# Telangana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 2.4

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

(1) (i) given, p(x) = x3 – x2 – 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) x4 – x3 + x2 – 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) = x4 + 2x3 + 2x2 + 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) x3 – x2 – (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) = 5x3 + x2 – 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) = x3 + 3x2 + 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) = x3 – 4x2 + x + 6, g(x) = x – 2

∴ Zero of g(x) = +2

Now,

f(x) = 23 – 4(2)2 + 2 + 6

= 8 – 16 + 2 + 6

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

(3) Given, f(x) = x3 – 3x2 – 10x + 24;

Let, g(x) = x – 2

L(x) = x + 3

P(x) = x – 4

Now, zero of g(x) = +2

Then, f(2) = 23 – 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) = 43 – 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) = x3 – 6x2 – 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) = 33 – 6 x 32 – 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) = 73 – 6 x 72 – 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) = Px2 + 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 22 + 5 x 2 + r

Or, 0 = p x 4 + 10 + r

Or, 4p + r = -10 ….(i)

(6) Let, f(x) = ax4 + bx3 + cx2 + d + e ;

g(x) = x2 -1

∴ zero of g(x)

X2 – 1 = 0

Or, x2 = 1

Or, x = √1

= 1

∴ by factor theorem,

f(1) = ax14 + bx13 + cx12 + dx1 + e

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

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

(7) (i) x3 – 2x2 – x + 2

= x3 – x2 – x2 + x – 2x + 2 [divide the two middle terms in this case -2x2 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]

= x2 (x – 1) – x(x – 1) -2(x – 1)

= (x-1) (x2 –x – 2)

Further,

X2 –x -2

= x2 + x -2x – 2

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

= (x+1) (x-2)

∴ The factors of x3 -2x2 –x+ 2 are (x – 1), (X+1), (X-2).

∴ x2 – 4x – 5 is another factor of f(x)

Now, x2 – 4x – 5

= x2 + x – 5x – 5

= x(x + 1) – 5(x+1)

= (x + 1) (x – 5)

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

(iii) x3 + 13x2 + 32x + 20

= x3 + x2 + 12x2 + 12x + 20x + 20

= x2(x+1) + 12x(x+1) + 20(x+1)

= (x+1) (x2 + 12x + 20)

Now,

X2 + 12x + 20

= x2 + 10x + 2x + 20

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

= (x + 10) (x + 2)

∴ (x  +1), (x + 10) and (x + 2) are the factors of x3 + 13x2 + 32x + 20.

(iv) y3 + y2 – y – 1

= y3 y2 + 2y2 – 2y + y – 1

= y2 (y – 1) + 2y (y – 1) + 1(y – 1)

= (y – 1) (y2 + 2y + 1)

Now, y2 + 2y + 1

= y2 + y + y + 1

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

= (y + 1)2

∴ The factors of y3 + y2 – y + 1 are (y – 1), (y + 1)2

(8) 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

P(x) = bx2 + 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 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

P(x) = x2 + 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) = a2 – a – 6

or, 0  = a2 – a – 6

or, a2 – a – 6 = 0

or, a2 + 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) = a2 + 3 x a – 18

Or, 0 = a2 + 3a – 18

Or, a2 – 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 y3 − 2y2 − 9y + 18 then find the other two factors

Updated: September 18, 2021 — 4:44 pm