# 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) x5 - x4 + 3

(ii) x2 + x - 5

(iii) 5

(iv) 3x6 + 6y3 - 7

(v) 4 - y2

(vi) 5t - √3

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

(i) The highest power of x in x5 - x4 + 3 is 5.

∴ The degree of x5 - x4 + 3 is 5.

(ii) The highest power of x in x2 + x - 5 is 2.

∴ The degree of x2 + 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 3x6 + 6y3 - 7 is 6.

∴ The degree of 3x6 + 6y3 - 7 is 6.

(v) The highest power of y in 4 - y2 is 2.

∴ The degree of 4 - y2 is 2.

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

∴ The degree of 5t – √3 is 1.

Question 2.

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

(i) 3x2 - 2x + 5

(ii) x2 + √2

(iii) p2 - 3p + q

(iv)

(v)

(vi) x100 + y100

(i) 3x2 - 2x + 5 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(ii) x2 + √2 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(iii) p2 – 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) x100 + y100 has two variables that are x and y.

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

Question 3.

Write the coefficient of x3 in each of the following

(i) x3 + x + 1 (ii) 2 - x3 + x2

(iii)  (iv) 2x3 + 5

(v)  (vi)

(vii) 2x2 + 5 (vi) 4

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 x3 in x3 + x + 1 is 1.

∴ The coefficient of x3 in x3 + x + 1 is 1.

(ii) The constant written before x3 in 2 – x3 + x2 is -1.

∴ The coefficient of x3 in 2 – x3 + x2 is -1.

(iii) The constant written before x3 in √2x3 + 5 is √2.

∴ The coefficient of x3 in √2x3 + 5 is √2.

(iv) The constant written before x3 in 2x3 + 5 is 2.

∴ The coefficient of x3 in 2x3 + 5 is 2.

(v) The constant written before x3 in  is.

∴ The coefficient of x3 in  is.

(vi) The constant written before x3 in  is.

∴ The coefficient of x3 in  is.

(vii) The term x3 does not exist in 2x2 + 5.

∴ The coefficient of x3 in 2x2 + 5 is 0.

(viii) The term x3 does not exist in 4.

∴ The coefficient of x3 in 4 is 0.

Question 4.

Classify the following as linear, quadratic and cubic polynomials

(i) 5x2 + x - 7 (ii) x - x3

(iii) x2 + x + 4 (iv) x - 1

(v) 3p (vi) Ï€r2

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

∵ the degree of 5x2 + x – 7 is 2

∴ 5x2 + x – 7 is a quadratic polynomial.

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

∵ the degree of 5x2 + x – 7 is 3

∴ 5x2 + x – 7 is a cubic polynomial.

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

∵ the degree of x2 + x + 4 is 2

∴ x2 + 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€r2 is 2

∴ Ï€r2 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 x2 + 2xy + y2 is 2

(vi) Ï€r2 is monomial.

(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 x2 + 2xy + y2 is 2, therefore its degree is 2.

∴ The statement is true.

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

∵ Ï€r2 has only one term

∴ The statement is true.

Question 6.

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

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 3x10.

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 3x10 + 2x2 + 5.

###### Exercise 2.2

Question 1.

Find the value of the polynomial 4x2 - 5x + 3, when

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

(iii) x = 2 (iv)

(i) p(x) = 4x2 – 5x + 3

⇒ p(0) = 4(0)2 – 5(0) + 3

⇒ p(0) = 0– 0 + 3

⇒ p(0) = 3

(ii) p(x) = 4x2 – 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) = 4x2 – 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) = 4x2 – 5x + 3

Question 2.

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

p(x) = x2 - x +1

p(x) = x2 – 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 + 2y2 - y3

p(y) = 2 + y + 2y2 – y3

⇒ 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) = z3

p(z) = z3

⇒ p(0) = 03

⇒ p(0) = 0

And,

⇒ p(1) = 13

⇒ p(1) = 1

And,

⇒ p(2) = 23

⇒ 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)

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

p(t) = t2 + t - t - 1

⇒ p(t) = t2 – 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) = x2 - 3x + 2

p(t) = x2 - 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;

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€;

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) = x2 - 1; x = ±1

p(x) = x2 - 1

⇒ p(-1) = (-1)2 – 1

⇒ p(-1) = 1 – 1

⇒ p(-1) = 0

∴ Yes x = -1 is the zero of polynomial x2 – 1.

And,

⇒ p(1) = (1)2 – 1

⇒ p(1) = 1 – 1

⇒ p(1) = 0

∴ Yes x = 1 is the zero of polynomial x2 – 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

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) = y2; y = 0

p(y) = y2

⇒ p(0) = 02

⇒ p(0) = 0

∴ Yes y = 0 is the zero of polynomial y2.

Question 12.

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

p(x) = ax + b ;

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) = 3x2 - 1; x = -

f(x) = 3x2 - 1

∴ Yes  is the zero of polynomial 3x2 - 1.

∴ No  is not the zero of polynomial 3x2 - 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 =

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

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

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

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

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) = x2

f(x) = x2

f(x) = 0

⇒ x2 = 0

⇒ x = 0

∴ x = 0 is the zero of the polynomial x2.

Question 20.

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

f(x) = px, p ≠ 0

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.

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) = 2x2 - 3x + 7a, find the value of a.

∵ 2 is the zeroes of the polynomial p(x) = 2x2 - 3x + 7a

∴ p(2) = 0

Now,

p(x) = 2x2 - 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) = 2x3 - 3x2 + ax + b, find the values of a and b.

∵ 0 and 1 are the zeroes of the polynomial f(x) = 2x3-3x2+ ax + b

∴ f(0) = 0 and f(1) = 1

Now,

f(x) = 2x3-3x2+ 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 x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x + 1

Let p(x) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 when divided by x+1 is 0

Question 2.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

Let p(x) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 when divided by  is

Question 3.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x

Let p(x) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 when divided by x is 1

Question 4.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

x + Ï€

Let p(x) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 when divided by x+ Ï€ is –Ï€3 + 3Ï€2 –3Ï€ + 1

Question 5.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials:

5 + 2x

Let p(x) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 when divided by 5 + 2x is

Question 6.

Find the remainder when x3 – px2 + 6x – p is divided by x – p.

Let q(x) = x3 – px2 + 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) = p3 – p3 + 6p – p

∴ Remainder of x3 – px2 + 6x – p when divided by x – p is 5p

Question 7.

Find the remainder when 2x2 – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial? State reason.

Let p(x) = 2x2 – 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 2x2 – 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 9x3 – 3x2 + x – 5 is divided by

Let p(x) = 9x3 – 3x2 + 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 9x3 – 3x2 + x – 5 when divided by  is –3

Question 9.

If the polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x + a leave the same remainder when divided by x – 2, find the value of a.

Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 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 x3 + ax2 + 5 and x3 – 2x2 + a are divided by (x + 2) leave the same remainder, find the value of a.

Let p(x) = x3 + ax2 + 5 and q(x) = x3 – 2x2 + 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) = x4 – 3x2 + 4 is divided by g(x)= x – 2 and verify the result by actual division.

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) = 24 – 3(2)2 + 4

⇒ f(2) = 16 – 12 + 4 = 8

∴ the remainder when x4 – 3x2 + 4 is divided by x – 2 is 8

Question 12.

Find the remainder when p(x) = x3 – 6x2 + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division.

Given: p(x) = x3 – 6x2 + 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 x3 – 6x2 + 14x – 3 is divided by 1 – 2x is
Result Verification:

We can see that the remainder is .

Hence, verified.

Question 13.

When a polynomial 2x3 +3x2 + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder –2. Find a and b.

Let p(x) = 2x3 +3x2 + 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.

x3 – x2 – x + 1

Let f(x) = x3 – x2 – 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 x3 – x2 – x + 1

Question 2.

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

x4 – x3 + x2 – x + 1

Let f(x) = x4 – x3 + x2 – 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 x4 – x3 + x2 – x + 1

Question 3.

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

x4 + 2x3 + 2x2 + x + 1

Let f(x) = x4 + 2x3 + 2x2 + 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 x4 + 2x3 + 2x2 + x + 1

Question 4.

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

x3 – x2 –(3 – √3 ) x + √3

Let f(x) = x3 – x2 – (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 x3 – x2 –(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) = 5x3 + x2 – 5x – 1, g(x) = x + 1

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) = x3 + 3x2 + 3x + 1, g(x) = x + 1

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) = x3 – 4x2 + x + 6, g(x) = x – 2

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) = 3x3 + x2 – 20x + 12, g(x) = 3x–2

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 (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) = 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3

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 (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 x3 – 3x2 – 10x + 24.

Let f(x) = x3 – 3x2 – 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 x3 – 3x2 – 10x + 24

Question 11.

Show that (x + 4), (x – 3) and (x – 7) are factors of x3 – 6x2 – 19x + 84.

Let f(x) = x3 – 6x2 – 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 x3 – 3x2 – 10x + 24

Question 12.

If both (x – 2) and  are factors of px2 + 5x + r, show that p = r.

Let f(x) = px2 + 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 (x2 – 1) is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0

Let f(x) = ax4 + bx3 + cx2 + 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, (x2 – 1) = (x + 1)(x – 1)

Since (x2 – 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

x3 – 2x2 – x + 2

Let p(x) = x3 – 2x2 – 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 x2 – x – 2.

Now, (x2 – x – 2) is a quadratic and can be solved by splitting the middle terms.

We have x2 – x – 2 = x2 – 2x + x – 2

⇒ x (x – 2) + 1 (x – 2)

⇒ (x + 1)(x – 2)

So, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)

Question 15.

Factorize

x3 – 3x2 – 9x – 5

Let p(x) = x3 – 3x2 – 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 x2 – 4x – 5.

Now, (x2 – 4x – 5) is a quadratic and can be solved by splitting the middle terms.

We have x2 – 4x – 5 = x2 – 5x + x – 5

⇒ x (x – 5) + 1 (x – 5)

⇒ (x + 1)(x – 5)

So, x3 – 3x2 – 9x – 5= (x + 1)(x + 1)(x – 5)

Question 16.

Factorize

x3 + 13x2 + 32x + 20

Let p(x) = x3 + 13x2 + 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 x2 + 12x + 20.

Now, (x2 + 12x + 20) is a quadratic and can be solved by splitting the middle terms.

We have x2 + 12x + 20= x2 + 10x + 2x + 20

⇒ x (x + 10) + 2 (x + 10)

⇒ (x + 2)(x + 10)

So, x3 + 13x2 + 32x + 20 = (x + 1)(x + 2)(x + 10)

Question 17.

Let p(y) = y3 + y2 – y – 1

On taking y2 common from first two terms in p(y), we get,

p (y) = y2(y + 1) –1(y + 1)

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

⇒ p(y) = (y2 – 1)(y + 1)

As we know the identity, (y2 – 1) = (y + 1)(y – 1)

⇒ p(y) = (y – 1)(y + 1)(y + 1)

Question 18.

If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that c = 0 and a = b.

Let f(x) = ax2 + bx + c and p(x) = bx2 + 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 x2 – x – 6 and x2 + 3x – 18 have a common factor (x – a) then find the value of a.

Let f(x) = x2 – x – 6 and p(x) = x2 + 3x – 18

As (x – a) 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(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 y3 – 2y2 – 9y + 18 then find the other two factors.

Let f(x) = y3 – 2y2 – 9y + 18

Taking y2 common from the first two terms of f(x) and 9 from the last two terms of f(x), we get,

⇒ f(x) = y2(y – 2) –9(y – 2)

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

⇒ f(x) = (y2 – 9)(y – 2) -------- (A)

We know the identity as,

a2 – b2 = (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 y3 – 2y2 – 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)

using the identity (x + a) × (x + b) = x2 + (a + b)x + ab

here a = 5 and b = 2

⇒ (x + 5) (x + 2) = x2 + (5 + 2)x + 5 × 2

Therefore (x + 5) (x + 2) = x2 + 7x + 10

Question 2.

Use suitable identities to find the following products

(x - 5) (x - 5)

(x - 5) (x - 5) = (x – 5)2

Using identity (a – b)2 = a2 – 2ab + b2

Here a = x and b = 5

⇒ (x - 5) (x - 5) = x2 – 2 × x × 5 + 52

⇒ (x - 5) (x - 5) = x2 – 10x + 25

Therefore (x - 5) (x - 5) = x2 – 10x + 25

Question 3.

Use suitable identities to find the following products

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

using the identity (a + b) × (a – b) = a2 – b2

here a = 3x and b = 2

⇒ (3x + 2)(3x - 2) = (3x)2 – 22

Therefore (3x + 2)(3x - 2) = 9x2 – 4

Question 4.

Use suitable identities to find the following products

using the identity (a + b) × (a – b) = a2 – b2

here a = x2 and b =

⇒  = (x2)2 -

Therefore  = x4 -

Question 5.

Use suitable identities to find the following products

(1 + x) (1 + x)

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

Using identity (a + b)2 = a2 + 2ab + b2

Here a = 1 and b = x

⇒ (1 + x) (1 + x) = 12 + 2(1)(x) + x2

Therefore (1 + x) (1 + x) = 1 + 2x + x2

Question 6.

Evaluate the following products without actual multiplication.

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) = a2 – b2

here a = 100 and b = 1

⇒ 101 × 99 = 1002 – 12

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

Question 7.

Evaluate the following products without actual multiplication.

999 × 999

999 can be written as (1000 – 1)

⇒ 999 × 999 = (1000 – 1) × (1000 – 1)

⇒ 999 × 999 = (1000 – 1)2

Using identity (a – b)2 = a2 – 2ab + b2

Here a = 1000 and b = 1

⇒ 999 × 999 = 10002 – 2(1000)(1) + 12

⇒ 999 × 999 = 1000000 – 2000 + 1

⇒ 999 × 999 = 998000 + 1

⇒ 999 × 999 = 998001

Question 8.

Evaluate the following products without actual multiplication.

⇒ 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) = a2 – b2

here a = 100 and b = 1

⇒ 101 × 99 = 1002 – 12

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

Therefore  =

Question 9.

Evaluate the following products without actual multiplication.

501 × 501

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 = a2 + 2ab + b2

Here a = 500 and b = 1

⇒ 501 × 501 = 5002 + 2(500)(1) + 12

⇒ 501 × 501 = 250000 + 1000 + 1

⇒ 501 × 501 = 251001

Question 10.

Evaluate the following products without actual multiplication.

30.5 × 29.5

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) = a2 – b2

here a = 60 and b = 1

⇒ 61 × 59 = 602 – 12

⇒ 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.

16x2 + 24xy + 9y2

16x2 can be written as (4x)2

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

9y2 can be written as (3y)2

⇒ 16x2 + 24xy + 9y2 = (4x)2 + 2(4x)(3y) + (3y)2

Using identity (a + b)2 = a2 + 2ab + b2

Here a = 4x and b = 3y

⇒ 16x2 + 24xy + 9y2 = (4x + 3y)2

Therefore 16x2 + 24xy + 9y2 = (4x + 3y) (4x + 3y)

Question 12.

Factorise the following using appropriate identities.

4y2 - 4y + 1

4y2 can be written as (2y)2

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

1 can be written as 12

⇒ 4y2 - 4y + 1 = (2y)2 - 2(1)(2y) + 12

Using identity (a - b)2 = a2 - 2ab + b2

Here a = 2y and b = 1

⇒ 4y2 - 4y + 1 = (2y - 1)2

Therefore 4y2 - 4y + 1 = (2y - 1) (2y - 1)

Question 13.

Factorise the following using appropriate identities.

4x2 can be written as (2x)2

can be written as

⇒ 4x2 -  = (2x)2 -

using the identity (a + b) × (a – b) = a2 – b2

here a = 2x and b =

Therefore 4x2 -  = (2x + ) (2x - )

Question 14.

Factorise the following using appropriate identities.

18a2 – 50

Take out common factor 2

⇒ 18a2 – 50 = 2 (9a2 - 25)

Now

9a2 can be written as (3a)2

25 can be written as 52

⇒ 18a2 – 50 = 2 ((3a)2 – 52)

using the identity (a + b) × (a – b) = a2 – b2

here a = 3a and b = 5

therefore 18a2 – 50 = 2 (3a + 5) (3a – 5)

Question 15.

Factorise the following using appropriate identities.

x2 + 5x + 6

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

⇒ x2 + 5x + 6 = x2 + 3x + 2x + 6

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

= (x + 3) (x + 2)

Therefore x2 + 5x + 6 = (x + 3) (x + 2)

Question 16.

Factorise the following using appropriate identities.

3p2 - 24p + 36

Take out common factor 3

⇒ 3p2 - 24p + 36 = 3 (p2 – 8p + 12)

Now splitting the middle term of quadratic p2 – 8p + 12 to factorise it

⇒ 3p2 - 24p + 36 = 3 (p2 – 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

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

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

⇒ (x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

Therefore

(x + 2y + 4z)2 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

Question 18.

Expand each of the following, using suitable identities

(2a - 3b)3

Using identity (x – y)3 = x3 - y3 – 3x2y + 3xy2

Here x = 2a and y = 3b

⇒ (2a - 3b)3 = (2a)3 – (3b)3 – 3(2a)2(3b) + 3(2a)(3b)2

= 8a3 – 27b3 – 18a2b + 18ab2

Therefore (2a - 3b)3 = 8a3 – 27b3 – 18a2b + 18ab2

Question 19.

Expand each of the following, using suitable identities

(-2a + 5b - 3c)2

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

Using (x + y + z)2 = x2 + y2 + z2 + 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 = 4a2 + 25b2 + 9c2 + (-20ab) + (-30bc) + 12ac

⇒ (-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12ac

Therefore

(-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12ac

Question 20.

Expand each of the following, using suitable identities

= [ +  + 1]2

Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Here x = , y =  and z = 1

⇒  =  +  + 12 + 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

Using identity (x + y)3 = x3 + y3 + 3x2y + 3xy2

Here x = p and y = 1

⇒ (p + 1)3 = p3 + 13 + 3(p)2(1) + 3(p)(1)2

= p3 + 13 + 3p2 + 3p

Therefore (p + 1)3 = p3 + 13 + 3p2 + 3p

Question 22.

Expand each of the following, using suitable identities

Using identity (a – b)3 = a3 - b3 – 3a2b + 3ab2

Here a = x and b = y

⇒  = x3 -  – 3(x)2() + 3(x)

⇒  = x3 -  – 2x2y + xy2

Therefore  = x3 -  – 2x2y + xy2

Question 23.

Factorise

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = 25x2 + 16y2 + 4z2 + (-40xy) + 16yz + (-20xz)

25x2 can be written as (-5x)2

16y2 can be written as (4y)2

4z2 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)

⇒ 25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (-5x)2 + (4y)2 +

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

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Comparing (-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) with a2 + b2 + c2 + 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)

25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (- 5x + 4y + 2z)2

Question 24.

Factorise

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = 9a2 + 4b2 + 16c2 + 12ab + (-16bc) + (-24ca)

9a2 can be written as (3a)2

4b2 can be written as (2b)2

16c2 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)

⇒ 9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)

Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Comparing (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x2 + y2 + z2 + 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)

9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a + 2b – 4c)2

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