Ncert Solutions For Class 9 Maths Ex 2.4

Ncert Solutions For Class 9 Maths Chapter 2 Ex 2.4

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


(i)
x3+x2+x+1

Answer:

Let p(x)= x3+x2+x+1

The zero of x+1 is -1.

p(1)=(1)3+(1)2+(1)+1=1+11+1=0

By factor theorem, x+1 is a factor of x3+x2+x+1

 

(ii) x4 + x3 + x2 + x + 1

Answer:

Let p(x)= x4+x3+x2+x+1

The zero of x+1 is -1.

p(1)=(1)4+(1)3+(1)2+(1)+1=11+11+1=10 By factor theorem, x+1 is a factor of x4+x3+x2+x+1

 

 (iii) x4 + 3x3 + 3x2 + x + 1 

Answer:

Let p(x)= x4+3x3+3x2+x+1

The zero of x+1 is -1.

p(1)=(1)4+3(1)3+3(1)2+(1)+1=13+31+1=10 By factor theorem, x+1 is a factor of x4+3x3+3x2+x+1

 

(iv) x3x2(2+2)x+2

Answer:

Let p(x)= x3x2(2+2)x+2

The zero of x+1 is -1.

p(1)=(1)3(1)2(2+2)(1)+2

By factor theorem, x+1 is not a factor of x3x2(2+2)x+2

 

Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:


(i)
p(x)=2x3+x22x1, g(x) = x + 1

Answer: p(x)=2x3+x22x1, g(x) = x + 1

g(x)=0

x+1=0x=1

Zero of g(x) is -1.

Now, p(1)=2(1)3+(1)22(1)1=2+1+21=0

By factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x3+3x2+3x+1, g(x) = x + 2

Answer: p(x)=x3+3x2+3x+1, g(x) = x + 2

g(x)=0

x+2=0x=2

Zero of g(x) is -2.

Now, p(2)=(2)3+3(2)2+3(2)+1=8+126+1=1=0

By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x34x2+x+6, g(x) = x – 3

Answer: p(x)=x34x2+x+6, g(x) = x -3

g(x)=0

x3=0x=3

Zero of g(x) is 3.

Now, p(3)=(3)34(3)2+(3)+6=2736+3+6=0

By factor theorem, g(x) is a factor of p(x).

 

Q.3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i)p(x)=x2+x+k

Answer: If x-1 is a factor of p(x), then p(1)=0   |By Factor Theorem (1)2+(1)+k=01+1+k=0

2+k=0k=2

 

(ii) p(x)=2x2+kx+2

Answer: If x-1 is a factor of p(x), then p(1)=0

2(1)2+k(1)+2=02+k+2=0k=(2+2)

 

(iii) p(x)=kx22x+1

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

k(1)22(1)+1=0k=21
(iv) p(x)=kx23x+k

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

k(1)23(1)+k=0k3+k=02k3=0k=32

 

Q.4. Factorize:

(i) 12x27x+1

Answer:

12x27x+1=12x24x3x+1=4x(3x1)1(3x1)=(4x1)(3x1)

Let p(x)= 12x27x+1

Then p(x)= 12(x2712x+112)=12q(x)

Where q(x)= x2712x+112

By trial, we find that

q(13)=(13)2712(13)+112=47+336=036=0

By Factor Theorem,

(x13) is a factor of q(x).

Similarly, by trial, we find that

q(14)=(14)2712(14)+112=37+418=048=0

By Factor Theorem,

(x14) is a factor of q(x).

Therefore, 12x27x+1=12(x13)(x14)=12(3x13)(4x14)=(3x1)(4x1)

 

(ii)2x2+7x+3

Answer: 

2x2+7x+3=2x2+6x+x+3=2x(x+3)+1(x+3)=(x+3)(2x+1)

Let p(x)= 2x2+7x+3

Then p(x)= 2(x2+72x+32)=2q(x)
Where q(x)= x2+72x+32

By trial, we find that

q(3)=(3)272(3)+32=9212+32=0

By Factor Theorem,

(x3),i.e.(x+3) is a factor of q(x).

Similarly, by trial, we find that

q(12)=(12)2+712(12)+32=1374+34=0

By Factor Theorem,

(x(12)),i.e.(x+12)isafactorofq(x).Therefore,2x2+7x+3=2(x+3)(x+12)=2(x+3)(2x+12)=(x+3)(2x+1)

By Simpler method,

(iii) \(6x^{2}+5x-6 =6x^{2}+ 9x – 4x – 6

=3x (2x + 3) – 2 (2x + 3) = (2x + 3) (3x – 2)\)

 

(iv)3x2 – x – 4 =3x2x4=3×24x+3x4=x(3x4)+1(3x4)=(3x4)(x+1)

 

Q.5. Factorize:


(i)
x32x2x+2

Answer: Let p(x)=x32x2x+2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x)=x32x2x+2

p(1)=(1)32(1)2(1)+2=11+1+2=0

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

1

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x23x+2)=(x+1)(x2x2x+2)=(x+1)(x(x1)2(x1))=(x+1)(x1)(x+2)

 

 (ii) x33x29x5
Answer: Let p(x) = x33x29x5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x33x29x5

p(5) = (5)33(5)29(5)5=12575455=0

Therefore, (x-5) is the factor of  p(x)

2
Now, Dividend = Divisor × Quotient + Remainder

(x5)(x2+2x+1)=(x5)(x2+x+x+1)=(x5)(x(x+1)+1(x+1))=(x5)(x+1)(x+1)

 

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

Answer: Let p(x) = x3+13x2+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x3+13x2+32x+20

p(-1) = (1)3+13(1)2+32(1)+20=1+1332+20=0

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

3

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2+12x+20)=(x+1)(x2+2x+10x+20)=(x5)x(x+2)+10(x+2)=(x5)(x+2)(x+10)

 

(iv) 2y3+y22y1
Answer: Let p(y) = 2y3+y22y1

Factors of ab = 2×(1)= -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) = 2y3+y22y1

p(1) = 2(1)3+(1)22(1)1=2+12=0

Therefore, (y-1) is the factor of  p(y)

4

Now, Dividend = Divisor × Quotient + Remainder

(y1)(2y2+3y+1)=(y1)(2y2+2y+y+1)=(y1)(2y(y+1)+1(y+1))=(y1)(2y+1)(y+1)

 

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