1 NCRTSOLUTIONS.BLOGSPOT.COM Class IX Chapter 2 Polynomials Maths Exercise 2.1 Question 1: Which of the following expressions are polynomials in one v...

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Class IX Chapter 2 Polynomials Maths Exercise 2.1 Question 1:

Which of the following expressions are polynomials in one variable and which are

No. It can be observed that the exponent of variable t in term a whole number. Therefore, this expression is not a polynomial.

(iv)

is , which is not

No. It can be observed that the exponent of variable y in term whole number. Therefore, this expression is not a polynomial.

is −1, which is not a

(v) No. It can be observed that this expression is a polynomial in 3 variables x, y, and t. Therefore, it is not a polynomial in one variable.

In the above expression, the coefficient of

is

.

(iv)

In the above expression, the coefficient of

is 0.

Question 3: Give one example each of a binomial of degree 35, and of a monomial of degree 100. Answer: Degree of a polynomial is the highest power of the variable in the polynomial. Binomial has two terms in it. Therefore, binomial of degree 35 can be written as NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM . Monomial has only one term in it. Therefore, monomial of degree 100 can be written as x100.

Question 4: Write the degree of each of the following polynomials:

(ii) (iv) 3

(i) (iii)

Answer: Degree of a polynomial is the highest power of the variable in the polynomial. (i) This is a polynomial in variable x and the highest power of variable x is 3. Therefore, the degree of this polynomial is 3.

(ii) This is a polynomial in variable y and the highest power of variable y is 2. Therefore, the degree of this polynomial is 2.

(iii) This is a polynomial in variable t and the highest power of variable t is 1. Therefore, the degree of this polynomial is 1.

(iv) 3

This is a constant polynomial. Degree of a constant polynomial is always 0. Question 5: Classify the following as linear, quadratic and cubic polynomial:

(i) (vi) Answer: Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

(i)

is a quadratic polynomial

as its degree is 2. (ii) is a cubic polynomial as its degree is 3. (iii) is a quadratic polynomial as its degree is 2. (iv)

1 + x is a linear polynomial as its degree is 1.

(v)

is a linear polynomial as its degree is 1.

(vi)

is a quadratic polynomial as its degree is 2.

(vii)

is a cubic polynomial as its degree is 3.

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NCRTSOLUTIONS.BLOGSPOT.COM

Exercise 2.2 Question 1: Find the value of the polynomial

at

(i) x = 0 (ii) x = −1 (iii) x = 2

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

(i) p(y) = y2 − y + 1 (ii) p(t) = 2 + t + 2t2 − t3 (iii) p(x) = x3 (iv) p(x) = (x − 1) (x + 1) Answer: (i) p(y) = y2 − y + 1 p(0) = (0)2 − (0) + 1 = 1 p(1) = (1)2 − (1) + 1 = 1 p(2) = (2)2 − (2) + 1 = 3 (ii) p(t) = 2 + t + 2t2 − t3 p(0) = 2 + 0 + 2 (0)2 − (0)3 = 2 p(1) = 2 + (1) + 2(1)2 − (1)3 = 2 + 1 + 2 − 1 = 4 p(2) = 2 + 2 + 2(2)2 − (2)3 =2+2+8−8=4 (iii) p(x) = x3 p(0) = (0)3 = 0 p(1) = (1)3 = 1 p(2) = (2)3 = 8 (iv) p(x) = (x − 1) (x + 1) p(0) = (0 − 1) (0 + 1) = (− 1)

(1) = − 1 p(1) = (1 − 1) (1 + 1) = 0 (2) = 0 p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3 Question 3:

NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM

should be 0.

Here, p(1) = (1)2 − 1 = 0, and p(− 1) = (− 1)2 − 1 = 0 Hence, x = 1 and −1 are zeroes of the given polynomial. (iv) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0 Therefore, x = −1 and x = 2 are zeroes of the given polynomial. (v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.

NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM

= 3x − 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, are real numbers.

Answer: Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = x + 5 p(x) =0x+5=0x=−5 Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial. (ii)

p(x) = x − 5 p(x) = 0 x − 5 =0x=5 Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial. (iii) p(x) = 2x + 5 p(x) = 0

2x + 5 = 0 2x = − 5

Therefore, for , the value of the polynomial is 0 and hence, given polynomial. (iv) p(x) = 3x − 2 p(x) = 0

is a zero of the

3x − 2 = 0

Therefore, for , the value of the polynomial is 0 and hence, given polynomial. (v) p(x) = 3x p(x) = 0 3x = 0 x = 0

is a zero of the

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial. (vi) p(x) = ax p(x) = 0 ax = 0 x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial. (vii) p(x) = cx + d p(x) = 0 cx+ d = 0

Therefore, for

, the value of the polynomial is 0 and hence,

given polynomial.

NCRTSOLUTIONS.BLOGSPOT.COM

is a zero of the

NCRTSOLUTIONS.BLOGSPOT.COM

Exercise 2.3 Question 1:

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

(i) x + 1 (ii)

(iii) x

(iv) x + π (v) 5 + 2x Answer:

NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM

Therefore, the remainder is (iii) x By long division,

Therefore, the remainder is 1.

NCRTSOLUTIONS.BLOGSPOT.COM

.

NCRTSOLUTIONS.BLOGSPOT.COM (iv) x + π By long division,

Therefore, the remainder is

(v) 5 + 2x By long division,

Therefore, the remainder is Question 2: Find the remainder when x3 − ax2 + 6x − a is divided by x − a. Answer: By long division,

Therefore, when x3 − ax2 + 6x − a is divided by x − a, the remainder obtained is 5a. NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM Question 3: Check whether 7 + 3x is a factor of 3x3 + 7x. Answer: Let us divide (3x3 + 7x) by (7 + 3x). If the remainder obtained is 0, then 7 + 3x will be a factor of 3x3 + 7x. By long division,

As the remainder is not zero, therefore, 7 + 3x is not a factor of 3x 3 + 7x.

Exercise 2.4

Question 1: Determine which of the following polynomials has (x + 1) a factor: (i) x3 + x2 + x + 1 (ii) x4 + x3 + x2 + x + 1 (iii) x4 + 3x3 + 3x2 + x + 1 (iv) Answer: (i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x3 + x2 + x + 1 p(−1) = (−1)3 + (−1)2 + (−1) + 1 =−1+1−1−1=0 Hence, x + 1 is a factor of this polynomial. (ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x). p(x) = x4 + x3 + x2 + x + 1 p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1 = 1 − 1 + 1 −1 + 1 = 1 As p ≠ 0, (− 1) Therefore, x + 1 is not a factor of this polynomial. (iii) If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial. NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM p(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1 =1−3+3−1+1=1 As p ≠ 0, (−1)

Question 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 + x2 − 2x − 1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 − 4 x2 + x + 6, g(x) = x − 3 Answer:

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero. p(x) = 2x3 + x2 − 2x − 1 p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1 = 2(−1) + 1 + 2 − 1 = 0 Hence, g(x) = x + 1 is a factor of the given polynomial. (ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must be 0.

p(x) = x3 +3x2 + 3x + 1 p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1 = − 8 + 12 − 6 + 1 = −1 As p ≠ 0, (−2)

Hence, g(x) = x + 2 is not a factor of the given polynomial. (iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then p(3) must be 0.

p(x) = x3 − 4 x2 + x + 6 p(3) = (3)3 − 4(3)2 + 3 + 6 = 27 − 36 + 9 = 0 Hence, g(x) = x − 3 is a factor of the given polynomial. Question 3: Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:

If x − 1 is a factor of polynomial p(x), then p(1) must be 0. (i) p(x) = x2 + x + k p(1) =0 (1)

2

+1+k=0

(2)

+k=0

k

= −2

NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM

+k=0 2k − 3 = 0

Therefore, the value of k is

.

Question 4: Factorise: (i) 12x2 − 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x − 6 (iv) 3x2 − x − 4 Answer: (i) 12x2 − 7x + 1 We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.

Here, 12x2 − 7x + 1 = 12x2 − 4x − 3x + 1 = 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1) (ii) 2x2 + 7x + 3 We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7. They are p = 6 and q = 1. Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3 = 2x (x + 3) + 1 (x + 3) = (x + 3) (2x+ 1) (iii) 6x2 + 5x − 6 We can find two numbers such that pq = −36 and p + q = 5. They are p = 9 and q = −4. Here, 6x2 + 5x − 6 = 6x2 + 9x − 4x − 6 = 3x (2x + 3) − 2 (2x + 3) = (2x + 3) (3x − 2) (iv) 3x2 − x − 4 We can find two numbers such that pq = 3 × (− 4) = −12 and p + q = −1.

They are p = −4 and q = 3. Here, 3x2 − x − 4 = 3x2 − 4x + 3x − 4 = x (3x − 4) + 1 (3x − 4) = (3x − 4) (x + 1) Question 5:

Factorise: (i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9x − 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1 Answer:

(i) Let p(x) = x3 − 2x2 − x + 2 All the factors of 2 have to be considered. These are ± 1, ± 2. By trial method, p(2) = (2)3 − 2(2)2 − 2 + 2 NCRTSOLUTIONS.BLOGSPOT.COM

NCRTSOLUTIONS.BLOGSPOT.COM =8−8−2+2=0 Therefore, (x − 2) is factor of polynomial p(x). Let us find the quotient on dividing x3 − 2x2 − x + 2 by x − 2. By long division,

It is known that, Dividend = Divisor × Quotient + Remainder

x3

− 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0 = (x + 1) [x2 − 2x − x + 2] = (x + 1) [x (x − 2) − 1 (x − 2)] = (x + 1) (x − 1) (x − 2) = (x − 2) (x − 1) (x + 1) (ii) Let p(x) = x3 − 3x2 − 9x − 5 All the factors of 5 have to be considered. These are ±1, ± 5. By trial method, p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5 =−1−3+9−5=0 Therefore, x + 1 is a factor of this polynomial. Let us find the quotient on dividing x3 + 3x2 − 9x − 5 by x + 1. By long division,

It is known that, Dividend = Divisor × Quotient + Remainder

x3

− 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0 = (x + 1) (x2 − 5x + x − 5) = (x + 1) [(x (x − 5) +1 (x − 5)] = (x + 1) (x − 5) (x + 1) = (x − 5) (x + 1) (x + 1) (iii) Let p(x) = x3 + 13x2 + 32x + 20 All the factors of 20 have to be considered. Some of them are ±1, ± 2, ± 4, ± 5 …… By trial method, p(−1) = (−1)3 + 13(−1)2 +

32(−1) + 20

= − 1 +13 − 32 + 20 = 33 − 33 = 0

NCRTSOLUTIONS.BLOGSPOT.COM

Cbse-spot.blogspot.com

As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x). Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).

It is known that, Dividend = Divisor × Quotient + Remainder x 3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0 = (x + 1) (x2 + 10x + 2x + 20) = (x + 1) [x (x + 10) + 2 (x + 10)] = (x + 1) (x + 10) (x + 2) = (x + 1) (x + 2) (x + 10) (iv) Let p(y) = 2y3 + y2 − 2y − 1 By trial method, p(1) = 2 ( 1)3 +

(1)2 − 2( 1) − 1 = 2 + 1 − 2 − 1= 0 Therefore, y − 1 is a factor of this polynomial. Let us find the quotient on dividing 2y3 + y2 − 2y − 1 by y − 1.

p(y) = 2y3 + y2 − 2y − 1 = (y − 1) (2y2 +3y + 1) = (y − 1) (2y2 +2y + y +1) = (y − 1) [2y (y + 1) + 1 (y + 1)] = (y − 1) (y + 1) (2y + 1) Question 5:

Factorise: (i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9x − 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1 Answer:

(i) Let p(x) = x3 − 2x2 − x + 2 All the factors of 2 have to be considered. These are ± 1, ± 2. By trial method, p(2) = (2)3 − 2(2)2 − 2 + 2

Cbse-spot.blogspot.com =8−8−2+2=0 Therefore, (x − 2) is factor of polynomial p(x). Let us find the quotient on dividing x3 − 2x2 − x + 2 by x − 2.

It is known that, Dividend = Divisor × Quotient + Remainder x3 − 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0

= (x + 1) [x2 − 2x − x + 2] = (x + 1) [x (x − 2) − 1 (x − 2)] = (x + 1) (x − 1) (x − 2) = (x − 2) (x − 1) (x + 1) (ii) Let p(x) = x3 − 3x2 − 9x − 5 All the factors of 5 have to be considered. These are ±1, ± 5. By trial method, p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5 =−1−3+9−5=0 Therefore, x + 1 is a factor of this polynomial. Let us find the quotient on dividing x3 + 3x2 − 9x − 5 by x + 1. By long division,

It is known that, Dividend = Divisor × Quotient + Remainder

x3

− 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0 = (x + 1) (x2 − 5x + x − 5) = (x + 1) [(x (x − 5) +1 (x − 5)] = (x + 1) (x − 5) (x + 1) = (x − 5) (x + 1) (x + 1) (iii) Let p(x) = x3 + 13x2 + 32x + 20 All the factors of 20 have to be considered. Some of them are ±1, ± 2, ± 4, ± 5 …… By trial method, p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20 = − 1 +13 − 32 + 20 = 33 − 33 = 0 As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x). Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).

Cbse-spot.blogspot.com

It is known that, Dividend = Divisor × Quotient + Remainder x 3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0 = (x + 1) (x2 + 10x + 2x + 20) = (x + 1) [x (x + 10) + 2 (x + 10)] = (x + 1) (x + 10) (x + 2) = (x + 1) (x + 2) (x + 10) (iv) Let p(y) = 2y3 + y2 − 2y − 1 By trial method, p(1) = 2 ( 1)3 + (1)2 − 2( 1) − 1 = 2 + 1 − 2 − 1= 0 Therefore, y − 1 is a factor of this polynomial. Let us find the quotient on dividing 2y3 + y2 − 2y − 1 by y − 1.

p(y) = 2y3 + y2 − 2y − 1 = (y − 1) (2y2 +3y + 1) = (y − 1) (2y2 +2y + y +1) = (y − 1) [2y (y + 1) + 1 (y + 1)] = (y − 1) (y + 1) (2y + 1) Exercise 2.5 Question 1: Use suitable identities to find the following products:

Cbse-spot.blogspot.com

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96 Answer: (i) 103 × 107 = (100 + 3) (100 + 7) = (100)2 + (3 + 7) 100 + (3) (7) [By using the identity 100, a = 3, and b = 7] = 10000 + 1000 + 21 = 11021

, where x =

(ii) 95 × 96 = (100 − 5) (100 − 4) = (100)2 + (− 5 − 4) 100 + (− 5) (− 4) [By using the identity 100, a = −5, and b = −4] = 10000 − 900 + 20 = 9120 (iii) 104 × 96 = (100 + 4) (100 − 4) = (100)2 − (4)2 = 10000 − 16 = 9984 Question 3: Factorise the following using appropriate identities:

Cbse-spot.blogspot.com

, where x =

Cbse-spot.blogspot.com

It is known that,

(i) (99)3 = (100 − 1)3 = (100)3 − (1)3 − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299

Cbse-spot.blogspot.com

(ii) (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100) (2) (100 + 2) = 1000000 + 8 + 600 (102) = 1000000 + 8 + 61200 = 1061208 (iii) (998)3= (1000 − 2)3 = (1000)3 − (2)3 − 3(1000) (2) (1000 − 2) = 1000000000 − 8 − 6000(998) = 1000000000 − 8 − 5988000 = 1000000000 − 5988008 = 994011992 Question 8:

Cbse-spot.blogspot.com

It is known that,

Cbse-spot.blogspot.com

Without actually calculating the cubes, find the value of each of the following:

It is known that if x + y + z = 0, then

Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given:

Answer: Area = Length × Breadth The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.

Cbse-spot.blogspot.com

(ii)

Therefore, possible length = 5y + 4 And, possible breadth = 7y − 3 Question 16:

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

Answer: Volume of cuboid = Length × Breadth × Height The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

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