POLYNOMIALS - Q&A

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

(i) 4x² - 3x + 7

Yes, this is a polynomial in one variable (x).
Reason: All exponents of x are whole numbers (2 and 1).

(ii) y² + √2

Yes, this is a polynomial in one variable (y).
Reason: The exponent of y is a whole number (2).

(iii) 3√t + t√2

No, this is not a polynomial.
Reason: The term 3√t can be written as 3t^1/2. Since the exponent 1/2 is not a whole number, it is not a polynomial.

(iv) y + 2/y

No, this is not a polynomial.
Reason: The term 2/y can be written as 2y^-1. Since the exponent -1 is not a whole number, it is not a polynomial.

(v) x¹⁰ + y³ + t⁵⁰

No, this is not a polynomial in one variable.
Reason: It contains three variables: x, y, and t.

2. Write the coefficients of x² in each of the following:

(i) 2 + x² + x

The coefficient of x² is 1 (since x² is the same as 1x²).

(ii) 2 - x² + x³

The coefficient of x² is -1 (since -x² is the same as -1x²).

(iii) (π/2)x² + x

The coefficient of x² is π/2.

(iv) √2x - 1

The coefficient of x² is 0.
Reason: There is no x² term in the expression, so we can think of it as 0x² + √2x - 1.

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

Binomial of degree 35: x³⁵ + 10 (It has two terms and the highest power is 35).
Monomial of degree 100: 5y¹⁰⁰ (It has one term and the highest power is 100).

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

(i) 5x³ + 4x² + 7x

Degree: 3
Reason: The highest power of the variable x is 3.

(ii) 4 - y²

Degree: 2
Reason: The highest power of the variable y is 2.

(iii) 5t - √7

Degree: 1
Reason: The highest power of the variable t is 1 (since t = t¹).

(iv) 3

Degree: 0
Reason: 3 can be written as 3x⁰. The degree of a non-zero constant polynomial is 0.

5. Classify the following as linear, quadratic and cubic polynomials:

(i) x² + x

Quadratic (Degree is 2).

(ii) x - x³

Cubic (Degree is 3).

(iii) y + y² + 4

Quadratic (Degree is 2).

(iv) 1 + x

Linear (Degree is 1).

(v) 3t

Linear (Degree is 1).

(vi) r²

Quadratic (Degree is 2).

(vii) 7x³

Cubic (Degree is 3).

1. Find the value of the polynomial 5x - 4x² + 3 at

(i) x = 0

Put x = 0 in p(x) = 5x - 4x² + 3:
p(0) = 5(0) - 4(0)² + 3
= 0 - 0 + 3
= 3.

(ii) x = -1

Put x = -1 in p(x):
p(-1) = 5(-1) - 4(-1)² + 3
= -5 - 4(1) + 3
= -5 - 4 + 3
= -6.

(iii) x = 2

Put x = 2 in p(x):
p(2) = 5(2) - 4(2)² + 3
= 10 - 4(4) + 3
= 10 - 16 + 3
= -3.

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

(i) p(y) = y² - y + 1

p(0) = (0)² - 0 + 1 = 1
p(1) = (1)² - 1 + 1 = 1 - 1 + 1 = 1
p(2) = (2)² - 2 + 1 = 4 - 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) = 0³ = 0
p(1) = 1³ = 1
p(2) = 2³ = 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

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

(i) p(x) = 3x + 1, x = -1/3

p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0.
Since p(-1/3) = 0, x = -1/3 is a zero of the polynomial.

(ii) p(x) = 5x - π, x = 4/5

p(4/5) = 5(4/5) - π = 4 - π ≠ 0.
So, x = 4/5 is not a zero.

(iii) p(x) = x² - 1, x = 1, -1

p(1) = (1)² - 1 = 0.
p(-1) = (-1)² - 1 = 1 - 1 = 0.
Both are zeroes.

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

p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0.
p(2) = (2 + 1)(2 - 2) = (3)(0) = 0.
Both are zeroes.

(v) p(x) = x², x = 0

p(0) = 0² = 0.
Yes, it is a zero.

(vi) p(x) = lx + m, x = -m/l

p(-m/l) = l(-m/l) + m = -m + m = 0.
Yes, it is a zero.

(vii) p(x) = 3x² - 1, x = -1/√3, 2/√3

p(-1/√3) = 3(-1/√3)² - 1 = 3(1/3) - 1 = 1 - 1 = 0. (Yes)
p(2/√3) = 3(2/√3)² - 1 = 3(4/3) - 1 = 4 - 1 = 3 ≠ 0. (No)

(viii) p(x) = 2x + 1, x = 1/2

p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0.
No, it is not a zero.

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

(i) p(x) = x + 5

To find the zero, set p(x) = 0.
x + 5 = 0 ⇒ x = -5.

(ii) p(x) = x - 5

x - 5 = 0 ⇒ x = 5.

(iii) p(x) = 2x + 5

2x + 5 = 0 ⇒ 2x = -5 ⇒ x = -5/2.

(iv) p(x) = 3x - 2

3x - 2 = 0 ⇒ 3x = 2 ⇒ x = 2/3.

(v) p(x) = 3x

3x = 0 ⇒ x = 0.

(vi) p(x) = ax, a ≠ 0

ax = 0 ⇒ x = 0.

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

cx + d = 0 ⇒ cx = -d ⇒ x = -d/c.

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

(i) x³ + x² + x + 1

Let p(x) = x³ + x² + x + 1.
The zero of x + 1 is -1.
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 - 1 + 1 = 0.
Since p(-1) = 0, (x + 1) is a factor.

(ii) x⁴ + x³ + x² + x + 1

p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 - 1 + 1 - 1 + 1 = 1 ≠ 0.
So, (x + 1) is not a factor.

(iii) x⁴ + 3x³ + 3x² + x + 1

p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
= 1 - 3 + 3 - 1 + 1 = 1 ≠ 0.
So, (x + 1) is not a factor.

(iv) x³ - x² - (2 + √2)x + √2

p(-1) = (-1)³ - (-1)² - (2 + √2)(-1) + √2
= -1 - 1 + (2 + √2) + √2
= -2 + 2 + √2 + √2 = 2√2 ≠ 0.
So, (x + 1) is not a factor.

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

Zero of g(x) is -1.
p(-1) = 2(-1)³ + (-1)² - 2(-1) - 1
= -2 + 1 + 2 - 1 = 0.
Yes, g(x) is a factor.

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

Zero of g(x) is -2.
p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1
= -8 + 12 - 6 + 1 = -1 ≠ 0.
No, g(x) is not a factor.

(iii) p(x) = x³ - 4x² + x + 6, g(x) = x - 3

Zero of g(x) is 3.
p(3) = (3)³ - 4(3)² + 3 + 6
= 27 - 36 + 3 + 6 = 0.
Yes, g(x) is a factor.

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

Since x - 1 is a factor, p(1) = 0.
(1)² + 1 + k = 0
1 + 1 + k = 0 ⇒ k = -2.

(ii) p(x) = 2x² + kx + √2

p(1) = 0
2(1)² + k(1) + √2 = 0
2 + k + √2 = 0 ⇒ k = -(2 + √2).

(iii) p(x) = kx² - √2x + 1

p(1) = 0
k(1)² - √2(1) + 1 = 0
k - √2 + 1 = 0 ⇒ k = √2 - 1.

(iv) p(x) = kx² - 3x + k

p(1) = 0
k(1)² - 3(1) + k = 0
k - 3 + k = 0
2k = 3 ⇒ k = 3/2.

4. Factorise:

(i) 12x² - 7x + 1

Find two numbers whose product is 12 × 1 = 12 and sum is -7. Numbers are -4 and -3.
= 12x² - 4x - 3x + 1
= 4x(3x - 1) - 1(3x - 1)
= (3x - 1)(4x - 1).

(ii) 2x² + 7x + 3

Product = 6, Sum = 7. Numbers are 6 and 1.
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1).

(iii) 6x² + 5x - 6

Product = -36, Sum = 5. Numbers are 9 and -4.
= 6x² + 9x - 4x - 6
= 3x(2x + 3) - 2(2x + 3)
= (2x + 3)(3x - 2).

(iv) 3x² - x - 4

Product = -12, Sum = -1. Numbers are -4 and 3.
= 3x² - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1).

5. Factorise:

(i) x³ - 2x² - x + 2

= x²(x - 2) - 1(x - 2)
= (x - 2)(x² - 1)
= (x - 2)(x - 1)(x + 1).

(ii) x³ - 3x² - 9x - 5

Let p(x) = x³ - 3x² - 9x - 5.
p(-1) = -1 - 3 + 9 - 5 = 0, so (x + 1) is a factor.
Dividing p(x) by (x + 1), we get x² - 4x - 5.
Now factorise x² - 4x - 5:
= x² - 5x + x - 5 = x(x - 5) + 1(x - 5) = (x - 5)(x + 1).
So, factors are (x + 1)(x + 1)(x - 5) or (x + 1)²(x - 5).

(iii) x³ + 13x² + 32x + 20

Let p(x) = x³ + 13x² + 32x + 20.
p(-1) = -1 + 13 - 32 + 20 = 0, so (x + 1) is a factor.
Dividing p(x) by (x + 1), we get x² + 12x + 20.
Now factorise x² + 12x + 20:
= (x + 10)(x + 2).
So, factors are (x + 1)(x + 2)(x + 10).

(iv) 2y³ + y² - 2y - 1

= 2y(y² - 1) + 1(y² - 1) [Regrouping terms: 2y³ - 2y + y² - 1]
Or grouping first two and last two:
= y²(2y + 1) - 1(2y + 1)
= (2y + 1)(y² - 1)
= (2y + 1)(y - 1)(y + 1).

1. Use suitable identities to find the following products:

(i) (x + 4)(x + 10)

Using (x + a)(x + b) = x² + (a + b)x + ab
= x² + (4 + 10)x + (4)(10)
= x² + 14x + 40.

(ii) (x + 8)(x - 10)

Using (x + a)(x + b) = x² + (a + b)x + ab
= x² + (8 - 10)x + (8)(-10)
= x² - 2x - 80.

(iii) (3x + 4)(3x - 5)

Using identity with y = 3x
= (3x)² + (4 - 5)(3x) + (4)(-5)
= 9x² - 3x - 20.

(iv) (y² + 3/2)(y² - 3/2)

Using (a + b)(a - b) = a² - b²
= (y²)² - (3/2)²
= y⁴ - 9/4.

(v) (3 - 2x)(3 + 2x)

Using (a - b)(a + b) = a² - b²
= 3² - (2x)²
= 9 - 4x².

2. Evaluate the following products without multiplying directly:

(i) 103 × 107

= (100 + 3)(100 + 7)
= (100)² + (3 + 7)(100) + (3)(7)
= 10000 + 1000 + 21
= 11021.

(ii) 95 × 96

= (100 - 5)(100 - 4)
= (100)² + (-5 - 4)(100) + (-5)(-4)
= 10000 - 900 + 20
= 9120.

(iii) 104 × 96

= (100 + 4)(100 - 4)
= (100)² - (4)²
= 10000 - 16
= 9984.

3. Factorise the following using appropriate identities:

(i) 9x² + 6xy + y²

= (3x)² + 2(3x)(y) + (y)²
= (3x + y)².

(ii) 4y² - 4y + 1

= (2y)² - 2(2y)(1) + (1)²
= (2y - 1)².

(iii) x² - y²/100

= x² - (y/10)²
= (x - y/10)(x + y/10).

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

(i) (x + 2y + 4z)²

= x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx.

(ii) (2x - y + z)²

= (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² - 4xy - 2yz + 4zx.

(iii) (-2x + 3y + 2z)²

= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9y² + 4z² - 12xy + 12yz - 8zx.

(iv) (3a - 7b - c)²

= (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a² + 49b² + c² - 42ab + 14bc - 6ca.

(v) (-2x + 5y - 3z)²

= (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x² + 25y² + 9z² - 20xy - 30yz + 12zx.

(vi) [(1/4)a - (1/2)b + 1]²

= (a/4)² + (-b/2)² + (1)² + 2(a/4)(-b/2) + 2(-b/2)(1) + 2(1)(a/4)
= a²/16 + b²/4 + 1 - ab/4 - b + a/2.

5. Factorise:

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
(Since yz and xz terms are negative, z term must be negative)
= (2x + 3y - 4z)².

(ii) 2x² + y² + 8z² - 2√2xy + 4√2yz - 8xz

= (-√2x)² + y² + (2√2z)² + 2(-√2x)(y) + 2(y)(2√2z) + 2(2√2z)(-√2x)
(Since xy and xz are negative, x term is negative)
= (-√2x + y + 2√2z)².

6. Write the following cubes in expanded form:

(i) (2x + 1)³

Using (a + b)³ = a³ + b³ + 3ab(a + b)
= (2x)³ + 1³ + 3(2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x² + 6x
= 8x³ + 12x² + 6x + 1.

(ii) (2a - 3b)³

Using (a - b)³ = a³ - b³ - 3ab(a - b)
= (2a)³ - (3b)³ - 3(2a)(3b)(2a - 3b)
= 8a³ - 27b³ - 18ab(2a - 3b)
= 8a³ - 27b³ - 36a²b + 54ab².

(iii) [(3/2)x + 1]³

= (3x/2)³ + 1³ + 3(3x/2)(1)(3x/2 + 1)
= 27x³/8 + 1 + (9x/2)(3x/2 + 1)
= 27x³/8 + 1 + 27x²/4 + 9x/2.

(iv) [x - (2/3)y]³

= x³ - (2y/3)³ - 3(x)(2y/3)(x - 2y/3)
= x³ - 8y³/27 - 2xy(x - 2y/3)
= x³ - 8y³/27 - 2x²y + 4xy²/3.

7. Evaluate the following using suitable identities:

(i) (99)³

= (100 - 1)³
= 100³ - 1³ - 3(100)(1)(100 - 1)
= 1000000 - 1 - 300(99)
= 999999 - 29700
= 970299.

(ii) (102)³

= (100 + 2)³
= 1000000 + 8 + 3(100)(2)(102)
= 1000008 + 600(102)
= 1000008 + 61200
= 1061208.

(iii) (998)³

= (1000 - 2)³
= 1000000000 - 8 - 3(1000)(2)(998)
= 999999992 - 6000(998)
= 999999992 - 5988000
= 994011992.

8. Factorise each of the following:

(i) 8a³ + b³ + 12a²b + 6ab²

= (2a)³ + b³ + 3(2a)²(b) + 3(2a)(b)²
= (2a + b)³.

(ii) 8a³ - b³ - 12a²b + 6ab²

= (2a)³ - b³ - 3(2a)²(b) + 3(2a)(b)²
= (2a - b)³.

(iii) 27 - 125a³ - 135a + 225a²

= 3³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²
= (3 - 5a)³.

(iv) 64a³ - 27b³ - 144a²b + 108ab²

= (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)²
= (4a - 3b)³.

(v) 27p³ - 1/216 - 9p²/2 + p/4

= (3p)³ - (1/6)³ - 3(3p)²(1/6) + 3(3p)(1/6)²
= (3p - 1/6)³.

9. Verify:

(i) x³ + y³ = (x + y)(x² - xy + y²)

RHS = x(x² - xy + y²) + y(x² - xy + y²)
= x³ - x²y + xy² + yx² - xy² + y³
= x³ + y³ = LHS. Verified.

(ii) x³ - y³ = (x - y)(x² + xy + y²)

RHS = x(x² + xy + y²) - y(x² + xy + y²)
= x³ + x²y + xy² - yx² - xy² - y³
= x³ - y³ = LHS. Verified.

10. Factorise each of the following:

(i) 27y³ + 125z³

= (3y)³ + (5z)³
Using identity from 9(i):
= (3y + 5z)((3y)² - (3y)(5z) + (5z)²)
= (3y + 5z)(9y² - 15yz + 25z²).

(ii) 64m³ - 343n³

= (4m)³ - (7n)³
Using identity from 9(ii):
= (4m - 7n)((4m)² + (4m)(7n) + (7n)²)
= (4m - 7n)(16m² + 28mn + 49n²).

11. Factorise: 27x³ + y³ + z³ - 9xyz

= (3x)³ + y³ + z³ - 3(3x)(y)(z)
Using a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
= (3x + y + z)((3x)² + y² + z² - 3xy - yz - z(3x))
= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx).

12. Verify that x³ + y³ + z³ - 3xyz = (1/2)(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

RHS = (1/2)(x + y + z)[x² - 2xy + y² + y² - 2yz + z² + z² - 2zx + x²]
= (1/2)(x + y + z)[2x² + 2y² + 2z² - 2xy - 2yz - 2zx]
= (1/2)(x + y + z) × 2[x² + y² + z² - xy - yz - zx]
= (x + y + z)(x² + y² + z² - xy - yz - zx)
This is the expansion of x³ + y³ + z³ - 3xyz. Hence verified.

13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

We know x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
If x + y + z = 0, then RHS = 0 × (...) = 0.
So, x³ + y³ + z³ - 3xyz = 0
⇒ x³ + y³ + z³ = 3xyz.

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

(i) (-12)³ + (7)³ + (5)³

Here, x = -12, y = 7, z = 5.
x + y + z = -12 + 7 + 5 = 0.
So, sum of cubes = 3xyz.
= 3(-12)(7)(5)
= -36 × 35
= -1260.

(ii) (28)³ + (-15)³ + (-13)³

Here, x = 28, y = -15, z = -13.
x + y + z = 28 - 15 - 13 = 28 - 28 = 0.
Value = 3(28)(-15)(-13)
= 3(28)(195)
= 16380.

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

(i) Area: 25a² - 35a + 12

We factorise the quadratic equation.
Product = 25 × 12 = 300, Sum = -35. Numbers are -20 and -15.
= 25a² - 20a - 15a + 12
= 5a(5a - 4) - 3(5a - 4)
= (5a - 4)(5a - 3).
Possible length and breadth are (5a - 4) and (5a - 3).

(ii) Area: 35y² + 13y - 12

Product = 35 × -12 = -420, Sum = 13. Numbers are 28 and -15.
= 35y² + 28y - 15y - 12
= 7y(5y + 4) - 3(5y + 4)
= (5y + 4)(7y - 3).
Possible length and breadth are (5y + 4) and (7y - 3).

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

(i) Volume: 3x² - 12x

= 3x(x - 4).
Dimensions are 3, x, and (x - 4).

(ii) Volume: 12ky² + 8ky - 20k

= 4k(3y² + 2y - 5)
Factorising 3y² + 2y - 5:
= 3y² + 5y - 3y - 5
= y(3y + 5) - 1(3y + 5) = (3y + 5)(y - 1).
So, Volume = 4k(3y + 5)(y - 1).
Dimensions are 4k, (3y + 5), and (y - 1).