# NCERT solution for class 9 maths polynomials ( Chapter 2) #### Solution for Exercise 2.1

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

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)$$\sqrt{2}x - 1$$

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.

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)$$3$$

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)$$7x^3$$

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

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)$$p(x) = (x - 1)(x + 1)$$

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

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.

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)$$5 + 2x$$

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.

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 $$3x^3 + 7x$$.

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)$$x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$$

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)$$p(x) = x^3 - 4x^2 + x + 6$$, $$g(x) = x - 3$$

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)$$p(x) = kx^2 - 3x + k$$

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)$$3x^2 - x - 4$$

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)$$2y^3 + y^2 - 2y - 1$$

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)$$(3 - 2x)(3 + 2x)$$

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

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)$$x^2 - y^2/100$$

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)$$[1/4a - 1/2b + 1]^2$$

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

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)$$(x - 2y/3)^3$$

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)$${998}^3$$

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)$$27p^3 - 1/216 - 9p^2/2 + p/4$$

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)$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$

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)$$64m^3 + 343n^3$$

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 $$27x^3 + y^3 + z^3 - 9xyz$$

$$27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$$
$$= (3x + y +z)((3x)^2 + y^2 + z^2 - 3xy - yz - z(3x)$$
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 $$x^3 + y^3 + z^3 - 3xyz = 1/2(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$$

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 $$x^3 + y^3 + z^3 = 3xyz$$.

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)$$(28)^3 + (-15)^3 + (-13)^3$$

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 $$35y^2 + 13y - 12$$

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 $$12ky^2 + 8ky - 20k$$

i)We have Volume of cuboid $$3x^2 - 12x = 3x(x - 4)$$
ii)We have Volume of cuboid $$12ky^2 + 8ky - 20k$$
$$= 12ky^2 + 20ky - 12ky - 20k$$
$$= 4ky(3y - 5) - 4k(3y - 5)$$
$$= (3y - 5)(4ky - 4k)$$
$$= (3y - 5)4k(y - 1)$$