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
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