##### Question 1:

Which of the following expressions are polynomials in one variable and which are not? State
reasons for your answer.

4x^{2} – 3x + 7

##### Answer:

4x^{2} – 3x + 7 is a polynomial in x because exponent of x in each of the terms is a
whole number.

##### Question 2:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

##### Answer:

##### Question 3:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

##### Answer:

##### Question 4:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

##### Answer:

##### Question 5:

Which of the following expressions are polynomials in one variable and which are not? State
reasons for your answer.

x^{10} + y^{3} + t^{50}

##### Answer:

x^{10} + y^{3} + t^{50} is a polynomial in three variables x, y and
t, because exponent of each variable is a whole number.

##### Question 6:

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

2 + x^{2} + x

##### Answer:

Coefficient of x^{2} in 2 + x^{2} + x is 1.

##### Question 7:

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

2 – x^{2} + x^{3}

##### Answer:

Coefficient of x^{2} in 2 – x^{2} + x^{3} is –1.

##### Question 8:

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

##### Answer:

##### Question 9:

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

##### Answer:

##### Question 10:

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

##### Answer:

Note: We can write some more polynomials with different coefficients.

##### Question 11:

Write the degree of each of the following polynomials:

5x^{3} + 4x^{2} + 7x

##### Answer:

Highest power term in p(x) is 5x^{3} and the exponent is 3. So, the degree is 3.

##### Question 12:

Write the degree of each of the following polynomials:

4 – y^{2}

##### Answer:

Highest power term in p(y) is – y^{2} and the exponent is 2. So, the degree is 2.

##### Question 13:

Write the degree of each of the following polynomials:

##### Answer:

Highest power term in f(t) is 5t and the exponent is 1. So, the degree is 1.

##### Question 14:

Write the degree of each of the following polynomials:

3

##### Answer:

The only term here is 3 which can be written as 3x^{0} and so the exponent is 0.
Therefore, the degree is 0.

##### Question 15:

Classify the following as linear, quadratic and cubic polynomials:

x^{2} + x

##### Answer:

The polynomial x^{2} + x is of degree 2. Therefore, it is a quadratic polynomial.

##### Question 16:

Classify the following as linear, quadratic and cubic polynomials:

x – x^{3}

##### Answer:

The polynomial x – x^{3} is of degree 3. Therefore, it is a cubic polynomial.

##### Question 17:

Classify the following as linear, quadratic and cubic polynomials:

y + y^{2} + 4

##### Answer:

The polynomial y + y^{2} + 4 is of degree 2. Therefore, it is a quadratic polynomial.

##### Question 18:

Classify the following as linear, quadratic and cubic polynomials:

1 + x

##### Answer:

Polynomial 1 + x is of degree 1. Therefore, it is a linear polynomial.

##### Question 19:

Classify the following as linear, quadratic and cubic polynomials:

3t

##### Answer:

3t is a polynomial of degree 1. Therefore, it is a linear polynomial.

##### Question 20:

Classify the following as linear, quadratic and cubic polynomials:

r^{2}

##### Answer:

r^{2} is a polynomial of degree 2. Therefore, it is a quadratic polynomial.

##### Question 21:

Write the degree of each of the following polynomials:

7x^{3}

##### Answer:

7x^{3} is a polynomial of degree 3. Therefore, it is a cubic polynomial.

##### Question 22:

Find the value of the polynomial 5x – 4x^{2} + 3 at

x = 0

##### Answer:

The value of polynomial

p(x) = 5x – 4x^{2} + 3 at x = 0 is given by

= 5(0) – 4(0)^{2} + 3 = 0 – 0 + 3 = 3

##### Question 23:

Find the value of the polynomial 5x – 4x^{2} + 3 at

x = –1

##### Answer:

The value of polynomial

p(x) = 5x – 4x^{2} + 3 at x = –1 is given by

= 5 (–1) – 4 (–1)^{2} + 3

= – 5 – 4 + 3 = –6

##### Question 24:

Find the value of the polynomial 5x – 4x^{2} + 3 at

x = 2

##### Answer:

The value of polynomial

p(x) = 5x – 4x^{2} + 3 at x = 2 is given by

= 5(2) – 4(2)^{2} + 3 = 10 – 4 × 4 + 3 = 10 – 16 + 3 = –3

##### Question 25:

Find p(0), p(1) and p(2) for each of the following polynomials:

p(y) = y^{2} – y + 1

##### Answer:

p(y) = y^{2} – y + 1

p(0) = (0)^{2} – 0 + 1

$\Rightarrow $ p(0) = 1

p(1) = (1)^{2} – 1 + 1

$\Rightarrow $ p(1) = 1 – 1 + 1

$\Rightarrow $ p(1) = 1

p(2) = (2)^{2}– 2 + 1

$\Rightarrow $ p(2) = 4 – 2 + 1

$\Rightarrow $ p(2) = 3

##### Question 26:

Find p(0), p(1) and p(2) for each of the following polynomials:

p(t) = 2 + t + 2t^{2} – t^{3}

##### Answer:

p(t) = 2 + t + 2t^{2} – t^{3}

p(0) = 2 + 0 + 2(0)^{2} – (0)^{3}

$\Rightarrow $ p(0) = 2 + 0 + 0 – 0

$\Rightarrow $ p(0) = 2

p(1) = 2 + 1 + 2(1)^{2} – (1)^{3}

$\Rightarrow $ p(1) = 2 + 1 + 2 – 1

$\Rightarrow $ p(1) = 4

p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

$\Rightarrow $ p(2) = 2 + 2 + 8 – 8

$\Rightarrow $ p(2) = 4

##### Question 27:

Find p(0), p(1) and p(2) for each of the following polynomials:

p(x) = x^{3}

##### Answer:

p(x) = x^{3}

p(0) = (0)3

$\Rightarrow $ p(0) = 0

p(1) = (1)^{3}

$\Rightarrow $ p(1) = 1

p(2) = (2)^{3}

$\Rightarrow $ p(2) = 8

##### Question 28:

Find p(0), p(1) and p(2) for each of the following polynomials:

p(x) = (x – 1) (x + 1).

##### Answer:

p(x) = (x – 1) (x + 1)

p(0) = (0 – 1) (0 + 1)

$\Rightarrow $ p(0) = (– 1) × (1)

$\Rightarrow $ p(0) = –1

p(1) = (1 – 1) (1 + 1)

$\Rightarrow $ p(1) = 0 × 2

$\Rightarrow $ p(1) = 0

p(2) = (2 – 1) (2 + 1)

$\Rightarrow $ p(2) = 1 × 3

$\Rightarrow $ p(2) = 3

##### Question 29:

Verify whether the following are zeros of the polynomial, indicated against them.

##### Answer:

##### Question 30:

Verify whether the following are zeros of the polynomial, indicated against them.

##### Answer:

##### Question 31:

Verify whether the following are zeros of the polynomial, indicated against them.

p(x) = x^{2} – 1, x = 1, – 1

##### Answer:

Here, p(x) = x^{2} – 1

Substitute x = 1 in p(x)

So, p(1) = (1)^{2} – 1

= 1 – 1

$\Rightarrow $ p(1) = 0

Therefore, x = 1 is the zero of the polynomial

x^{2} – 1. Now substitute x = – 1, in the given

polynomial.

So, p( – 1) = ( – 1)^{2} – 1

= 1 – 1

$\Rightarrow $ p( – 1) = 0

Therefore, x = – 1 is the zero of the

polynomial x^{2} – 1.

##### Question 32:

Verify whether the following are zeros of the polynomial, indicated against them.

p(x) = (x + 1) (x – 2), x = – 1, 2

##### Answer:

Here, p(x) = (x + 1) (x – 2)

Substitute x = – 1 in p(x)

So, p( – 1) = ( – 1 + 1) (– 1 – 2)

= 0 ( – 3)

= 0

Therefore, it is verified that x = – 1

is the zero of the polynomial

(x + 1) (x – 2)

Now substitute x = 2 in p(x)

So, p(2) = (2 + 1) (2 – 2)

= 3 × 0

$\Rightarrow $ p(2) = 0

Therefore, it is verified that x = 2 is the zero

of the polynomial (x + 1) (x – 2)

##### Question 33:

Verify whether the following are zeros of the polynomial, indicated against them.

p(x) = x^{2}, x = 0

##### Answer:

Here, p(x) = x^{2}

Substitute x = 0 in p(x)

So, p(0) = (0)^{2}

$\Rightarrow $ p(0) = 0

Therefore, it is verified that x = 0 is the zero

of the polynomial x^{2}.

##### Question 34:

Verify whether the following are zeros of the polynomial, indicated against them.

##### Answer:

##### Question 35:

Verify whether the following are zeros of the polynomial, indicated against them.

##### Answer:

##### Question 36:

Verify whether the following are zeros of the polynomial, indicated against them.

##### Answer:

##### Question 37:

Find the zero of the polynomial in each of the following cases:

p(x) = x + 5

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0,

We have, x + 5 = 0

$\Rightarrow $ x = – 5

So, – 5 is the zero of the polynomial x + 5

##### Question 38:

Find the zero of the polynomial in each of the following cases:

p(x) = x – 5

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0

We have, x – 5 = 0

$\Rightarrow $ x = 5

So, 5 is the zero of the polynomial x – 5

##### Question 39:

Find the zero of the polynomial in each of the following cases:

p(x) = 2x + 5

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.

We have,

2x + 5 = 0

$\Rightarrow $ 2x = – 5

##### Question 40:

Find the zero of the polynomial in each of the following cases:

p(x) = 3x – 2

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.

We have,

3x – 2 = 0

$\Rightarrow $ 3x = 2

##### Question 41:

Find the zero of the polynomial in each of the following cases:

p(x) = 3x

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.

We have, 3x = 0

$\Rightarrow $ x = 0

So, 0 is the zero (root) of the polynomial 3x.

##### Question 42:

Find the zero of the polynomial in each of the following cases:

p(x) = ax, a $\ne $ 0

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.

We have, ax = 0

So, 0 is the zero (root) of the polynomial ax.

##### Question 43:

Find the zero of the polynomial in each of the following cases:

p(x) = cx + d, c $\ne $ 0, c, d are real numbers.

##### Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0

We have, cx + d = 0

where c $\ne $ 0, c, d are real numbers

$\Rightarrow $ cx = – d

$\Rightarrow $ x = 0

So, 0 is the zero (root) of the polynomial ax.

##### Question 44:

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

x + 1

##### Answer:

Let p(x) be x^{3} + 3x^{2} + 3x + 1 and divisor is x + 1.

Remainder by long division method is as follows:

ALITER. Finding of remainder by using

remainder theorem.

Let p (x) = x

^{3}+ 3x

^{2}+ 3x + 1

Divisor is x + 1

So, take x + 1 = 0

$\Rightarrow $ x = – 1

Put x = – 1 in p(x), we get:

p( – 1) = (–1)

^{3}+ 3 (–1)

^{2}+ 3( – 1) + 1

= – 1 + 3 – 3 + 1

$\Rightarrow $ p ( – 1) = 0

Hence, remainder is 0.

##### Question 45:

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

x

##### Answer:

##### Question 46:

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

x + $\mathrm{\pi}$

##### Answer:

Let p(x)= x^{3} + 3x^{2} + 3x + 1

Here, divisor is x + $\mathrm{\pi}$.

We find the remainder by using remainder

theorem.

So, take x + $\mathrm{\pi}$ = 0

$\Rightarrow $ x = – $\mathrm{\pi}$

Now put x = – $\mathrm{\pi}$ in p(x), we get:

p($\mathrm{\pi}$) = (– $\mathrm{\pi}$)^{3} + 3(– $\mathrm{\pi}$)^{2} + 3(–$\mathrm{\pi}$) + 1

= – $\mathrm{\pi}$^{3} + 3$\mathrm{\pi}$^{2} – 3$\mathrm{\pi}$ + 1

##### Question 47:

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

5 + 2x

##### Answer:

Let p(x) = x^{3} + 3x^{2} + 3x + 1

Here, divisor is 5 + 2x

We find the remainder by using the remainder theorem

##### Question 48:

Find the remainder when

x^{3} – ax^{2} + 6x – a is divided by x – a.

##### Answer:

Let p(x) = x^{3} – ax^{2} + 6x – a

Divisor is x – a

So, take x – a = 0

$\Rightarrow $ x = a

put x = a in p(x), we get:

p(a) = a^{3} – a(a)^{2} + 6a – a

= a^{3} – a^{3} + 6a – a

= 5a

$\Rightarrow $ p(a) = 5a

Hence, remaninder is 5a.

##### Question 49:

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

##### Answer:

##### Question 50:

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

x^{3} + x^{2} + x + 1

##### Answer:

Let p (x) = x^{3} + x^{2} + x + 1

Put x = – 1 in p (x), we get:

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

= – 1 + 1 – 1 + 1

= 0

Hence, by factor theorem x + 1 is a factor of

x^{3} + x^{2} + x + 1.

##### Question 51:

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

x^{4} + x^{3} + x^{2} + x + 1

##### Answer:

Let p (x) = x^{4} + x^{3} + x^{2} + x + 1

Put x = –1 in p (x), we get:

p (–1) = (–1)^{4} + (–1)^{3} + (–1)^{2} + (–1) + 1

= 1 – 1 + 1 – 1 + 1

= 1 ≠ 0

Hence, by factor theorem x + 1 is not a factor

of x^{4} + x^{3} + x^{2} + x + 1.

##### Question 52:

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

x^{4} + 3x^{3} + 3x^{2} + x + 1

##### Answer:

Let p (x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

Put x = –1 in p (x), we get:

p (–1) = (–1)^{4} + 3(–1)^{3} + 3(–1)^{2} + (–1) + 1

= 1 – 3 + 3 – 1 + 1

= 1 ≠ 0

Hence, by factor theorem x + 1 is not a factor

of x^{4} + 3x^{3} + 3x^{2} + x + 1.

##### Question 53:

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

##### Answer:

##### Question 54:

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in ach of the following
cases:

p(x) = 2x^{3} + x^{2} – 2x – 1, g(x) = x + 1

##### Answer:

We have,

p (x) = 2x^{3} + x^{2} – 2x – 1

and divisor; g (x) = x + 1

Take (x + 1) = 0 $\Rightarrow $ x = –1

Put x = –1 in p (x), we get:

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

= – 2 + 1 + 2 – 1 = 0

Hence, by factor theorem x + 1 i.e. g (x) is

a factor of 2x^{3} + x^{2} – 2x – 1.

##### Question 55:

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in ach of the following
cases:

p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

##### Answer:

We have,

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

and divisor; g (x) = x + 2

Take x + 2 = 0

$\Rightarrow $ x = –2

Put x = – 2 in p (x), we get:

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

= – 8 + 12 – 6 + 1

= – 1 ≠ 0

Hence, by factor theorem x + 2 i.e. g (x) not

a factor of x^{3} + 3x^{2} + 3x + 1.

##### Question 56:

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in ach of the following
cases:

p(x) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

##### Answer:

We have,

p (x) = x^{3} – 4x^{2} + x + 6

and divisor; g(x) = x – 3

Take x – 3 = 0

$\Rightarrow $ x = 3

Put x = 3 in p (x), we get:

p (3) = (3)^{3} – 4 (3)^{2} + 3 + 6

= 27 – 36 + 3 + 6

= 36 – 36

= 0

Hence, by factor theorem x – 3, i.e. g (x) is a

factor of x^{3} – 4x^{2} + x + 6.

##### Question 57:

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

p (x) = x^{2} + x + k

##### Answer:

As x – 1 is a factor of

p (x) = x^{2} + x + k

$\therefore $ By factor theorem, p (1) = 0

$\Rightarrow $ (1)^{2} + 1 + k = 0

$\Rightarrow $ 1 + 1 + k = 0

$\Rightarrow $ 2 + k = 0

$\Rightarrow $ k = –2

##### Question 58:

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

##### Answer:

##### Question 59:

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

##### Answer:

##### Question 60:

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

p (x) = kx^{2} – 3x + k.

##### Answer:

As x – 1 is a factor of

p (x) = kx^{2} – 3x + k

$\therefore $ By factor theorem,

p (1) = 0

$\Rightarrow $ k (1)2 – 3(1) + k = 0

$\Rightarrow $ k – 3 + k = 0

$\Rightarrow $ 2k = 3

##### Question 61:

Factorise:

12x^{2} – 7x + 1

##### Answer:

##### Question 62:

Factorise:

2x^{2} + 7x + 3

##### Answer:

2x^{2} + 7x + 3 = 2x^{2} + 6x + x + 3

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

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

##### Question 63:

Factorise:

6x^{2} + 5x – 6

##### Answer:

6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

##### Question 64:

Factorise:

3x^{2} – x – 4

##### Answer:

3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

##### Question 65:

Factorise:

x^{3} – 2x^{2} – x + 2

##### Answer:

Let us denote the given polynomial as p(x)
So, p(x) = x^{3} – 2x^{2} – x + 2

All factors of +2 are $\pm $ 1, $\pm $ 2

By inspection,

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

= – 1 – 2 + 1 + 2

$\Rightarrow $ p(–1) = 0

So, by factor theorem (x + 1) is a factor of p(x)

Now divide p(x) by (x + 1)

Thus x^{3} – 2x^{2}– x + 2

= (x + 1) (x^{2} – 3x + 2)

The factors of x^{2} – 3x + 2 can be solved by

splitting the middle term or using factor

theorem.

By splitting the middle term,

we have

x^{2} – 3x + 2 = x^{2} – x – 2x + 2

= x(x – 1) –2(x – 1)

= (x – 1) (x – 2)

So, x^{3} – 2x^{2} – x + 2 = (x + 1) (x – 1) (x –2)

##### Question 66:

Factorise:

x^{3} – 3x^{2} – 9x – 5

##### Answer:

Let us denote the given polynomial as

p(x) = x^{3} – 3x^{2} – 9x – 5

All factors of –5 are $\pm $ 1, $\pm $ 5

By inspection,

p(–1) = (–1)3 –3 (–1)2 –9 (–1) – 5

= –1 – 3 + 9 – 5

= 9 – 9

= 0

So, by factor theorem x + 1 is a factor of p(x)

Now divide p(x) by x + 1.

Thus x^{3} – 3x^{2} – 9x – 5

= (x + 1) (x^{2} – 4x – 5)

Factors of x^{2} – 4x – 5 by splitting the middle

terms are as follows:

x^{2} – 4x – 5 = x^{2} – 5x + x – 5

= x(x – 5) + 1(x –5)

= (x – 5) (x + 1)

So, x^{3} – 3x^{2} – 9x – 5 = (x + 1) (x – 5) (x +1)

or = (x + 1) (x + 1) (x – 5)

##### Question 67:

Factorise:

x^{3} + 13x^{2} + 32x + 20

##### Answer:

Let us denote the given polynomial as

p(x) = x^{3} + 13x^{2} + 32x + 20

All factors of 20 are

$\pm $ 1, $\pm $ 2, $\pm $ 4, $\pm $ 5, $\pm $ 10, $\pm $ 20

By inspection,

p(–1) = (–1)^{3} + 13(–1)^{2} + 32(–1) + 20

= – 1 + 13 – 32 + 20

= 33 – 33

$\Rightarrow $ p(–1) = 0

So, by factor theorem x + 1 is a factor of p(x).

Now divide p(x) by (x + 1)

Thus x^{3} + 13x^{2} + 32x + 20

= (x + 1) (x^{2} + 12x + 20)

Factors of x^{2} + 12x + 20 by splitting the

middle terms are as follows:

x^{2} + 12x + 20 = x^{2} + 2x + 10x + 20

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

= (x + 2) (x + 10)

So, x^{3} + 13x^{2} + 32 x + 20

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

##### Question 68:

Factorise:

2y^{3} + y^{2} – 2y – 1

##### Answer:

Let us denote the given polynomial as

p(y) = 2y^{3} + y^{2} – 2y – 1

All factors of –1 are $\pm $ 1

By inspection,

p(1) = 2(1)^{3} + (1)^{2} – 2(1) – 1

= 2 + 1 – 2 – 1

= 3 – 3

$\Rightarrow $ p(1) = 0

So, by factor theorem; y – 1 is a factor of

p(y).

Now divide p(y) by y – 1.

Thus 2y^{3} + y^{2} – 2y – 1

= (y – 1) (2y^{2} + 3y + 1)

Factors of 2y^{2} + 3y + 1 by splitting the middle

terms are as follows:

2y^{2} + 3y + 1 = 2y^{2} + 2y + y + 1

= 2y (y + 1) + 1 (y + 1)

= (y + 1) (2y + 1)

So, 2y^{3} + y^{2} – 2y – 1

= (y – 1) (y + 1) (2y + 1)

##### Question 69:

Use the suitable identity to find the following products:

(x + 4) (x + 10)

##### Answer:

(x + 4) (x + 10)

= x^{2} + (4 + 10)x + 4 × 10

[Using the identity (x + a) (x + b) = x^{2} +

(a + b)x + ab. Here a = 4, b = 10]

= x^{2} + 14x + 40

##### Question 70:

Use the suitable identity to find the following products:

(x + 8) (x – 10)

##### Answer:

(x + 8) (x – 10)
=x^{2} + {8 + ( – 10)} x + 8 × ( – 10)
=x^{2} – 2x – 80
[Using the identity (x + a) (x + b)= x^{2}
+ (a + b) x + ab. Here a = 8, b = – 10]

##### Question 71:

Use the suitable identity to find the following products:

(3x + 4) (3x – 5)

##### Answer:

(3x + 4) (3x – 5)

Put 3x = y, we get:

(y + 4) (y – 5)

= y^{2} + {4 + ( – 5)} y + 4 (–5)

[Using the identity(x + a) (x + b) = x^{2}

+ (a + b) x + ab. Here a = 4, b = – 5]

= y^{2} – y – 20

= (3x)^{2} – 3x – 20 [$\because $ 3x = y]

= 9x^{2} – 3x – 20

##### Question 72:

Use the suitable identity to find the following products:

##### Answer:

##### Question 73:

Use the suitable identity to find the following products:

(3 – 2x) (3 + 2x)

##### Answer:

(3 – 2x) (3 + 2x)

or = – (2x – 3) (2x + 3)

or = – (2x + 3) (2x – 3)

Put 2x = y, we get:

– (y + 3) (y – 3)

= – [y2 + (3 – 3)y + 3 ( – 3)]

[Using the identity (x + a) (x + b) = x^{2} +

(a + b)x + ab. Here a = 3, b = – 3]

= – (y^{2} + 0y – 9)

= – (y^{2} – 9)

= – [(2x)^{2} – 9]

(where 2x = y as substituted above)

= – (4x^{2} – 9) = 9 – 4x^{2}

##### Question 74:

Evaluate the following products without multiplying directly:

103 × 107

##### Answer:

103 × 107

= (100 + 3) (100 + 7)

[Writing 103 as 100 + 3 and

107 as 100 + 7]

= (100)^{2} + (3 + 7) 100 + 3 × 7

= 10000 + 10 × 100 + 21

= 10000 + 1000 + 21

= 11021 [Using the identity (x + a) (x + b)

= x^{2} + (a + b)x + ab. Here

x = 100, a = 3, b = 7]

##### Question 75:

Evaluate the following products without multiplying directly:

95 × 96

##### Answer:

95 × 96 = (100 – 5) (100 – 4)

[Writing 95 as 100 – 5 and 96 as 100 – 4]

= (100)^{2} + [( – 5) + ( – 4)] 100 + ( – 5) ( – 4)

[Using the identity (x + a) (x + b) = x^{2} +

(a + b) x + ab. Here x = 100, a = – 5, b = – 4]

= 10000 + (– 9) × 100 + 20

= 10000 – 900 + 20

= 9120

##### Question 76:

Evaluate the following products without multiplying directly:

104 × 96

##### Answer:

104 × 96

= (100 + 4) (100 – 4)

[Writing 104 as 100 + 4 and 96 as 100 – 4]

(100)^{2} – (4)^{2}

[Using identity a^{2} – b^{2} = (a + b) (a – b).

Here a = 100 and b = 4]

= 10000 – 16

= 9984

##### Question 77:

Factorise the following using appropriate identities:

9x^{2} + 6xy + y^{2}

##### Answer:

9x^{2} + 6xy + y^{2}

=(3x)^{2} + 2(3x)y + y^{2}

=(3x + y)^{2}

[Using identity (a + b)^{2} = a^{2} + 2ab + b^{2}.

Here a = 3x and b = y]

##### Question 78:

Factorise the following using appropriate identities:

4y^{2} – 4y + 1

##### Answer:

4y^{2} – 4y + 1

= (2y)^{2} – 2(2y) 1 + 1^{2}

= (2y – 1)^{2}

[Using the identity (a – b)^{2} = a^{2} – 2ab + b^{2}.

Here a = 2y and b = 1]

##### Question 79:

Factorise the following using appropriate identities:

##### Answer:

##### Question 80:

Expand each of the following, using suitable identities:

(x + 2y + 4z)^{2}

##### Answer:

(x + 2y + 4z)^{2}

Comparing the given expression with

(a + b + c)^{2} we find that a = x, b = 2y

and c = 4z

Therefore using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +
2ca

we write (x + 2y + 4z)^{2} = x^{2} + (2y)^{2} + (4z)^{2}
+

2x(2y) + 2(2y) (4z) + 2(4z)x

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8xz

##### Question 81:

Expand each of the following, using suitable identities:

(2x – y + z)^{2}

##### Answer:

(2x – y + z)^{2}

= [2x + ( – y) + z]^{2}

Comparing the given expression with

(a + b + c)^{2} we find that a = 2x, b = – y

and c = z.

Therefore using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +
2ca

we write

{(2x + (– y) + z)^{2}} = (2x)^{2} + (–y)^{2} + z^{2}

+ 2(2x) (– y) + 2( –y) z + 2z (2x)

= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4zx

##### Question 82:

Expand each of the following, using suitable identities:

(– 2x + 3y + 2z)^{2}

##### Answer:

(– 2x + 3y + 2z)^{2}

Comparing the given expression with

(a + b + c)^{2} we find that a = – 2x, b = 3y and

c = 2z

Therefore using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +
2ca

we write

(– 2x + 3y + 2z)^{2} = {(– 2x)^{2} + (3y)^{2} + (2z)^{2}

+ 2( – 2x) (3y) + 2 (3y) (2z) + 2(2z) (– 2x)}

= 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8zx.

##### Question 83:

Expand each of the following, using suitable identities:

(3a – 7b – c)^{2}

##### Answer:

(3a – 7b – c)^{2}

= [(3a) + (–7b) + (–c)]^{2}

Comparing the given expression with

(a + b + c)^{2} we find that a = 3a, b = –7b, and

c = –c

Using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +
2ca

we get:

[3a + (–7b) + (–c)]^{2} = (3a)^{2} + (–7b)^{2} +
(–c)^{2}

+ 2 (3a) (–7b) + 2 (–7b) (–c) + 2(–c) (3a)

= 9a^{2} + 49b^{2} + c^{2} – 42ab + 14bc – 6ca

##### Question 84:

Expand each of the following, using suitable identities:

(– 2x + 5y – 3z)^{2}

##### Answer:

(–2x + 5y – 3z)^{2} = [–2x + 5y + (–3z)]^{2}

Comparing the given expression with

(a + b + c)^{2} we find that a = –2x, b = 5y and c = –3z

Therefore using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +
2ca

we write

[–2x + 5y + (–3z)]^{2}

= (–2x)^{2} + (5y)^{2} + (–3z)^{2} + 2(–2x) (5y)

+ 2(5y) (–3z) + 2 (–3z) (–2x)

= 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12zx

##### Question 85:

Expand each of the following, using suitable identities:

##### Answer:

##### Question 86:

Factorise:

4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

##### Answer:

##### Question 87:

Factorise:

##### Answer:

##### Question 88:

Write the following cube in expanded form:

(2x + 1)^{3}

##### Answer:

(2x + 1)^{3}

Compare the expression (2x + 1)^{3} with (a + b)^{3},

we find that a = 2x, b = 1

Therefore using the identity

(a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)

we have:

(2x + 1)^{3} = (2x)^{3} + (1)^{3} + 3(2x) 1 (2x + 1)

= 8x^{3} + 1 + 12x^{2} + 6x

= 8x^{3} + 12x^{2} + 6x + 1

[Re-arranging in descending power of x.]

##### Question 89:

Write the following cube in expanded form:

(2a – 3b)^{3}

##### Answer:

(2a – 3b)^{3}

Comparing the expression (2a – 3b)^{3}

with (x – y)^{3} we find that x = 2a, y = 3b

Therefore using the identity

(x – y)^{3}=x^{3} – y^{3} – 3xy (x – y), we get:

(2a – 3b)^{3}=(2a)^{3} – (3b)^{3} –3 (2a) (3b) (2a – 3b)

=8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}

##### Question 90:

Write the following cube in expanded form:

##### Answer:

##### Question 91:

Write the following cube in expanded form:

##### Answer:

##### Question 92:

Evaluate the following using suitable identities:

(99)^{3}

##### Answer:

(99)^{3} = (100 – 1)^{3}

= (100)^{3} – 1^{3} – 3(100) 1(100 – 1)

[Using the identity

(a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]
= 1000000 – 1 – 30000 + 300

= 970299

##### Question 93:

Evaluate the following using suitable identities:

(102)^{3}

##### Answer:

(102)^{3}= (100 + 2)^{3}

= (100)^{3} + 2^{3} + 3(100) (2) (100 + 2)

[Using the identity

(a + b) ^{3} = a^{3} + b^{3} + 3ab (a + b)]

= 1000000 + 8 + 60000 + 1200

= 1061208

##### Question 94:

Evaluate the following using suitable identities:

(998)^{3}

##### Answer:

(998)^{3}= (1000 – 2)^{3}

= (1000)^{3} – 2^{3} – 3(1000) (2)(1000 – 2)

[Using the identity

(a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]

= 1000000000 – 8 – 6000000 + 12000

= 1000012000 – 6000008

= 994011992

##### Question 95:

Factorise each of the following:

8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

##### Answer:

8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

= (2a)^{3} + b^{3} + 3 (2a) b (2a + b)

= (2a + b)^{3}

[Using the identity

(x + y)^{3} = x^{3} + y^{3} + 3xy (x + y).

Here x = 2a and y = b]

##### Question 96:

Factorise each of the following:

8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

##### Answer:

8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

= (2a)^{3} – b^{3} – 3 (2a) b (2a – b)

= (2a – b)^{3}

[Using the identity

(x – y)^{3} = x^{3} – y^{3} – 3xy (x – y).

Here x = 2a and y = b]

##### Question 97:

Factorise each of the following:

27 – 125a^{3} – 135a + 225a^{2}

##### Answer:

27 – 125a^{3} – 135a + 225a^{2}

= (3)^{3} – (5a)^{3} – 3 (3) (5a) (3 – 5a)

= (3 – 5a)^{3}

[Using the identity

(x – y)^{3} = x^{3} – y^{3} – 3xy (x – y).

Here x = 3, y = 5a]

##### Question 98:

Factorise each of the following:

64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

##### Answer:

64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

= (4a)^{3} – (3b)^{3} – 3(4a) (3b)

(4a – 3b)

= (4a – 3b)^{3}

[Using the identity

(x – y)^{3} = x^{3} – y^{3} – 3xy (x – y).

Here x = 4a, y = 3b]

##### Question 99:

Factorise each of the following:

##### Answer:

##### Question 100:

Verify: x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

##### Answer:

##### Question 101:

Verify: x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

##### Answer:

##### Question 102:

Factorise each of the following:

27y^{3} + 125z^{3}

##### Answer:

27y^{3} + 125z^{3}

= (3y)^{3} + (5z)^{3}

= (3y + 5z) [(3y)^{2} – (3y) (5z) + (5z)^{2}]

[Using identity

a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})

Here a = 3y, b = 5z]

= (3y + 5z) (9y^{2} – 15yz + 25z^{2})

##### Question103 :

Factorise each of the following:

64m^{3} – 343n^{3}

##### Answer:

64m^{3} – 343n^{3}

= (4m)^{3} – (7n)^{3}

= (4m – 7n) [(4m)^{2} + (4m) (7n) + (7n)^{2}]

[Using identity

a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2}).

Here a = 4m, b = 7n]

= (4m – 7n) (16m^{2} + 28mn + 49n^{2})

##### Question 104:

Factorise: 27x^{3} + y^{3} + z^{3} – 9xyz

##### Answer:

##### Question 105:

##### Answer:

##### Question 106:

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz

##### Answer:

##### Question 107:

Without actually calculating the cubes, find the value of the following:

(–12)^{3} + (7)^{3} + (5)sup>3

##### Answer:

Let a = –12, b = 7 and c = 5

When a + b + c = 0

then a^{3} + b^{3} + c^{3} = 3abc

Here a + b + c = – 12 + 7 + 5 = 0

$\therefore $(–12)^{3} + (7)^{3} + (5)^{3}

= 3(–12) (7) (5)

= –1260

##### Question 108:

Without actually calculating the cubes, find the value of the following:

(28)^{3} + (–15)^{3} + (–13)^{3}

##### Answer:

Let a = 28, b = –15 and c = –13

Now a + b + c = 28 + (–15) + (–13)

= 28 – 15 – 13

= 28 – 28

= 0

$\therefore $a^{3} + b^{3} + c^{3} = 3abc

$\Rightarrow $ (28)^{3} + (–15)^{3} + (–13)^{3}

= 3(28) (–15) (–13)

= 16380

##### Question 109:

Give possible expressions for the length and breadth of the following rectangle, in which
area is given:

Area: 25a^{2} – 35a + 12

##### Answer:

Area of rectangle

= 25a^{2} – 35a + 12 (given)

$\Rightarrow $ length × breadth = 25a^{2} – 15a – 20a + 12

= 5a (5a – 3) – 4 (5a – 3)

= (5a – 4) (5a – 3)

If length = (5a – 4)

then breadth = 5a – 3

and If length = 5a – 3

then breadth = (5a – 4)

##### Question 110:

Give possible expressions for the length and breadth of the following rectangle, in which
area is given:

Area: 35y^{2} + 13y – 12

##### Answer:

Area of rectangle = 35y^{2} + 13y – 12

$\Rightarrow $ length × breadth = 35y^{2} + 28y – 15y – 12

= 7y(5y + 4) – 3(5y + 4)

$\Rightarrow $ Length × Breadth= (5y + 4) (7y – 3)

If length = 5y + 4

then breadth = 7y – 3

and If length = 7y – 3

then breadth = 5y + 4

##### Question 111:

What are the possible expressions for the dimensions of the cuboids whose volumes are given
below?

Volume: 3x^{2} – 12x

##### Answer:

Volume of cuboid = 3x^{2} – 12x (given)

$\Rightarrow $ Length × Breadth × Height = 3x (x – 4)

$\therefore $ One possible dimensions of cuboid are

3, x and x – 4

##### Question 112:

What are the possible expressions for the dimensions of the cuboids whose volumes are given
below?

Volume: 12ky^{2} + 8ky – 20k

##### Answer:

Volume of cuboid = 12ky^{2} + 8ky – 20k

$\Rightarrow $ Length × Breadth × Height

= 4k (3y^{2} + 2y – 5)

= 4k [3y^{2} + 5y – 3y – 5]

= 4k [y(3y + 5) – 1 (3y + 5)]

= 4k [(3y + 5) (y – 1)]

$\Rightarrow $ Length × Breadth × Height

= 4k (3y + 5) (y – 1)

$\therefore $ One possible dimensions of cuboid are

4k, (3y + 5) and (y –1).