**Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.4**

**(1) (i) given, p(x) = x ^{3} – x^{2} – x – 1 ; g(x) = (x + 1)**

∴ using remainder theorem we get

P(-1) = (-1)^{3} – (-1)^{2} – (-1) + 1

= -1 – 1 + 1 + 1

= 0

Hence,

∴ (x + 1) is a factor of p(x) since the remainder is 0.

**(ii) x ^{4} – x^{3} + x^{2} – x + 1 = p(x) ; g(x) = x + 1**

∴ using remainder theorem

P(-1_ = (-1)^{4} – (-1)^{3} + (-1)^{2} – (-1) + 1

= 1 + 1 + 1 + 1 + 1

= 5

Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.

**(iii) given, p(x) = x ^{4} + 2x^{3} + 2x^{2} + x + 1 ; g(x) = x + 1**

∴ by remainder theorem,

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

= 1 – 2 + 2 -1 + 1

= 1

Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.

**(iv) x ^{3} – x^{2} – (3 – **

**√3)x +**

**√3**∴ by remainder theorem,

P(-1) = (-1)^{3} – (-1)^{2} – (3 – **√***3) (-1) + √3*

*= -1 – 1 + 3 – √3 + √3*

= 1

Hence, x + 1 is not a factor of p(x) since the remainder is not equal to 0.

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

∴ zero of g(x) = -1

Now, f(-1) = 5(-1)^{3} + (-1)^{2} – 5(-1) – 1

= -5 + 1 + 5 – 1

= 0

∴ By factor theorem g(x) is a factor of f(x)

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

Zero of g(x) = -1

Now, f(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

∴ By factor theorem g(x) is a factor of f(x)

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

∴ Zero of g(x) = +2

Now,

f(x) = 2^{3} – 4(2)^{2} + 2 + 6

= 8 – 16 + 2 + 6

∴ By factor theorem g(x) is a factor of f(x).

**(3) Given, f(x) = x ^{3} – 3x^{2} – 10x + 24; **

Let, g(x) = x – 2

L(x) = x + 3

P(x) = x – 4

Now, zero of g(x) = +2

Then, f(2) = 2^{3} – 3x(2)^{2} – 10×2 + 24

= 8 – 12 – 20 + 24

= 32 – 32

= 0

∴ By factor theorem we can say that g(x) is a factor of f(x).

Now, zero of L(x) = -3

Then, f(-3) = (-3)^{3} – 3 x (-3)^{2} – 10 x -3 + 24

= -27 – 27 + 30 + 24

= 54 – 54

= 0

Hence, by factor theorem L(x) is a factor of f(x).

Now, the zero of p(x) = + 4

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

= 64 – 48 – 40 + 24

= 88 – 88

= 0

Hence, by factor theorem p(x) is the factor of p(x).

**(4) Let, f(x) = x ^{3} – 6x^{2} – 19x + 84**; P(x) = x + 4, q(x) = x – 3, R(x) = x – 7

Now, zero of p(x) = -4

Then, f(-4) = (-4)^{3} – 6x(-4)^{2} – 19 x -4 + 84

= -64 – 96 + 76 + 84

= 160 – 160

= 0

Hence by factorisation, p(x) is the factor of f(x).

Now, zero of f(x) = 3

Then, f(3) = 3^{3} – 6 x 3^{2} – 19 x 3 + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0

Hence by factor theorem, q(x) is a factor of f(x)

Now, zero of R(x) = 7

Then, f(7) = 7^{3} – 6 x 7^{2} – 19 x 7 + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

Hence by factor theorem, R(x) is a factor of f(x)

**(5) Let, f(x) = Px ^{2} + 5x + r** , g(x) = x – 2, p(x) = x – ½

Given, that both g(x) and p(x) are factors of f(x).

∴ By factor theorem,

Zero of g(x) = 2

F(2) = p x 2^{2} + 5 x 2 + r

Or, 0 = p x 4 + 10 + r

Or, 4p + r = -10 ….(i)

**(6) Let, f(x) = ax ^{4} + bx^{3} + cx^{2} + d + e ; **

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

∴ zero of g(x)

X^{2} – 1 = 0

Or, x^{2} = 1

Or, x = √1

= 1

∴ by factor theorem,

f(1) = ax1^{4} + bx1^{3} + cx1^{2} + dx1 + e

or, 0 = a + b + c + d + e

or, a + b+ c + d + e = 0 (Hence proved)

**(7) (i) x ^{3} – 2x^{2} – x + 2**

= x^{3} – x^{2} – x^{2} + x – 2x + 2 [divide the two middle terms in this case -2x^{2} and –x into two parts such that each pair out of the six terms when divided by power of x or a number gives the same value for all three pairs]

= x^{2} (x – 1) – x(x – 1) -2(x – 1)

= (x-1) (x^{2} –x – 2)

Further,

X^{2} –x -2

= x^{2} + x -2x – 2

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

= (x+1) (x-2)

∴ The factors of x^{3} -2x^{2} –x+ 2 are (x – 1), (X+1), (X-2).

∴ x^{2} – 4x – 5 is another factor of f(x)

Now, x^{2} – 4x – 5

= x^{2} + x – 5x – 5

= x(x + 1) – 5(x+1)

= (x + 1) (x – 5)

∴ (x+1) and (x+5) are the factors of (x).

**(iii) x ^{3} + 13x^{2} + 32x + 20**

= x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

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

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

Now,

X^{2} + 12x + 20

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

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

= (x + 10) (x + 2)

∴ (x +1), (x + 10) and (x + 2) are the factors of x^{3} + 13x^{2} + 32x + 20.

**(iv) y ^{3} + y^{2} – y – 1**

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

= y^{2} (y – 1) + 2y (y – 1) + 1(y – 1)

= (y – 1) (y^{2} + 2y + 1)

Now, y^{2} + 2y + 1

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

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

= (y + 1)^{2}

∴ The factors of y^{3} + y^{2} – y + 1 are (y – 1), (y + 1)^{2}

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

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

P(x) = bx^{2} + ax + c

g(x) = x + 1

∴ zero of g(x) = -1

Now, given that g(x) is a common factor of both f(x) and p(x).

∴ Using factor theorem we can say that

f(-1) = a(-1)^{2} + b x -1 + c

or, 0 = a – b – c

or, a – b + c = 0….(i)

similarly,

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

or, 0 = b – a + c

or, -a + b + c = 0 ….(ii)

Now, adding both equation (i) and (ii) we get

a – b + c = 0

– a + b + c = 0

____________

2c = 0

or, c = 0 (proved)

Putting the value of c in equation (i) we get

a – b + c = 0

or, a – b + 0 = 0

or, a = b (proved).

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

Let, f(x) = x^{2} – x – 6

P(x) = x^{2} + 3x – 18

g(x) = x – a

zero of g(x) = a

given that g(x) is the common factor of f(x) and p(x)

∴ By factor theorem

f(a) = a^{2} – a – 6

or, 0 = a^{2} – a – 6

or, a^{2} – a – 6 = 0

or, a^{2} + 2a – 3a – 6

or, a (a + 2) – 3 (a + 2)

or, (a+2) (a – 3) = 0

∴ either a = -2 or, a = 3.

Now also p(x) = a^{2} + 3 x a – 18

Or, 0 = a^{2} + 3a – 18

Or, a^{2} – 3a + 6a – 18 = 0

Or, a(a – 3) + 6(a – 3) = 0

Or, (a – 3) (a+6) = 0

∴ either a = 3 or, a = -6

∴ The values are -2, 3 , -6.

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

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