NCERT Solutions For Class 9 Maths Chapter 2

NCERT Solutions Class 9 Maths Polynomials

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NCERT Solutions For Class 9 Maths Chapter 2 Exercises

Exercise – 1

Q.1.Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i)4x23x+7

Answer: It is a polynomial in one variable.

 

(ii) y2+2

Answer: It is a polynomial in one variable.

 

(iii) 3t+t2

Answer: It is not a polynomial since the power of the variable is not a whole number.

 

(iv) y+2y

Answer: It is not a polynomial since the power of the variable is not a whole number.

 

(v) x10+y3+t50

Answer: It is a polynomial in three variables.

 

Q.2. Write the coefficients of a2 in each of the following:

(i) 2+a2+a

Answer: Coefficient of a2 is 1.

 

 (ii) 2a2+a3

Answer: Coefficient of a2 is -1.

 

 (iii) π2x2+x

Answer: Coefficient of a2 is π2.


(iv)
2x1

Answer: Coefficient of a2 is 0.

 

Q.3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer: 3x35+5and4x100

 

Q.4. Write the degree of each of the following polynomials:

(i) 5a3+4a2+7a

Answer : Degree is 3.
(ii) 3b2

Answer : Degree is 2.
(iii) 5t8

Answer : Degree is 1.

 

(iv)8

Answer : No Degree.

 

Q.5. Classify the following as linear, quadratic and cubic polynomial.

(i) a2+a

Answer: Quadratic Polynomial

 

(ii)aa3

Answer: Cubic Polynomial

 

(iii) y+y2+4

Answer: Quadratic Polynomial

 

(iv)1 + x

Answer: Linear Polynomial

 

(v) 3a

Answer: Linear Polynomial

 

(vi) a2

Answer: Quadratic Polynomial
(vii) 6a3

Answer: Cubic Polynomial

 

Exercise – 2

Q.1. Find the value of the polynomial atf(x)=5a4a2+3 at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Answer:

Let f(x)=5a4a2+3

 

(i) When a=0

f(0)=5(0)+4(0)2+3=3

 

(ii) When a=-1

f(a)=5a+4a2+3f(1)=5(1)+4(1)2+3=54+3=6

 

(iii) When a=2

f(a)=5a+4a2+3f(2)=5(2)+4(2)2+3=1016+3=3

 

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

 (i) p(y)=y2y+1

Answer: p(y)=y2y+1p(0)=(0)2(0)+1=1p(1)=(1)2(1)+1=1p(2)=(2)2(2)+1=3

 

 (ii) p(a)=2+a+2a2a3

Answer: p(a)=2+a+2a2a3p(0)=2+0+2(0)2(0)3=2p(1)=2+1+2(1)2(1)3=2+1+21=4p(2)=2+2+2(2)2(2)3=2+2+88=4

 

(iii) p(x)=x3

Answer: p(x)=x3p(0)=(0)3=0p(1)=(1)3=1p(2)=(2)3=8

 

(iv) p(a)=(a1)(a+1)

Answer: p(a)=(a1)(a+1)p(0)=(01)(0+1)=(1)(1)=1p(1)=(11)(1+1)=0(2)=0p(2)=(21)(2+1)=1(3)=3

 

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x)=3x+1,x=13

Answer: For, x=13

p(x)=3x+1p(13)=3(13)+1=1+1=0

13 is a zero of p(x).

 

 (ii) p(x)=5xπ,x=45

Answer: For, x=45

p(x)=5xπp(45)=5(45)π=4π

45 is not a zero of p(x).

 

 (iii) p(x)=x21,x=1,1

Answer: For, x=1,1

p(x)=x21p(1)=x11=11=0p(1)=x11=11=0

1,1 are zeros of p(x).

 

(iv) p(x)=(x+1)(x2),x=1,2

Answer: For, x=1,2

p(x)=(x+1)(x2)p(1)=(1+1)(12)=((0)(3))=0p(2)=(2+1)(22)=(3)(0)=0

1,2 are zeros of p(x).

 

(v) p(x)=x2,x=0

Answer: For, x=0

p(x)=02=0

0 is a  zero of p(x).

 

(vi) p(x)=lx+m,x=ml

Answer: For, x=ml

p(x)=lx+mp(ml)=l(ml)+m=m+m=0

ml is a zero of p(x).

 

(vii) p(x)=3x21,x=13,23

Answer: For, x=13,23
p(x)=3x21p(13)=3(13)21=3(13)1=11=0p(23)=3(23)21=3(43)1=41=30

13 is a zero of p(x) but 23 is not a zero of p(x).

 

(viii) p(x)=2x+1,x=12
Answer: For, x=12

p(x)=2x+1p(12)=2(12)+1=1+1=20

12 is not a  zero of p(x).

 

Q.4. Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5 

Answer: p(x)=x+5x+5=0x=5

-5 is a zero polynomial of the polinomila p(x).

 

(ii) p(x) = x – 5

Answer: p(x)=x5x5=0x=5

5 is a zero polynomial of the polinomila p(x).

 

(iii) p(x) = 2x + 5

Answer: p(x)=2x+52x+5=02x=5x=52

x=52 is a zero polynomial of the polynomial p(x).

 

(iv)p(x) = 3x – 2 

Answer: p(x)=3x23x2=03x=2x=23

x=23 is a zero polynomial of the polynomial p(x).

 

 (v)p(x) = 3x 

Answer: p(x)=3x3x=0x=0

0 is a zero polynomial of the polynomial p(x).

 

Exercise – 3

Q.1.Find the remainder when x3+3x2+3x+1 is divided by

(i) x+1

Answer:

x+1=0x=1 Remainder=p(1)=(1)3+3(1)2+3(1)+1=1+33+1=0

 

(ii) x12

Answer:

x12=0x=12 Remainder(12)3+3(12)2+3(12)+1=18+34+32+1=278

 

(iii) x

Answer:

Remainder=(0)3+3(0)2+3(0)+1=1

 

(iv) x+π

Answer:

x+π=0x=π Remainder=(π)3+3(π)2+3(π)+1=π3+3π23π+1

 

(v) 5+2x

Answer:

5+2x=02x=5x=52 Remainder=(52)3+3(52)2+3(52)+1=1258+754152+1=278

 

Q.2.Find the remainder when x3ax2+6xa is divided by x-a.

Answer:

Let p(x)=x3ax2+6xaxa=0

x=a Remainder=(a)2a(a2)+6(a)a=a3a3+6aa=5a

 

Q.3.Check whether 7+3x is a factor of 3x3+7x.

Answer:

7+3x=03x=7 only if 7+3x divides 3x3+7x leaving no reaminder.

Let p(x)=3x3+7x

7+3x=03x=7x=73 Remainder=3(73)3+7(73)=3439493=49090

7+3x is not a factor of 3x3+7x

 

Exercise – 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)

 

Exercise – 5

Q.1.Use suitable identities to find the following products:

 

(i)(x + 4) (x + 10) 

Answer: (x+4)(x+10)=x2+(4+10)x+(4×10)=x2+14x+40
(ii)(x + 8) (x – 10)     

Answer: (x+8)(x10)=x2+(8+(10))x+(8×(10))=x2+(810)x80=x22x80

 

(iii)(3x + 4) (3x – 5)

Answer: (3x+4)(3x5)=(3x)2+4+(5)3x+4×(5)=9x2+3x(45)20=9x23x20
(iv)(y2+32)(y232)

Answer: (y2+32)(y232)=(y2)2(32)2=y494

 

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   103×107=(100+3)×(100+7)=(100)2+(3+7)(100+(3×7))=10000+1000+21=11021

 

(ii) 95 × 96  

Answer: 95×96=(90+5)×(90+6)=(90)2+90(5+6)+(5×6)=8100+990+30=9120

 

(iii) 104 × 96
Answer: 104×96=(100+4)×(1004)=(100)2(4)2=1000016=9984

 

Q.3. Factorize the following using appropriate identities:

(i)9x2+6xy+y2

Answer: 9x2+6xy+y2=(3x)2+(2×3x×y)+y2=(3x+y)2=(3x+y)(3x+y)

 

(ii) 4y24y+1

Answer: 4y24y+1=(2y)2(2×2y×1)+12=(2y1)2=(2y1)(2y1)

 

 (iii) x2y2100

Answer: x2(y10)2=(xy10)(x+y10)

 

Q.4. Expand each of the following, using suitable identities:

(i) (x+2y+4z)2

(ii) (2xy+z)2

(iii) (2x+3y+2z)2

 (iv) [14a12b+1]2

Answer:

(i) (x+2y+4z)2

Using identity,

(x+2y+4z)2=x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)=x2+4y2+16z2+4xy+16yz+8xz
(ii)  (2xy+z)2
Using identity,

(2xy+z)2