# Polynomials And Factorisation Solution of TS & AP Board Class 9 Mathematics

###### Exercise 2.1

**Question 1.**

Find the degree of each of the polynomials given below

(i) x^{5} - x^{4} + 3

(ii) x^{2} + x - 5

(iii) 5

(iv) 3x^{6} + 6y^{3} - 7

(v) 4 - y^{2}

(vi) 5t - √3

**Answer:**

Degree of p(x) is the highest power of x in p(x).

(i) The highest power of x in x^{5} - x^{4} + 3 is 5.

∴ The degree of x^{5} - x^{4} + 3 is 5.

(ii) The highest power of x in x^{2} + x - 5 is 2.

∴ The degree of x^{2} + x - 5 is 2.

(iii) The highest power of x in 5 is 0(∵ there is no term of x).

∴ The degree of 5 is 0.

(iv) The highest power of x in 3x^{6} + 6y^{3} - 7 is 6.

∴ The degree of 3x^{6} + 6y^{3} - 7 is 6.

(v) The highest power of y in 4 - y^{2} is 2.

∴ The degree of 4 - y^{2} is 2.

(vi) The highest power of t in 5t – √3 is 1.

∴ The degree of 5t – √3 is 1.

**Question 2.**

**Answer:**

(i) 3x^{2} - 2x + 5 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(ii) x^{2} + √2 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(iii) p^{2} – 3p + q has two variables that are p and q.

∴ no, it is not a polynomial in one variable.

(iv) has a negative exponent of y.

∴ no, it is not a polynomial.

(v) The exponent of x in 5√x + x√5 is 1/2 which is not a non-negative integer

∴ no, it is not a polynomial.

(vi) x^{100} + y^{100} has two variables that are x and y.

∴ no, it is not a polynomial in one variable.

**Question 3.**

Write the coefficient of x^{3} in each of the following

(i) x^{3} + x + 1 (ii) 2 - x^{3} + x^{2}

(iii) (iv) 2x^{3} + 5

(v) (vi)

(vii) 2x^{2} + 5 (vi) 4

**Answer:**

A coefficient is a multiplicative factor in some term of a polynomial. It is the constant written before the variable.

Therefore,

(i) The constant written before x^{3} in x^{3} + x + 1 is 1.

∴ The coefficient of x^{3} in x^{3} + x + 1 is 1.

(ii) The constant written before x^{3} in 2 – x^{3} + x^{2} is -1.

∴ The coefficient of x^{3} in 2 – x^{3} + x^{2} is -1.

(iii) The constant written before x^{3} in √2x^{3} + 5 is √2.

∴ The coefficient of x^{3} in √2x^{3} + 5 is √2.

(iv) The constant written before x^{3} in 2x^{3} + 5 is 2.

∴ The coefficient of x^{3} in 2x^{3} + 5 is 2.

(v) The constant written before x^{3} in is.

∴ The coefficient of x^{3} in is.

(vi) The constant written before x^{3} in is.

∴ The coefficient of x^{3} in is.

(vii) The term x^{3} does not exist in 2x^{2} + 5.

∴ The coefficient of x^{3} in 2x^{2} + 5 is 0.

(viii) The term x^{3} does not exist in 4.

∴ The coefficient of x^{3} in 4 is 0.

**Question 4.**

Classify the following as linear, quadratic and cubic polynomials

(i) 5x^{2} + x - 7 (ii) x - x^{3}

(iii) x^{2} + x + 4 (iv) x - 1

(v) 3p (vi) Ï€r^{2}

**Answer:**

(i) A quadratic polynomial is a polynomial of degree 2

∵ the degree of 5x^{2} + x – 7 is 2

∴ 5x^{2} + x – 7 is a quadratic polynomial.

(ii) A cubic polynomial is a polynomial of degree 3

∵ the degree of 5x^{2} + x – 7 is 3

∴ 5x^{2} + x – 7 is a cubic polynomial.

(iii) A quadratic polynomial is a polynomial of degree 2

∵ the degree of x^{2} + x + 4 is 2

∴ x^{2} + x + 4 is a quadratic polynomial.

(iv) A linear polynomial is a polynomial of degree 1

∵ the degree of x – 1 is 1

∴ x – 1 is a linear polynomial.

(v) A linear polynomial is a polynomial of degree 1

∵ the degree of 3p is 1

∴ 3p is a linear polynomial.

(vi) A quadratic polynomial is a polynomial of degree 2

∵ the degree of Ï€r^{2} is 2

∴ Ï€r^{2} is a quadratic polynomial.

**Question 5.**

Write whether the following statements are True or False. Justify your answer

(i) A binomial can have at the most two terms

(ii) Every polynomial is a binomial

(iii) A binomial may have degree 3

(iv) Degree of zero polynomial is zero

(v) The degree of x^{2} + 2xy + y^{2} is 2

(vi) Ï€r^{2} is monomial.

**Answer:**

(i) A polynomial with two terms is called a binomial.

∴ The statement is true.

(ii) A polynomial can have more than two terms.

∴ The statement is false.

(iii) A binomial should have two terms, the degree of those terms can be any integer.

∴ The statement is true.

(iv) The constant polynomial whose coefficients are all equal to 0, is called a zero polynomial. Its degree can be any integer.

∴ The statement is false.

(v) The highest power in x^{2} + 2xy + y^{2} is 2, therefore its degree is 2.

∴ The statement is true.

(vi) A monomial is a polynomial which has only one term.

∵ Ï€r^{2} has only one term

∴ The statement is true.

**Question 6.**

Give one example each of a monomial and trinomial of degree 10.

**Answer:**

A monomial is a polynomial which has only one term, and the degree is the highest power of the variable. Therefore, an example of a monomial of degree 10 is 3x^{10}.

A trinomial is a polynomial which has three terms, and the degree is the highest power of the variable. Therefore, example of a trinomial of degree 10 is 3x^{10} + 2x^{2} + 5.

###### Exercise 2.2

**Question 1.**

Find the value of the polynomial 4x^{2} - 5x + 3, when

(i) x = 0 (ii) x = -1

(iii) x = 2 (iv)

**Answer:**

(i) p(x) = 4x^{2} – 5x + 3

⇒ p(0) = 4(0)^{2} – 5(0) + 3

⇒ p(0) = 0– 0 + 3

⇒ p(0) = 3

(ii) p(x) = 4x^{2} – 5x + 3

⇒ p(-1) = 4(-1)^{2} – 5(-1) + 3

⇒ p(-1) = 4 × 1– (-5) + 3

⇒ p(-1) = 4 +5 + 3

⇒ p(-1) = 12

(iii) p(x) = 4x^{2} – 5x + 3

⇒ p(2) = 4(2)^{2} – 5(2) + 3

⇒ p(2) = 4 × 4– 10 + 3

⇒ p(2) = 16– 10 + 3

⇒ p(2) = 9

(iv) p(x) = 4x^{2} – 5x + 3

**Question 2.**

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

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

**Answer:**

p(x) = x^{2} – x + 1

⇒ p(0) = (0)^{2} – 0 + 1

⇒ p(0) = 1

And,

⇒ p(1) = (1)^{2} – 1 + 1

⇒ p(1) = 1– 1 + 1

⇒ p(1) = 1

And,

⇒ p(2) = (2)^{2} – 2 + 1

⇒ p(2) = 4– 2 + 1

⇒ p(2) = 3

**Question 3.**

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

p(y) = 2 + y + 2y^{2} - y^{3}

**Answer:**

p(y) = 2 + y + 2y^{2} – y^{3}

⇒ p(0) = 2 + 0 + 2(0)^{2} – (0)^{3}

⇒ p(0) = 2 + 0 + 0– 0

⇒ p(0) = 2

And,

⇒ p(1) = 2 + 1 + 2(1)^{2} – (1)^{3}

⇒ p(1) = 2 + 1 + 2– 1

⇒ p(1) = 4

And,

⇒ p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}

⇒ p(2) = 2 + 2 + 8– 8

⇒ p(2) = 4

**Question 4.**

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

p(z) = z^{3}

**Answer:**

p(z) = z^{3}

⇒ p(0) = 0^{3}

⇒ p(0) = 0

And,

⇒ p(1) = 1^{3}

⇒ p(1) = 1

And,

⇒ p(2) = 2^{3}

⇒ p(2) = 8

**Question 5.**

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

p(t) = (t - 1) (t + 1)

**Answer:**

p(t) = (t - 1) (t + 1)

p(t) = t

^{2}+ t - t - 1

⇒ p(t) = t^{2} – 1

⇒ p(0) = (0)^{2} – 1

⇒ p(0) = 0– 1

⇒ p(0) = -1

And,

⇒ p(1) = (1)^{2} – 1

⇒ p(1) = 1– 1

⇒ p(1) = 0

And,

⇒ p(2) = (2)^{2} – 1

⇒ p(2) = 4– 1

⇒ p(2) = 3

**Question 6.**

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

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

**Answer:**

p(t) = x^{2} - 3x + 2

⇒ p(0) = (0)^{2} – 3(0) + 2

⇒ p(0) = 0– 0 + 2

⇒ p(0) = 2

And,

⇒ p(1) = (1)^{2} – 3(1) + 2

⇒ p(1) = 1– 3 + 2

⇒ p(0) = 0

And,

⇒ p(2) = (2)^{2} – 3(2) + 2

⇒ p(2) = 4– 6 + 2

⇒ p(2) = 0

**Question 7.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = 2x + 1;

**Answer:**

p(x) = 2x + 1

⇒ p(-1/2) = 2(-1/2) + 1

⇒ p(-1/2) = -1 + 1

⇒ p(-1/2) = 0

∴ Yes x = -1/2 is the zero of polynomial 2x + 1.

**Question 8.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = 5x - Ï€;

**Answer:**

p(x) = 5x - Ï€

∴ No x = - is not the zero of polynomial 5x – Ï€.

**Question 9.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = x^{2} - 1; x = ±1

**Answer:**

p(x) = x^{2} - 1

⇒ p(-1) = (-1)^{2} – 1

⇒ p(-1) = 1 – 1

⇒ p(-1) = 0

∴ Yes x = -1 is the zero of polynomial x^{2} – 1.

And,

⇒ p(1) = (1)^{2} – 1

⇒ p(1) = 1 – 1

⇒ p(1) = 0

∴ Yes x = 1 is the zero of polynomial x^{2} – 1.

**Question 10.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = (x - 1)(x + 2); x = -1, -2

**Answer:**

p(x) = (x - 1)(x + 2)

⇒ p(-1) = (-1 - 1)(-1 + 2);

⇒ p(-1) = -2 × 1

⇒ p(-1) = -2

∴ No x = -1 is not the zero of polynomial (x - 1)(x + 2).

And,

⇒ p(-2) = (-2 - 1)(-2 + 2);

⇒ p(-2) = -3 × 0

⇒ p(-2) = 0

∴ Yes x = -2 is not the zero of polynomial (x - 1)(x + 2).

**Question 11.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(y) = y^{2}; y = 0

**Answer:**

p(y) = y^{2}

⇒ p(0) = 0^{2}

⇒ p(0) = 0

∴ Yes y = 0 is the zero of polynomial y^{2}.

**Question 12.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

p(x) = ax + b ;

**Answer:**

p(x) = ax + b

∴ Yes is the zero of polynomial ax + b.

**Question 13.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

f(x) = 3x^{2} - 1; x = -

**Answer:**

f(x) = 3x^{2} - 1

∴ Yes is the zero of polynomial 3x^{2} - 1.

∴ No is not the zero of polynomial 3x^{2} - 1.

**Question 14.**

Verify whether the values of x given in each case are the zeroes of the polynomial or not?

f (x) = 2x - 1, x =

**Answer:**

f(x) = 2x - 1

∴ Yes is the zero of polynomial 2x - 1.

∴ No is not the zero of polynomial 2x - 1.

**Question 15.**

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

f(x) = x + 2

**Answer:**

f(x) = x + 2

f(x) = 0

⇒ x + 2 = 0

⇒ x = 0 – 2

⇒ x = -2

∴ x = -2 is the zero of the polynomial x + 2.

**Question 16.**

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

f(x) = x - 2

**Answer:**

f(x) = x - 2

f(x) = 0

⇒ x – 2 = 0

⇒ x = 0 + 2

⇒ x = 2

∴ x = 2 is the zero of the polynomial x – 2.

**Question 17.**

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

f(x) = 2x + 3

**Answer:**

f(x) = 2x + 3

f(x) = 0

⇒ 2x + 3 = 0

⇒ 2x = 0 – 3

⇒ 2x = -3

∴ is the zero of the polynomial 2x + 3.

**Question 18.**

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

f(x) = 2x - 3

**Answer:**

f(x) = 2x – 3

f(x) = 0

⇒ 2x – 3 = 0

⇒ 2x = 0 + 3

⇒ 2x = 3

∴ is the zero of the polynomial 2x – 3.

**Question 19.**

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

f(x) = x^{2}

**Answer:**

f(x) = x^{2}

f(x) = 0

⇒ x^{2} = 0

⇒ x = 0

∴ x = 0 is the zero of the polynomial x^{2}.

**Question 20.**

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

f(x) = px, p ≠ 0

**Answer:**

f(x) = px, p ≠ 0

f(x) = 0

⇒ px = 0

⇒ x = 0

∴ x = 0 is the zero of the polynomial px.

**Question 21.**

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

f(x) = px + q, p ≠ 0, p q are real numbers.

**Answer:**

f(x) = px + q, p ≠ 0, p q are real numbers.

f(x) = 0

⇒ px + q = 0

⇒ px = -q

∴ is the zero of the polynomial px + q.

**Question 22.**

If 2 is a zero of the polynomial p(x) = 2x^{2} - 3x + 7a, find the value of a.

**Answer:**

∵ 2 is the zeroes of the polynomial p(x) = 2x^{2} - 3x + 7a

∴ p(2) = 0

Now,

p(x) = 2x^{2} - 3x + 7a

⇒p(2) = 2(2)^{2} – 3(2)+ 7a

⇒ 2 × 4– 3 × 2 + 7a = 0

⇒ 8– 6 + 7a = 0

⇒ 2 + 7a = 0

⇒ 7a = -2

**Question 23.**

If 0 and 1 are the zeroes of the polynomial f(x) = 2x^{3} - 3x^{2} + ax + b, find the values of a and b.

**Answer:**

∵ 0 and 1 are the zeroes of the polynomial f(x) = 2x^{3}-3x^{2}+ ax + b

∴ f(0) = 0 and f(1) = 1

Now,

f(x) = 2x^{3}-3x^{2}+ ax + b

⇒ f(0) = 2(0)^{3} – 3(0)^{2}+ a(0) + 0

⇒ 2 × 0– 3 × 0 + a × 0 + b = 0

⇒ 0– 0 + 0 + b = 0

⇒ b = 0

And,

⇒ f(1) = 2(1)^{3} – 3(1)^{2}+ a(1) +1

⇒ 2 × 1– 3 × 1 + a × 1 + b = 0

⇒ 2– 3 + a + b = 0

⇒ 2– 3 + a + 0 = 0 [∵ b = 0]

⇒ -1 + a = 0

⇒ a = 1

###### Exercise 2.3

**Question 1.**

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

x + 1

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x+1 is p(–1)

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

⇒ p(–1) = –1 + 3 – 3 + 1 = 0

∴ Remainder of x^{3} + 3x^{2} + 3x + 1 when divided by x+1 is 0

**Question 2.**

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

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by is p(1/2)

∴ Remainder of x^{3} + 3x^{2} + 3x + 1 when divided by is

**Question 3.**

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

x

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x is p(0)

P(0) = (0)^{3}+ 3(0)^{2}+ 3(0) + 1

= 1

∴ Remainder of x^{3} + 3x^{2} + 3x + 1 when divided by x is 1

**Question 4.**

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

x + Ï€

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x + Ï€ is p(–Ï€)

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

⇒ p(–Ï€)= –Ï€^{3} + 3Ï€^{2} –3Ï€ + 1

∴ Remainder of x^{3} + 3x^{2} + 3x + 1 when divided by x+ Ï€ is –Ï€^{3} + 3Ï€^{2} –3Ï€ + 1

**Question 5.**

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

5 + 2x

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by 5 + 2x is

∴ Remainder of x^{3} + 3x^{2} + 3x + 1 when divided by 5 + 2x is

**Question 6.**

Find the remainder when x^{3} – px^{2} + 6x – p is divided by x – p.

**Answer:**

Let q(x) = x^{3} – px^{2} + 6x – p

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of q(x) when divided by x – p is q(p)

q(p) = (p)^{3}– p(p)^{2}+ 6(p) – p

⇒ q(p) = p^{3} – p^{3} + 6p – p

∴ Remainder of x^{3} – px^{2} + 6x – p when divided by x – p is 5p

**Question 7.**

Find the remainder when 2x^{2} – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial? State reason.

**Answer:**

Let p(x) = 2x^{2} – 3x + 5

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by 2x – 3 is

= 5

⇒ Remainder of 2x^{2} – 3x + 5 when divided by 2x – 3 is 5.

As on dividing the given polynomial by 2x – 3, we get a non–zero remainder, therefore, 2x – 3 does not completely divide the polynomial.

∴ It is not a factor.

**Question 8.**

Find the remainder when 9x^{3} – 3x^{2} + x – 5 is divided by

**Answer:**

Let p(x) = 9x^{3} – 3x^{2} + x – 5

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by is

⇒ Remainder of 9x^{3} – 3x^{2} + x – 5 when divided by is –3

**Question 9.**

If the polynomials 2x^{3} + ax^{2} + 3x – 5 and x^{3} + x^{2} – 4x + a leave the same remainder when divided by x – 2, find the value of a.

**Answer:**

Let p(x) = 2x^{3} + ax^{2} + 3x – 5 and q(x) = x^{3} + x^{2} – 4x + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)

⇒ p(2) = 2(2)^{3} +a(2)^{2} + 3(2) – 5

⇒p(2) = 16 + 4a +6 – 5

⇒p(2) = 17 + 4a

Similarly, q(2) = (2)^{3} + (2)^{2} + –4(2) + a

⇒ q(2) = 8 + 4 –8 + a

⇒ q(2) = 4 + a

Since they both leave the same remainder, so p(2) = q(2)

⇒ 17 + 4a = 4 + a

⇒ 13 = 3a

∴ The value of a is –13/3

**Question 10.**

If the polynomials x^{3} + ax^{2} + 5 and x^{3} – 2x^{2} + a are divided by (x + 2) leave the same remainder, find the value of a.

**Answer:**

Let p(x) = x^{3} + ax^{2} + 5 and q(x) = x^{3} – 2x^{2} + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x + 2 is p(–2). Similarly, Remainder of q(x) when divided by x + 2 is q(–2)

⇒ p(–2) = (–2)^{3} +a(–2)^{2} + 5

⇒p(–2) = –8 + 4a + 5

⇒p(–2) = –3 + 4a

Similarly, q(–2) = (–2)^{3} – 2(–2)^{2} + a

⇒ q(–2) = –8 –8 + a

⇒ q(–2) = –16 + a

Since they both leave the same remainder, so p(–2) = q(–2)

⇒ –3 + 4a = –16 + a

⇒ –13 = 3a

∴ The value of a is –13/3

**Question 11.**

Find the remainder when f (x) = x^{4} – 3x^{2} + 4 is divided by g(x)= x – 2 and verify the result by actual division.

**Answer:**

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

Therefore, remainder when f(x) is divided by g(x) is f(2)

f(2) = 2^{4} – 3(2)^{2} + 4

⇒ f(2) = 16 – 12 + 4 = 8

∴ the remainder when x^{4} – 3x^{2} + 4 is divided by x – 2 is 8

**Question 12.**

Find the remainder when p(x) = x^{3} – 6x^{2} + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division.

**Answer:**

Given: p(x) = x^{3} – 6x^{2} + 14x – 3 and g(x) = 1 – 2x

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

Therefore, remainder when p(x) is divided by g(x) is

∴ the remainder when x^{3} – 6x^{2} + 14x – 3 is divided by 1 – 2x is

Result Verification:

We can see that the remainder is .

Hence, verified.

**Question 13.**

When a polynomial 2x^{3} +3x^{2} + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder –2. Find a and b.

**Answer:**

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

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2)

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

⇒ p(2) = 16 + 12 + 2a + b

Also, it is given that p(2) = 2, on substituting value above, wev get,

2 = 28 + 2a + b

⇒ 2a + b = –26 ---------- (A)

Similarly,

Remainder of p(x) when divided by x + 2 is p(–2)

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

⇒ p(–2) = –16 + 12 – 2a + b

Also, it is given that p(–2) = –2, on substituting value above, we get,

–2 = –4 – 2a + b

⇒ – 2a + b = 2 ---------- (B)

On solving the above two equ. (A) and (b), we get,

a = –7 and b = –12

∴ Value of a and b is –7 and –12 respectively.

###### Exercise 2.4

**Question 1.**

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

x^{3} – x^{2} – x + 1

**Answer:**

Let f(x) = x^{3} – x^{2} – x + 1

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)^{3} – (–1)^{2} – (–1) + 1

⇒ f(–1) = –1 –1 +1 +1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore (x+1) is a factor x^{3} – x^{2} – x + 1

**Question 2.**

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

x^{4} – x^{3} + x^{2} – x + 1

**Answer:**

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

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

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

⇒ f(–1) = 1 + 1 + 1 + 1 + 1

⇒ f(–1) = 5

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x^{4} – x^{3} + x^{2} – x + 1

**Question 3.**

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

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

**Answer:**

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

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

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

⇒ f(–1) = 1 – 2 + 2 – 1 + 1

⇒ f(–1) = 1

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x^{4} + 2x^{3} + 2x^{2} + x + 1

**Question 4.**

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

x^{3} – x^{2} –(3 – √3 ) x + √3

**Answer:**

Let f(x) = x^{3} – x^{2} – (3 –√3 ) x + √3

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)^{3} – (–1)^{2} – (3 –√3)(–1) + √3

⇒ f(–1) = –1 – 1 + 3 – √3 + √3

⇒ f(–1) = 1

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x^{3} – x^{2} –(3 –√3) x + √3

**Question 5.**

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

f(x) = 5x^{3} + x^{2} – 5x – 1, g(x) = x + 1

**Answer:**

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = 5 (–1)^{3} + (–1)^{2} – 5(–1) – 1

⇒ f(–1) = –5 + 1 + 5 – 1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)

**Question 6.**

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

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

**Answer:**

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)^{3} + 3(–1)^{2} + 3(–1) + 1

⇒ f(–1) = – 1 + 3 – 3 + 1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)

**Question 7.**

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

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

**Answer:**

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x – 2) to be a factor, we will find f(2)

⇒ f(2) = (2)^{3} – 4(2)^{2} + (2) + 6

⇒ f(–1) = 8 – 16 + 2 + 6

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x – 2) is a factor of f(x)

**Question 8.**

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

f(x) = 3x^{3} + x^{2} – 20x + 12, g(x) = 3x–2

**Answer:**

By Factor Theorem, we know that,

For checking (3x – 2) to be a factor, we will find f(2/3)

As, f(–1) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)

**Question 9.**

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

f(x) = 4x^{3} + 20x^{2} + 33x + 18, g(x) = 2x + 3

**Answer:**

By Factor Theorem, we know that,

For checking (2x + 3) to be a factor, we will find f(–3/2)

As, f(–3/2) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)

**Question 10.**

Show that (x – 2), (x + 3) and (x – 4) are factors of x^{3} – 3x^{2} – 10x + 24.

**Answer:**

Let f(x) = x^{3} – 3x^{2} – 10x + 24

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0

For checking (x – 2) to be a factor, we will find f(2)

⇒ f(2) = (2)^{3} – 3(2)^{2} – 10(2) + 24

⇒ f(2) = 8 – 12 – 20 + 24

⇒ f(2) = 0

So, (x–2) is a factor.

For checking (x + 3) to be a factor, we will find f(–3)

⇒ f(–3) = (–3)^{3} – 3(–3)^{2} – 10(–3) + 24

⇒ f(–3) = –27 – 27 + 30 + 24

⇒ f(–3) = 0

So, (x+3) is a factor.

For checking (x – 4) to be a factor, we will find f(4)

⇒ f(4) = (4)^{3} – 3(4)^{2} – 10(4) + 24

⇒ f(4) = 64 – 48 – 40 + 24

⇒ f(4) = 0

So, (x–4) is a factor.

∴ (x – 2), (x + 3) and (x – 4) are factors of x^{3} – 3x^{2} – 10x + 24

**Question 11.**

Show that (x + 4), (x – 3) and (x – 7) are factors of x^{3} – 6x^{2} – 19x + 84.

**Answer:**

Let f(x) = x^{3} – 6x^{2} – 19x + 84

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0

For checking (x + 4) to be a factor, we will find f(–4)

⇒ f(–4) = (–4)^{3} – 6(–4)^{2} – 19(–4) + 84

⇒ f(–4) = –64 – 96 + 76 + 84

⇒ f(–4) = 0

So, (x+4) is a factor.

For checking (x – 3) to be a factor, we will find f(3)

⇒ f(3) = (3)^{3} – 6(3)^{2} – 19(3) + 84

⇒ f(3) = 27 – 54 – 57 + 84

⇒ f(3) = 0

So, (x–3) is a factor.

For checking (x – 7) to be a factor, we will find f(7)

⇒ f(7) = (7)^{3} – 6(7)^{2} – 19(7) + 84

⇒ f(7) = 343 – 294 – 133 + 84

⇒ f(7) = 0

So, (x–7) is a factor.

∴ (x + 4), (x – 3) and (x – 7) are factors of x^{3} – 3x^{2} – 10x + 24

**Question 12.**

If both (x – 2) and are factors of px^{2} + 5x + r, show that p = r.

**Answer:**

Let f(x) = px^{2} + 5x + r

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

So, if (x – 2) is a factor of f(x)

⇒ f(2) = 0

⇒ p(2)^{2} + 5(2) + r = 0

⇒ 4p + r = –10 -------- (A)

Also as is also a factor,

⇒ p + 4r = –10 ----- (B)

Subtract B from A to get,4p + r - (p + 4r)= –10 - (-10)

4p + r - p - 4r = -10 + 10

3p - 3r = 0

3p = 3r

p = r

**Question 13.**

If (x^{2} – 1) is a factor of ax^{4} + bx^{3} + cx^{2} + dx + e, show that a + c + e = b + d = 0

**Answer:**

Let f(x) = ax^{4} + bx^{3} + cx^{2} + dx + e

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

Also we can write, (x^{2} – 1) = (x + 1)(x – 1)

Since (x^{2} – 1) is a factor of f(x), this means (x + 1) and (x – 1) both are factors of f(x).

So, if (x – 1) is a factor of f(x)

⇒ f(1) = 0

⇒ a(1)^{4} + b(1)^{3} + c(1)^{2} + d(1) + e = 0

⇒ a + b + c + d + e = 0 ----- (A)

Also as (x + 1) is also a factor,

⇒ f(–1) = 0

⇒ a(–1)^{4} + b(–1)^{3} + c(–1)^{2} + d(–1) + e = 0

⇒ a – b + c – d + e = 0

⇒ a + c + e = b + d ---- (B)

On solving equations (A) and (B), we get,

a + c + e = b + d = 0

**Question 14.**

Factorize

x^{3} – 2x^{2} – x + 2

**Answer:**

Let p(x) = x^{3} – 2x^{2} – x + 2

By trial, we find that p(1) = 0, so by Factor theorem,

(x – 1) is the factor of p(x)

When we divide p(x) by (x – 1), we get x^{2} – x – 2.

Now, (x^{2} – x – 2) is a quadratic and can be solved by splitting the middle terms.

We have x^{2} – x – 2 = x^{2} – 2x + x – 2

⇒ x (x – 2) + 1 (x – 2)

⇒ (x + 1)(x – 2)

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

**Question 15.**

Factorize

x^{3} – 3x^{2} – 9x – 5

**Answer:**

Let p(x) = x^{3} – 3x^{2} – 9x – 5

By trial, we find that p(–1) = 0, so by Factor theorem,

(x + 1) is the factor of p(x)

When we divide p(x) by (x + 1), we get x^{2} – 4x – 5.

Now, (x^{2} – 4x – 5) is a quadratic and can be solved by splitting the middle terms.

We have x^{2} – 4x – 5 = x^{2} – 5x + x – 5

⇒ x (x – 5) + 1 (x – 5)

⇒ (x + 1)(x – 5)

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

**Question 16.**

Factorize

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

**Answer:**

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

By trial, we find that p(–1) = 0, so by Factor theorem,

(x + 1) is the factor of p(x)

When we divide p(x) by (x + 1), we get x^{2} + 12x + 20.

Now, (x^{2} + 12x + 20) is a quadratic and can be solved by splitting the middle terms.

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

⇒ x (x + 10) + 2 (x + 10)

⇒ (x + 2)(x + 10)

**So, x ^{3} + 13x^{2} + 32x + 20 = (x + 1)(x + 2)(x + 10)**

**Question 17.**

**Answer:**

Let p(y) = y^{3} + y^{2} – y – 1

On taking y^{2} common from first two terms in p(y), we get,

p (y) = y^{2}(y + 1) –1(y + 1)

Now, taking (y + 1) common, we get,

⇒ p(y) = (y^{2} – 1)(y + 1)

As we know the identity, (y^{2} – 1) = (y + 1)(y – 1)

⇒ p(y) = (y – 1)(y + 1)(y + 1)

**Question 18.**

If ax^{2} + bx + c and bx^{2} + ax + c have a common factor x + 1 then show that c = 0 and a = b.

**Answer:**

Let f(x) = ax^{2} + bx + c and p(x) = bx^{2} + ax + c

As (x + 1) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

⇒ f(–1) = p (–1) = 0

⇒ a(–1)^{2} + b(–1) + c = b(–1)^{2} + a(–1) + c

⇒ a – b + c = b – a + c

⇒ 2a = 2b

⇒ a = b ------ (A)

Also, we discussed that,

f(–1) = 0

⇒ a(–1)^{2} + b(–1) + c = 0

⇒ a – b + c = 0

From equation (A), we see that a = b,

⇒ c = 0 -------- (B)

∴ Equations (A) and (B) show us the required result.

**Question 19.**

If x^{2} – x – 6 and x^{2} + 3x – 18 have a common factor (x – a) then find the value of a.

**Answer:**

Let f(x) = x^{2} – x – 6 and p(x) = x^{2} + 3x – 18

As (x – a) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,

⇒ f(a) = p (a)

⇒ (a)^{2} – (a) – 6 = (a)^{2} + 3(a) – 18

⇒ 4a = 12

⇒ a = 3

∴ The value of a is 3.

**Question 20.**

If (y – 3) is a factor of y^{3} – 2y^{2} – 9y + 18 then find the other two factors.

**Answer:**

Let f(x) = y^{3} – 2y^{2} – 9y + 18

Taking y^{2} common from the first two terms of f(x) and 9 from the last two terms of f(x), we get,

⇒ f(x) = y^{2}(y – 2) –9(y – 2)

Now, taking (y – 2) common from above,

⇒ f(x) = (y^{2} – 9)(y – 2) -------- (A)

We know the identity as,

a^{2} – b^{2} = (a – b)(a + b)

So, using above identity on equation (A), we get,

⇒ f(x) = (y + 3)(y – 3)(y – 2)

∴ the other two factors of y^{3} – 2y^{2} – 9y + 18 besides (y – 3) are (y + 3) and (y – 2).

###### Exercise 2.5

**Question 1.**

Use suitable identities to find the following products

(x + 5) (x + 2)

**Answer:**

using the identity (x + a) × (x + b) = x^{2} + (a + b)x + ab

here a = 5 and b = 2

⇒ (x + 5) (x + 2) = x^{2} + (5 + 2)x + 5 × 2

Therefore (x + 5) (x + 2) = x^{2} + 7x + 10

**Question 2.**

Use suitable identities to find the following products

(x - 5) (x - 5)

**Answer:**

(x - 5) (x - 5) = (x – 5)^{2}

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

Here a = x and b = 5

⇒ (x - 5) (x - 5) = x^{2} – 2 × x × 5 + 5^{2}

⇒ (x - 5) (x - 5) = x^{2} – 10x + 25

Therefore (x - 5) (x - 5) = x^{2} – 10x + 25

**Question 3.**

Use suitable identities to find the following products

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

**Answer:**

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 3x and b = 2

⇒ (3x + 2)(3x - 2) = (3x)^{2} – 2^{2}

Therefore (3x + 2)(3x - 2) = 9x^{2} – 4

**Question 4.**

Use suitable identities to find the following products

**Answer:**

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = x^{2} and b =

⇒ = (x^{2})^{2} -

Therefore = x^{4} -

**Question 5.**

Use suitable identities to find the following products

(1 + x) (1 + x)

**Answer:**

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

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

Here a = 1 and b = x

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

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

**Question 6.**

Evaluate the following products without actual multiplication.

101 × 99

**Answer:**

101 can be written as (100 + 1) and

99 can be written as (100 - 1)

⇒ 101 × 99 = (100 + 1) × (100 - 1)

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 100 and b = 1

⇒ 101 × 99 = 100^{2} – 1^{2}

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

**Question 7.**

Evaluate the following products without actual multiplication.

999 × 999

**Answer:**

999 can be written as (1000 – 1)

⇒ 999 × 999 = (1000 – 1) × (1000 – 1)

⇒ 999 × 999 = (1000 – 1)^{2}

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

Here a = 1000 and b = 1

⇒ 999 × 999 = 1000^{2} – 2(1000)(1) + 1^{2}

⇒ 999 × 999 = 1000000 – 2000 + 1

⇒ 999 × 999 = 998000 + 1

⇒ 999 × 999 = 998001

**Question 8.**

Evaluate the following products without actual multiplication.

**Answer:**

⇒ 50 = and 49 =

⇒ 50 = and 49 =

⇒ 50 = and 49 =

⇒ = ×

⇒ =

Consider 101 × 99

101 can be written as (100 + 1) and

99 can be written as (100 - 1)

⇒ 101 × 99 = (100 + 1) × (100 - 1)

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 100 and b = 1

⇒ 101 × 99 = 100^{2} – 1^{2}

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

Therefore =

**Question 9.**

Evaluate the following products without actual multiplication.

501 × 501

**Answer:**

501 can be written as (500 + 1)

⇒ 501 × 501 = (500 + 1) × (500 + 1)

⇒ 501 × 501 = (500 + 1)^{2}

⇒ 501 × 501 = (500 + 1) × (500 + 1)

⇒ 501 × 501 = (500 + 1)^{2}

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

Here a = 500 and b = 1

⇒ 501 × 501 = 500^{2} + 2(500)(1) + 1^{2}

⇒ 501 × 501 = 250000 + 1000 + 1

⇒ 501 × 501 = 251001

**Question 10.**

Evaluate the following products without actual multiplication.

30.5 × 29.5

**Answer:**

30.5 = and 29.5 =

⇒ 30.5 × 29.5 = ×

⇒ 30.5 × 29.5 = ×

⇒ 30.5 × 29.5 = …(i)

Consider 61 × 59

61 = (60 + 1)

59 = (60 – 1)

⇒ 61 × 59 = (60 + 1)(60 – 1)

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 60 and b = 1

⇒ 61 × 59 = 60^{2} – 1^{2}

⇒ 61 × 59 = 3600 – 1

⇒ 61 × 59 = 3599

From (i)

⇒ 30.5 × 29.5 =

Therefore 30.5 × 29.5 = 899.75

**Question 11.**

Factorise the following using appropriate identities.

16x^{2} + 24xy + 9y^{2}

**Answer:**

16x^{2} can be written as (4x)^{2}

24xy can be written as 2(4x)(3y)

9y^{2} can be written as (3y)^{2}

⇒ 16x^{2} + 24xy + 9y^{2} = (4x)^{2} + 2(4x)(3y) + (3y)^{2}

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

Here a = 4x and b = 3y

⇒ 16x^{2} + 24xy + 9y^{2} = (4x + 3y)^{2}

Therefore 16x^{2} + 24xy + 9y^{2} = (4x + 3y) (4x + 3y)

**Question 12.**

Factorise the following using appropriate identities.

4y^{2} - 4y + 1

**Answer:**

4y^{2} can be written as (2y)^{2}

4y can be written as 2(1)(2y)

1 can be written as 1^{2}

⇒ 4y^{2} - 4y + 1 = (2y)^{2} - 2(1)(2y) + 1^{2}

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

Here a = 2y and b = 1

⇒ 4y^{2} - 4y + 1 = (2y - 1)^{2}

Therefore 4y^{2} - 4y + 1 = (2y - 1) (2y - 1)

**Question 13.**

Factorise the following using appropriate identities.

**Answer:**

4x^{2} can be written as (2x)^{2}

can be written as

⇒ 4x^{2} - = (2x)^{2} -

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 2x and b =

Therefore 4x^{2} - = (2x + ) (2x - )

**Question 14.**

Factorise the following using appropriate identities.

18a^{2} – 50

**Answer:**

Take out common factor 2

⇒ 18a^{2} – 50 = 2 (9a^{2} - 25)

Now

9a^{2} can be written as (3a)^{2}

25 can be written as 5^{2}

⇒ 18a^{2} – 50 = 2 ((3a)^{2} – 5^{2})

using the identity (a + b) × (a – b) = a^{2} – b^{2}

here a = 3a and b = 5

therefore 18a^{2} – 50 = 2 (3a + 5) (3a – 5)

**Question 15.**

Factorise the following using appropriate identities.

x^{2} + 5x + 6

**Answer:**

Given is quadratic equation which can be factorised by splitting the middle term as shown

⇒ x^{2} + 5x + 6 = x^{2} + 3x + 2x + 6

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

= (x + 3) (x + 2)

Therefore x^{2} + 5x + 6 = (x + 3) (x + 2)

**Question 16.**

Factorise the following using appropriate identities.

3p^{2} - 24p + 36

**Answer:**

Take out common factor 3

⇒ 3p^{2} - 24p + 36 = 3 (p^{2} – 8p + 12)

Now splitting the middle term of quadratic p^{2} – 8p + 12 to factorise it

⇒ 3p^{2} - 24p + 36 = 3 (p^{2} – 6p – 2p + 12)

= 3 [p (p – 6) – 2 (p – 6)]

= 3 (p – 2) (p – 6)

**Question 17.**

Expand each of the following, using suitable identities

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

**Answer:**

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

Here a = x, b = 2y and c = 4z

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

Therefore

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

**Question 18.**

Expand each of the following, using suitable identities

(2a - 3b)^{3}

**Answer:**

Using identity (x – y)^{3} = x^{3} - y^{3} – 3x^{2}y + 3xy^{2}

Here x = 2a and y = 3b

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

= 8a^{3} – 27b^{3} – 18a^{2}b + 18ab^{2}

Therefore (2a - 3b)^{3} = 8a^{3} – 27b^{3} – 18a^{2}b + 18ab^{2}

**Question 19.**

Expand each of the following, using suitable identities

(-2a + 5b - 3c)^{2}

**Answer:**

(-2a + 5b - 3c)^{2} = [(-2a) + (5b) + (-3c)]^{2}

Using (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx

Here x = -2a, y = 5b and z = -3c

⇒ (-2a + 5b - 3c)^{2} = (-2a)^{2} + (5b)^{2} + (-3c)^{2} + 2(-2a)(5b) + 2(5b)(-3c) + 2(-3c)(-2a)

⇒ (-2a + 5b - 3c)^{2} = 4a^{2} + 25b^{2} + 9c^{2} + (-20ab) + (-30bc) + 12ac

⇒ (-2a + 5b - 3c)^{2} = 4a^{2} + 25b^{2} + 9c^{2} - 20ab - 30bc + 12ac

Therefore

(-2a + 5b - 3c)^{2} = 4a^{2} + 25b^{2} + 9c^{2} - 20ab - 30bc + 12ac

**Question 20.**

Expand each of the following, using suitable identities

**Answer:**

= [ + + 1]^{2}

Using (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx

Here x = , y = and z = 1

⇒ = + + 1^{2} + 2 + 2(1) + 2(1)

⇒ = + + 1 + + (-b) +

⇒ = + + 1 - – b +

Therefore = + + 1 - – b +

**Question 21.**

Expand each of the following, using suitable identities

(p + 1)^{3}

**Answer:**

Using identity (x + y)^{3} = x^{3} + y^{3} + 3x^{2}y + 3xy^{2}

Here x = p and y = 1

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

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

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

**Question 22.**

Expand each of the following, using suitable identities

**Answer:**

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

Here a = x and b = y

⇒ = x^{3} - – 3(x)^{2}() + 3(x)

⇒ = x^{3} - – 2x^{2}y + xy^{2}

Therefore = x^{3} - – 2x^{2}y + xy^{2}

**Question 23.**

Factorise

25x^{2} + 16y^{2} + 4z^{2} - 40xy + 16yz - 20xz

**Answer:**

25x^{2} + 16y^{2} + 4z^{2} - 40xy + 16yz - 20xz

25x^{2} + 16y^{2} + 4z^{2} - 40xy + 16yz - 20xz = 25x^{2} + 16y^{2} + 4z^{2} + (-40xy) + 16yz + (-20xz)

25x^{2} can be written as (-5x)^{2}

16y^{2} can be written as (4y)^{2}

4z^{2} can be written as (2z)^{2}

-40xy can be written as 2(-5x)(4y)

16yz can be written as 2(4y)(2z)

-20xz can be written as 2(-5x)(2z)

⇒ 25x^{2} + 16y^{2} + 4z^{2} - 40xy + 16yz - 20xz = (-5x)^{2} + (4y)^{2} +

(2z)^{2} + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) …(i)

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

Comparing (-5x)^{2} + (4y)^{2} + (2z)^{2} + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) with a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca we get

a = -5x, b = 4y and c = 2z

therefore

(-5x)^{2} + (4y)^{2} + (2z)^{2} + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) = (- 5x + 4y + 2z)^{2}

From (i)

25x^{2} + 16y^{2} + 4z^{2} - 40xy + 16yz - 20xz = (- 5x + 4y + 2z)^{2}

**Question 24.**

Factorise

9a^{2} + 4b^{2} + 16c^{2} + 12ab - 16bc - 24ca

**Answer:**

9a^{2} + 4b^{2} + 16c^{2} + 12ab - 16bc - 24ca

9a^{2} + 4b^{2} + 16c^{2} + 12ab - 16bc - 24ca = 9a^{2} + 4b^{2} + 16c^{2} + 12ab + (-16bc) + (-24ca)

9a^{2} can be written as (3a)^{2}

4b^{2} can be written as (2b)^{2}

16c^{2} can be written as (-4c)^{2}

12ab can be written as 2(3a)(2b)

-16bc can be written as 2(2b)(-4c)

-24ca can be written as 2(-4c)(3a)

⇒ 9a^{2} + 4b^{2} + 16c^{2} + 12ab - 16bc - 24ca = (3a)^{2} + (2b)^{2} + (-4c)^{2} + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)

Using (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx

Comparing (3a)^{2} + (2b)^{2} + (-4c)^{2} + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx we get

x = 3a, y = 2b and z = -4c

therefore

(3a)^{2} + (2b)^{2} + (-4c)^{2} + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) = (3a + 2b + (-4c))^{2}

From (i)

9a^{2} + 4b^{2} + 16c^{2} + 12ab - 16bc - 24ca = (3a + 2b – 4c)^{2}