = 0, a \ne 0\)
Hence, 0 is the zero of given ploynomial.

vii)We have, \(p(x) = cx + d\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(cx + d = 0\)
i.e.\( x = -d/c, c \ne 0\)
Hence, -d/c is the zero of given ploynomial.

Solution for Exercise 2.3

1. Find the remainder when \(x^3 + 3x^2 + 3x + 1\) is divided by
i)\(x + 1\)
ii)\(x - 1/2\)
iii)\(x\)
iv)\(x + \pi\)
v)\(5 + 2x\)

Open in new tab link

Answer :

Let \(p(x) = x^3 + 3x^2 + 3x + 1\)
i)The zero of \(x + 1\) is x = -1 i.e. (x + 1 = 0)
So, \(p(-1) = {-1}^3 + 3{-1}^2 + 3{-1} + 1\)
i.e. \(p(-1) = -1 + 3 - 3 +1\)
i.e.\(p(-1) = 0\)
Hence, By remainder theorem, required remainder = 0.

ii)The zero of \(x - 1/2\) is x = 1/2 i.e. (x - 1/2 = 0)
So, \(p(1/2) = {1/2}^3 + 3{1/2}^2 + 3{1/2} + 1\)
i.e. \(p(1/2) = 1/8 + 3/4 + 3/2 +1\)
i.e.\(p(1/2) = 27/8\)
Hence, By remainder theorem, required remainder = 27/8.

iii)The zero of \(x\) is x = 0 i.e. (x = 0)
So, \(p(0) = {0}^3 + 3{0}^2 + 3{0} + 1\)
i.e. \(p(-1) = 1\)
Hence, By remainder theorem, required remainder = 1.

iv)The zero of \(x + \pi\) is \(x = -\pi\) i.e. \((x + \pi = 0)\)
So, \(p(-\pi) = {-\pi}^3 + 3{-\pi}^2 + 3{-\pi} + 1\)
i.e. \(p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1\)
Hence, By remainder theorem, required remainder = \(-\pi^3 + 3\pi^2 - 3\pi + 1\).

v)The zero of \(5 + 2x\) is x = -5/2 i.e. (5 + 2x = 0)
So, \(p(-5/2) = {-5/2}^3 + 3{-5/2}^2 + 3{-5/2} + 1\)
i.e. \(p(-5/2) = -125/8 + 75/4 - 15/2 + 1\)
i.e. \(p(-5/2) = -27/8\)
Hence, By remainder theorem, required remainder = -27/8.

2. Find the remainder when \(x^3 - ax^2 + 6x - a\) is divided by x - a.

Open in new tab link

Answer :

Let \(p(x) = x^3 - ax^2 + 6x - a\)
The zero of \(x - a\) is x = a i.e. (x - a = 0)
So, \(p(a) = {a}^3 - a{a}^2 + 6{a} - a\)
i.e. \(p(a) = a^3 - a^3 + 6a - a\)
i.e.\(p(a) = 5a\)
Hence, By remainder theorem, required remainder = 5a.

3. Check whether 7 + 3x is a factor of \(3x^3 + 7x\).

Open in new tab link

Answer :

Let \(p(x) = 3x^3 + 7x\)
To check, whether 7 + 3x is a factor of \(3x^3 + 7x\).By factor theorem, \(7 + 3x = 0\) i.e. x = -7/3
So, \(p(-7/3) = 3(-7/3)^3 + 7(-7/3)\)
i.e. \(p(-7/3) = - 3×(343/27) - 49/3\)
i.e.\(p(-7/3) = -1470/27\)
i.e.\(p(-7/3) = -490/9\)
Hence, 7 + 3x is not the factor of \(3x^3 + 7x\) since the remainder is not zero.

Solution for Exercise 2.4

1.Determine which of the following polynomials has a factor :
i)\(x^3 + x^2 + x + 1\)
ii)\(x^4 + x^3 + x^2 + x + 1\)
iii)\(x^4 + 3x^3 + 3x^2 + x + 1\)
iv)\(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)

Open in new tab link

Answer :

The zero of x + 1 is –1.
i)Let \(p(x) = x^3 + x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^3 + x^2 + x + 1\).
By factor theorem,
\(p(–1) = {–1}^3 + {–1}^2 + {-1} + 1\)
i.e. \(p(–1) = –1 + 1 -1 + 1\)
i.e.\(p(-1) = 0\)
Hence, x + 1 is the factor of \(x^3 + x^2 + x + 1\)

ii)Let \(p(x) = x^4 + x^3 + x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^4 + x^3 + x^2 + x + 1\).
By factor theorem,
\(p(–1) = {-1}^4 + {–1}^3 + {–1}^2 + {-1} + 1\)
i.e. \(p(–1) = 1 – 1 + 1 - 1 + 1\)
i.e.\(p(-1) = 1\)
Hence, x + 1 is not the factor of \(x^4 + x^3 + x^2 + x + 1\)

iii)Let \(p(x) = x^4 + 3x^3 + 3x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^4 + 3x^3 + 3x^2 + x + 1\).
By factor theorem,
\(p(–1) = {-1}^4 + 3{–1}^3 + 3{–1}^2 + {-1} + 1\)
i.e. \(p(–1) = 1 – 3 + 3 - 1 + 1\)
i.e.\(p(-1) = 1\)
Hence, x + 1 is not the factor of \(x^4 + 3x^3 + 3x^2 + x + 1\)

iv)Let \(p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)
So, To check, whether x + 1 is a factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\).
By factor theorem,
\(p(–1) = {–1}^3 - {–1}^2 - (2 + \sqrt{2}){-1} + \sqrt{2}\)
i.e. \(p(–1) = – 1 - 1 + 2 + \sqrt{2} + \sqrt{2}\)
i.e.\(p(-1) = 2\sqrt{2}\)
Hence, x + 1 is not the factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)

2.Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
i) \(p(x) = 2x^3 + x^2 - 2x - 1\), \(g(x) = x + 1\)
ii)\(p(x) = x^3 + 3x^2 + 3x + 1\), \(g(x) = x + 2\)
iii)\(p(x) = x^3 - 4x^2 + x + 6\), \(g(x) = x - 3\)

Open in new tab link

Answer :

i)The zero of g(x) = x + 1 is –1.
Let \(p(x) = 2x^3 + x^2 - 2x - 1\)
So, To check, whether x + 1 is a factor of \(2x^3 + x^2 - 2x - 1\).
By factor theorem,
\(p(–1) = 2{–1}^3 + {–1}^2 - 2{-1} - 1\)
i.e. \(p(–1) = –2 + 1 + 2 - 1\)
i.e.\(p(-1) = 0\)
Hence, g(x) = x + 1 is the factor of \(2x^3 + x^2 - 2x - 1\)

ii)The zero of \(g(x) = x + 2\) is –2.
Let \(p(x) = x^3 + 3x^2 + 3x + 1\)
So, To check, whether x + 2 is a factor of \(x^3 + 3x^2 + 3x + 1\).
By factor theorem,
\(p(–2) = {–2}^3 + 3{–2}^2 + 3{–2} + 1\)
i.e. \(p(–2) = –8 + 12 - 6 + 1\)
i.e.\(p(-2) = -1\)
Hence, \(g(x) = x + 2\) is not the factor of \(x^3 + 3x^2 + 3x + 1\)

iii)The zero of \(g(x) = x - 3\) is 3.
Let \(p(x) = x^3 - 4x^2 + x + 6\)
So, To check, whether x - 3 is a factor of \(p(x) = x^3 - 4x^2 + x + 6\).
By factor theorem,
\(p(3) = {3}^3 - 4{3}^2 + {3} + 6\)
i.e. \(p(3) = 27 - 36 + 3 + 6\)
i.e.\(p(3) = 0\)
Hence, \(g(x) = x - 3\) is the factor of \(p(x) = x^3 - 4x^2 + x + 6\)

3.Find the value of k, if x – 1 is a factor of p(x) in each of the following cases :
i)\(p(x) = x^2 + x + k\)
ii)\(p(x) = 2x^2 + kx + \sqrt{2}\)
iii)\(p(x) = kx^2 - \sqrt{2}x + 1\)
iv)\(p(x) = kx^2 - 3x + k\)

Open in new tab link

Answer :

The zero of x - 1 is 1.
i)So, as x - 1 is a factor of \(p(x) = x^2 + x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = (1)^2 + 1 + k\)
i.e.\(0 = 1 + 1 + k\)
i.e.\(0 = 2 + k\)
i.e.\(k = -2\)

ii)So, as x - 1 is a factor of \(p(x) = 2x^2 + kx + \sqrt{2}\), therefore, \(p(1) = 0\)
i.e. \(0 = 2(1)^2 + k(1) + \sqrt{2}\)
i.e.\(0 = 2 + k + \sqrt{2}\)
i.e.\(k = -2 - \sqrt{2}\)

iii)So, as x - 1 is a factor of \(p(x) = kx^2 - \sqrt{2}x + 1\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - ( \sqrt{2})1 + 1\)
i.e.\(0 = k - \sqrt{2} + 1\)
i.e.\(k = \sqrt{2} - 1\)

iv)So, as x - 1 is a factor of \(p(x) = kx^2 - 3x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - 3(1) + k\)
i.e.\(0 = k - 3 + k\)
i.e.\(3 = 2k\)
i.e.\(k = 3/2\)

4.Factorize :
i)\(12x^2 - 7x + 1\)
ii)\(2x^2 + 7x + 3\)
iii)\(6x^2 + 5x - 6\)
iv)\(3x^2 - x - 4\)

Open in new tab link

Answer :

i)\(12x^2 - 7x + 1 = 12x^2 - 4x -3x + 1\)...(by splitting middle term)
i.e. \(= 4x(3x - 1) - 1(3x - 1)\)
i.e.\(=(3x - 1)(4x - 1)\)

ii)\(2x^2 + 7x + 3 = 2x^2 + 6x + x + 3\)...(by splitting middle term)
i.e. \(= 2x(x + 3) + 1(x + 3)\)
i.e.\(=(x + 3)(2x + 1)\)

iii)\(6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6\)...(by splitting middle term)
i.e. \(= 3x(2x + 3) - 2(2x + 3)\)
i.e.\(=(2x + 3)(3x - 2)\)

iv)\(3x^2 - x - 4 = 3x^2 - 4x + 3x - 4\)...(by splitting middle term)
i.e. \(= x(3x - 4) + 1(3x - 4)\)
i.e.\(=(3x - 4)(x + 1)\)

5.Factorize :
i)\(x^3 - 2x^2 - x + 2\)
ii)\(x^3 - 3x^2 -9x - 5\)
iii)\(x^3 + 13x^2 + 32x + 20\)
iv)\(2y^3 + y^2 - 2y - 1\)

Open in new tab link

Answer :

i)Let \(p(x) = x^3 - 2x^2 - x + 2\)
Here, constant term is 2. So, finding the factors of +2, we get, ±1 and ±2.
Now, \(p(1) = (1)^3 - 2(1)^2 - (1) + 2\)
\(= 1 - 2 - 1 + 2\)
= 0
Therefore, by trial method, we find that, p(1) = 0.
Hence, x - 1 is a factor of p(x).
So, \(x^3 - 2x^2 - x + 2 = x^3 - x^2 - x^2 + x - 2x + 2\)
\(= x^2(x - 1) -x(x - 1) - 2(x - 1)\)
\(= (x - 1)(x^2 - x - 2)\)
now, splitting the middle term,
\(= (x - 1)(x^2 - 2x + x - 2)\)
\(= (x - 1)[x(x - 2) + 1(x - 2)]\)
\(= (x - 1)(x - 2)(x + 1)\)

ii)Let \(p(x) = x^3 - 3x^2 -9x - 5\)
Here, constant term is -5. So, finding the factors of -5, we get, 1 and -5 or vice versa.
Now, \(p(5) = (5)^3 - 3(5)^2 - 9(5) - 5\)
\(= 125 - 75 - 45 - 5\)
= 0
Therefore, by trial method, we find that, p(5) = 0.
Hence, x - 5 is a factor of p(x).
So, \(x^3 - 3x^2 -9x - 5 = x^3 - 5x^2 + 2x^2 - 10x + x - 5\)
\(= x^2(x - 5) + 2x(x - 5) + 1(x - 5)\)
\(= (x - 5)(x^2 + 2x + 1)\)
now, splitting the middle term,
\(= (x - 5)(x^2 + x + x + 1)\)
\(= (x - 5)[x(x + 1) + 1(x + 1)]\)
\(= (x - 5)(x + 1)^2\)

iii)Let \(p(x) = x^3 + 13x^2 + 32x + 20\)
Here, constant term is +20. So, finding the factors of 20, we get, ±1, ±2, ±4 and ±5 or vice versa.
Now, \(p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20\)
\(= -1 + 13 - 32 + 20\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.
Hence, x + 1 is a factor of p(x).
So, \(x^3 + 13x^2 + 32x + 20 = x^3 + 2x^2 + 11x^2 + 22x + 10x + 20\)
\(= x^2(x + 2) + 11x(x + 2) + 10(x + 2)\)
\(= (x + 2)(x^2 + 11x + 10)\)
now, splitting the middle term,
\(= (x + 2)(x^2 + 10x + x + 10)\)
\(= (x + 2)[x(x + 10) + 1(x + 10)]\)
\(= (x + 2)(x + 1)(x + 10)\)

iv)Let \(p(x) = 2y^3 + y^2 - 2y - 1\)
Here, constant term is -1. So, finding the factors of -1, we get, -1.
Now, \(p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1\)
\(= -2 + 1 + 2 - 1\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.
Hence, x + 1 is a factor of p(x).
So, \(2y^3 + y^2 - 2y - 1 = 2y^3 + 2y^2 - y^2 - y - y -1\)
\(= 2y^2(y + 1) - y(x + 1) - 1(y + 1)\)
\(= (y + 1)(2y^2 - y - 1)\)
now, splitting the middle term,
\(= (y + 1)(2y^2 - 2y + y - 1)\)
\(= (y + 1)[2y(y - 1) + 1(y - 1)]\)
\(= (y - 1)(y + 1)(2y + 1)\)

Solution for Exercise 2.5

1.Use suitable identities to find the following products :
i)\((x + 4)(x + 10)\)
ii)\((x + 8)(x - 10)\)
iii)\((3x - 4)(3x - 5)\)
iv)\((y^2 + 3/2)(y^2 - 3/2)\)
v)\((3 - 2x)(3 + 2x)\)

Open in new tab link

Answer :

i)\((x + 4)(x + 10)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((x + 4)(x + 10) = x^2 + (4 + 10)x + (4)(10)\)
\(= x^2 + 14x + 40\)

ii)\((x + 8)(x - 10)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((x + 8)(x - 10) = x^2 + (8 - 10)x + (8)(-10)\)
\(= x^2 - 2x - 80\)

iii)\((3x + 4)(3x - 5)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((3x + 4)(3x - 5) = (3x)^2 + (4 - 5)x + (4)(5)\)
\(= 9x^2 - x + 20\)

iv)\((y^2 + 3/2)(y^2 - 3/2)\)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((y^2 + 3/2)(y^2 - 3/2) = (y^2)^2 - (3/2)^2\)
\(= y^4 - 9/4\)

v)\((3 - 2x)(3 + 2x)\)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((3 - 2x)(3 + 2x) = (3)^2 - (2x)^2)\)
\(= 9 - 4x^2\)

2.Evaluate the following products without multiplying directly
i)103 × 107
ii)95 × 96
iii)104 × 96

Open in new tab link

Answer :

i)103 × 107 = (100 + 3)(100 + 7)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((100 + 3)(100 + 7) = {100}^2 + (3 + 7)100 + (3)(7)\)
\(= 10000 + 1000 + 21\)
\(= 11021\)

ii)95 × 96 = (100 - 5)(100 - 4)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((100 - 5)(100 - 4) = {100}^2 - (5 + 4)100 + (4)(5)\)
\(= 10000 - 900 + 20\)
\(= 9120\)

iii)104 × 96 = (100 + 4)(100 - 4)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((100 + 4)(100 - 4) = {100}^2 - {4}^2\)
\(= 10000 - 16\)
\(= 9984\)

3.Factorize the following using appropriate identities :
i)\(9x^2 + 6xy + y^2\)
ii)\(4y^2 - 4y + 1\)
iii)\(x^2 - y^2/100\)

Open in new tab link

Answer :

i)\(9x^2 + 6xy + y^2 = {3x}^2 + 2(3x)(y) + {y}^2\).....by using identity, \((a + b)^2 = a^2 + 2ab + b^2\)
= (3x + y)^2
Therefore, the factors are (3x + y)(3x + y).

ii)\(4y^2 - 4y + 1 = {2y}^2 - 2(2y)(1) + {1}^2\).....by using identity, \((a + b)^2 = a^2 + 2ab + b^2\)
= (2y - 1)^2
Therefore, the factors are (2y - 1)(2y - 1).

iii)\(x^2 - y^2/100 = (x + y/10)(x - y/10)\).....by using identity, \((a^2 - b^2) = (a + b)(a - b))\)
= (x + y/10)(x - y/10)
Therefore, the factors are (x + y/10)(x - y/10).

4.Factorize the following using appropriate 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)\([1/4a - 1/2b + 1]^2\)

Open in new tab link

Answer :

i)\((x + 2y + 4z)^2 = (x)^2 + (2y)^2 +(4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz\)

ii)\((2x - y + z)^2 = (2x)^2 + (-y)^2 + (z)^2 - 2(2x)(y) - 2(y)(z) + 2(2x)(z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz\)

iii)\((-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 +(2z)^2 - 2(2x)(3y) + 2(3y)(2z) - 2(2x)(2z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz\)

iv)\((3a - 7b - c)^2 = (3a)^2 + (-7b)^2 + (-c)^2 - 2(3a)(7b) + 2(-7b)(-c) - 2(3a)(c)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac\)

v)\((-2x + 5y - 3z)^2 = (-2x)^2 + (5y)^2 +(-3z)^2 - 2(2x)(5y) - 2(5y)(3z) + 2(2x)(3z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz\)

vi)\([1/4a - 1/2b + 1]^2 = (1/4a)^2 + (-1/2b)^2 + (1)^2 - 2(1/4a)(1/2b) - 2(-1/2b)(1) + 2(1/4a)(1)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= a^2/16 + b^2/4 + 1 - ab/4 - b + a/2\)

5.Factorize :
i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz\)
ii)\(2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz\)

Open in new tab link

Answer :

i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz \)
\(= (2x)^2 + (3y)^2 - (4z)^2 + 2(2x)(3y) - 2(3y)(4z) - 2(2x)(4z)\)
\(=(2x + 3y - 4z)^2\)

ii)\(2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz \)
= \(-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 - 2(\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) - 2(\sqrt{2}x)(2\sqrt{2}z)\)
\(=(-\sqrt{2}x + y + 2\sqrt{2}z)^2\)

6.Write the following cubes in expanded form.
i)\((2x + 1)^3\)
ii)\((2a - 3b)^3\)
iii)\([3x/2 + 1]^3\)
iv)\((x - 2y/3)^3\)

Open in new tab link

Answer :

i)\((2x + 1)^3 = (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)\)
by using identity\((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 8x^3 + 1 + 6x(2x + 1))\)
\(= 8x^3 + 1 + 12x^2 + 6x\)
\(= 8x^3 + 12x^2 + 6x + 1\)

ii)\((2a - 3b)^3 = (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 8a^3 - 27b^3 - 18ab(2a - 3b))\)
\(= 8a^3 - 27b^3 - 36a^2b + 54ab^2)\)
\(= 8a^3 - 36a^2b + 54ab^2 - 27b^3\)

iii)\([3x/2 + 1]^3 = (3x/2)^3 + (1)^3 + 3(3x/2)(1)(3x/2 + 1)\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 27x^3/8 + 1 + 9x/2(3x/2 + 1))\)
\(= 27x^3/8 + 1 + 27x^2/4 + 9x/2\)
\(= 27x^3/8 + 27x^2/4 + 9x/2 + 1\)

iv)\((x - 2y/3)^3 = (x)^3 - (2y/3)^3 - 3(x)(2y/3)(x - 2y/3)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= x^3 - 8y^3/27 - 2xy(x - 2y/3)\)
\(= x^3 - 8y^3/27 - 2x^2y + 4xy^2/3)\)
\(= x^3 - 2x^2y + 4xy^2/3 + - 8y^3/27\)

7.Evaluate the following using suitable identities.
i)\({99}^3\)
ii)\({102}^3\)
iii)\({998}^3\)

Open in new tab link

Answer :

i)\({99}^3 = (100 - 1)^3 = (100)^3 - (1)^3 - 3(100)(1)(100 - 1)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 1000000 - 1 - 300(99)\)
\(= 1000000 - 1 - 2700)\)
\(= 970299\)

ii)\({102}^3 = (100 + 2)^3 = (100)^3 + (2)^3 + 3(100)(2)(100 + 2)\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 1000000 + 8 + 600(102)\)
\(= 1000000 + 8 + 61200)\)
\(= 1061208\)

iii)\({998}^3 = (1000 - 2)^3 = (1000)^3 - (2)^3 - 3(1000)(2)(1000 - 2)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 1000000000 - 8 - 6000(998)\)
\(= 1000000000 - 1 - 5988000)\)
\(= 994011992\)

8.Factorize each of the following :
i)\(8a^3 + b^3 + 12a^2b + 6ab^2\)
ii)\(8a^3 - b^3 - 12a^2b + 6ab^2\)
iii)\(27 - 125a^3 - 135a + 225a^2\)
iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
v)\(27p^3 - 1/216 - 9p^2/2 + p/4\)

Open in new tab link

Answer :

i)\(8a^3 + b^3 + 12a^2b + 6ab^2 \)
i.e.,\(= (2a)^3 + b^3 +3(2a)(b)(2a + b))\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= (2a)^3 + b^3\)

ii)\(8a^3 - b^3 - 12a^2b + 6ab^2 \)
i.e.,\(= (2a)^3 - b^3 -3(2a)(b)(2a - b)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (2a)^3 - b^3\)

iii)\(27 - 125a^3 - 135a + 225a^2 \)
i.e.,\(= (3)^3 - (5a)^3 -3(3)(5a)(3 - 5a)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (3)^3 - 5a^3\)

iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
i.e.,\( = (4a)^3 - (3b)^3 -3(4a)(4b)(4a - 3b))\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (4a)^3 - (3b)^3\)

v)\((27p^3 - 1/216 - 9p^2/2 + p/4\)
i.e.,\( = (3p)^3 - (1/6)^3 -3(3p)(1/6)(3p - 1/6))\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (3p)^3 - {1/6}^3\)

9.Verify :
i)\(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
ii)\(x^3 - y^3 = (x - y)(x^2 + xy + y^2) \)

Open in new tab link

Answer :

i)We know that,
\((x + y)^3 = x^3 + y^3 + 3xy(x + y)\)
\(x^3 + y^3 = (x + y)^3 - 3xy(x + y)\)
\(= (x + y)[(x + y)^2 - 3xy]\)
\(= (x + y)[(x^2 + y^2 +2xy - 3xy]\)
\(= (x + y)[(x^2 + y^2 - xy]\)
\(= R.H.S\)
Hence, Proved.

ii)We know that,
\((x - y)^3 = x^3 - y^3 - 3xy(x - y)\)
\(x^3 - y^3 = (x - y)^3 + 3xy(x - y)\)
\(= (x - y)[(x - y)^2 + 3xy]\)
\(= (x - y)[(x^2 + y^2 - 2xy + 3xy]\)
\(= (x - y)[(x^2 + y^2 + xy]\)
\(= R.H.S\)
Hence, Proved.

10.Factorize each of the following :
i)\(27y^3 + 125z^3\)
ii)\(64m^3 + 343n^3\)

Open in new tab link

Answer :

i)\(27y^3 + 125z^3 = (3y)^3 + (5z)^3\)
\(= (3y + 5z)[((3y)^2 + (5y)^2 - (3y)(5z)]\)
\(= (3y + 5z)[(9y^2 - 15yz + 25z^2)]\)

ii)\(64m^3 + 343n^3 = (4m)^3 + (7n)^3\)
\(= (4m + 7n)[((4m)^2 + (7n)^2 - (4m)(7n)]\)
\(= (4m + 7n)[(16m^2 - 48mn + 49n^2)]\)

11.Factorize \(27x^3 + y^3 + z^3 - 9xyz\)

Open in new tab link

Answer :

\(27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)\)
\(= (3x + y +z)((3x)^2 + y^2 + z^2 - 3xy - yz - z(3x)\)
i.e., by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
We get,
\(= (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)\)
Therefore, \(27x^3 + y^3 + z^3 - 9xyz = (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)\)

12.Verify that \(x^3 + y^3 + z^3 - 3xyz = 1/2(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)

Open in new tab link

Answer :

We know that, \(x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 - xy - yz - zx]\)
\( = 1/2(x + y + z)[2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx]\)
\( = 1/2(x + y + z)[x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx]\)
\( = 1/2(x + y + z)[x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx]\)
\( = 1/2(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)

13.If show that x + y + z = 0, show that \(x^3 + y^3 + z^3 = 3xyz\).

Open in new tab link

Answer :

We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
As it is given that (x + y + z = 0)
i.e., \(= (0)(x^2 + y^2 + z^2 - xy - yz - zx)\)
Therefore, \(x^3 + y^3 + z^3 = 3xyz\)
Hence, proved.

14.Without actually calculating the cubes, find the value of each of the following :
i)\((-12)^3 + (5)^3 + (7))^3\)
ii)\((28)^3 + (-15)^3 + (-13)^3\)

Open in new tab link

Answer :

We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
Also, if x + y + z = 0 then, \(x^3 + y^3 + z^3 = 3xyz\)

i)Here, -12 + 7 + 5 = 0
So, \((-12)^3 + (5)^3 + (7))^3 = 3(-12)(7)(5)\)
\(= -1260\)

ii)Here, 28 - 15 - 13 = 0
So, \((28)^3 + (-15)^3 + (-13))^3 = 3(-13)(28)(-15)\)
\(= 16380\)

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 \(35y^2 + 13y - 12\)

Open in new tab link

Answer :

i)We have area of rectangle \(= 25a^2 - 35a + 12\)
\(= 25a^2 - 20a - 15a + 12\)
\(= 5a(5a - 4) - 3(5a - 4)\)
\(= (5a - 4)(5a - 3)\)
One possible answers : Length = 5a - 3, Breadth = 5a - 4.

ii)We have area of rectangle \(= 35y^2 + 13y - 12\)
\(= 35y^2 - 15y + 28y - 12\)
\(= 5y(7y - 3) + 4(7y - 3)\)
\(= (7y - 3)(5y + 4)\)
One possible answers : Length = 7y - 3, Breadth = 5y + 4.

16.What are the possible expressions for the dimensions of the cuboids whose volumes are given below ?
i)Volume \(3x^2 - 12x\)
ii)Volume \(12ky^2 + 8ky - 20k\)

Open in new tab link

Answer :

i)We have Volume of cuboid \(3x^2 - 12x = 3x(x - 4)\)
One possible expressions for the dimensions of the cuboid is 3, x and x-4.

ii)We have Volume of cuboid \(12ky^2 + 8ky - 20k\)
\(= 12ky^2 + 20ky - 12ky - 20k\)
\(= 4ky(3y - 5) - 4k(3y - 5)\)
\(= (3y - 5)(4ky - 4k)\)
\(= (3y - 5)4k(y - 1)\)
One possible expressions for the dimensions of the cuboid is 4k, 3y - 5 and y - 1.