(i) 4x² - 3x + 7 - Yes, it is a polynomial in one variable x.
(ii) y² + √2 - Yes, it is a polynomial in one variable y.
(iii) 3√t + t√2 - No, because the exponent of t in 3√t is 1/2, which is not a whole number.
(iv) y + 2/y - No, because 2/y can be written as 2y⁻¹, and -1 is not a whole number.
(v) x¹⁰ + y³ + t⁶⁰ - No, because it has three variables x, y, and t.
(i) 2 + x² + x - Coefficient of x² is 1
(ii) 2 - x² + x³ - Coefficient of x² is -1
(iii) (π/2)x² + x - Coefficient of x² is π/2
(iv) √2x - 1 - Coefficient of x² is 0
Binomial of degree 35: x³⁵ + 1
Monomial of degree 100: 5x¹⁰⁰
(i) 5x³ + 4x² + 7x - Degree is 3
(ii) 4 - y² - Degree is 2
(iii) 5t - √7 - Degree is 1
(iv) 3 - Degree is 0
(i) x² + x - Quadratic polynomial
(ii) x - x³ - Cubic polynomial
(iii) y + y² + 4 - Quadratic polynomial
(iv) 1 + x - Linear polynomial
(v) 3t - Linear polynomial
(vi) r² - Quadratic polynomial
(vii) 7x³ - Cubic polynomial
(i) x = 0: 5(0) - 4(0)² + 3 = 3
(ii) x = -1: 5(-1) - 4(-1)² + 3 = -5 - 4 + 3 = -6
(iii) x = 2: 5(2) - 4(2)² + 3 = 10 - 16 + 3 = -3
(i) p(y) = y² - y + 1
p(0) = 0² - 0 + 1 = 1
p(1) = 1² - 1 + 1 = 1
p(2) = 2² - 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² - t³
p(0) = 2 + 0 + 0 - 0 = 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
(i) p(x) = 3x + 1, x = -1/3
p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Yes, 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
No, x = 4/5 is not a zero of the polynomial.
(iii) p(x) = x² - 1, x = 1, -1
p(1) = 1² - 1 = 0
p(-1) = (-1)² - 1 = 1 - 1 = 0
Yes, both x = 1 and x = -1 are zeroes of the polynomial.
(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
Yes, both x = -1 and x = 2 are zeroes of the polynomial.
(v) p(x) = x², x = 0
p(0) = 0² = 0
Yes, x = 0 is a zero of the polynomial.
(vi) p(x) = lx + m, x = -m/l
p(-m/l) = l(-m/l) + m = -m + m = 0
Yes, x = -m/l is a zero of the polynomial.
(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
p(2/√3) = 3(2/√3)² - 1 = 3(4/3) - 1 = 4 - 1 = 3 ≠ 0
Only x = -1/√3 is a zero of the polynomial.
(viii) p(x) = 2x + 1, x = 1/2
p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0
No, x = 1/2 is not a zero of the polynomial.
(i) p(x) = x + 5
Set p(x) = 0: x + 5 = 0 → x = -5
Zero is -5
(ii) p(x) = x - 5
Set p(x) = 0: x - 5 = 0 → x = 5
Zero is 5
(iii) p(x) = 2x + 5
Set p(x) = 0: 2x + 5 = 0 → 2x = -5 → x = -5/2
Zero is -5/2
(iv) p(x) = 3x - 2
Set p(x) = 0: 3x - 2 = 0 → 3x = 2 → x = 2/3
Zero is 2/3
(v) p(x) = 3x
Set p(x) = 0: 3x = 0 → x = 0
Zero is 0
(vi) p(x) = ax, a ≠ 0
Set p(x) = 0: ax = 0 → x = 0
Zero is 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers
Set p(x) = 0: cx + d = 0 → cx = -d → x = -d/c
Zero is -d/c
(i) x³ + x² + x + 1
Let p(x) = x³ + x² + x + 1
p(-1) = (-1)³ + (-1)² + (-1) + 1 = -1 + 1 - 1 + 1 = 0
Yes, (x+1) is a factor.
(ii) x⁴ + x³ + x² + x + 1
Let p(x) = x⁴ + x³ + x² + x + 1
p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 ≠ 0
No, (x+1) is not a factor.
(iii) x⁴ + 3x³ + 3x² + x + 1
Let p(x) = x⁴ + 3x³ + 3x² + x + 1
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1 ≠ 0
No, (x+1) is not a factor.
(iv) x³ - x² - (2+√2)x + √2
Let p(x) = x³ - x² - (2+√2)x + √2
p(-1) = (-1)³ - (-1)² - (2+√2)(-1) + √2 = -1 - 1 + 2 + √2 + √2 = 0 + 2√2 ≠ 0
No, (x+1) is not a factor.
(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 of p(x).
(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 of p(x).
(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 of p(x).
(i) p(x) = x² + x + k
If x - 1 is a factor, then p(1) = 0
p(1) = 1² + 1 + k = 1 + 1 + k = 2 + k = 0
So, k = -2
(ii) p(x) = 2x² + kx + √2
If x - 1 is a factor, then p(1) = 0
p(1) = 2(1)² + k(1) + √2 = 2 + k + √2 = 0
So, k = -2 - √2
(iii) p(x) = kx² - √2x + 1
If x - 1 is a factor, then p(1) = 0
p(1) = k(1)² - √2(1) + 1 = k - √2 + 1 = 0
So, k = √2 - 1
(iv) p(x) = kx² - 3x + k
If x - 1 is a factor, then p(1) = 0
p(1) = k(1)² - 3(1) + k = k - 3 + k = 2k - 3 = 0
So, k = 3/2
(i) 12x² - 7x + 1
We need to find two numbers whose sum is -7 and product is 12×1 = 12
The numbers are -3 and -4
12x² - 7x + 1 = 12x² - 3x - 4x + 1
= 3x(4x - 1) - 1(4x - 1)
= (4x - 1)(3x - 1)
(ii) 2x² + 7x + 3
We need to find two numbers whose sum is 7 and product is 2×3 = 6
The numbers are 1 and 6
2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
(iii) 6x² + 5x - 6
We need to find two numbers whose sum is 5 and product is 6×(-6) = -36
The numbers are 9 and -4
6x² + 5x - 6 = 6x² + 9x - 4x - 6
= 3x(2x + 3) - 2(2x + 3)
= (2x + 3)(3x - 2)
(iv) 3x² - x - 4
We need to find two numbers whose sum is -1 and product is 3×(-4) = -12
The numbers are -4 and 3
3x² - x - 4 = 3x² - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1)
(i) x³ - 2x² - x + 2
Let p(x) = x³ - 2x² - x + 2
By trial, p(1) = 1 - 2 - 1 + 2 = 0
So, (x - 1) is a factor
Dividing p(x) by (x - 1), we get x² - x - 2
Now, x² - x - 2 = (x - 2)(x + 1)
So, x³ - 2x² - x + 2 = (x - 1)(x - 2)(x + 1)
(ii) x³ - 3x² - 9x - 5
Let p(x) = x³ - 3x² - 9x - 5
By trial, 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, x² - 4x - 5 = (x - 5)(x + 1)
So, x³ - 3x² - 9x - 5 = (x + 1)(x - 5)(x + 1) = (x + 1)²(x - 5)
(iii) x³ + 13x² + 32x + 20
Let p(x) = x³ + 13x² + 32x + 20
By trial, 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, x² + 12x + 20 = (x + 2)(x + 10)
So, x³ + 13x² + 32x + 20 = (x + 1)(x + 2)(x + 10)
(iv) 2y³ + y² - 2y - 1
Let p(y) = 2y³ + y² - 2y - 1
By trial, p(1) = 2 + 1 - 2 - 1 = 0
So, (y - 1) is a factor
Dividing p(y) by (y - 1), we get 2y² + 3y + 1
Now, 2y² + 3y + 1 = (2y + 1)(y + 1)
So, 2y³ + y² - 2y - 1 = (y - 1)(2y + 1)(y + 1)
(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 (x + a)(x + b) = x² + (a + b)x + ab
= (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
= (3)² - (2x)² = 9 - 4x²
(i) 103 × 107
103 × 107 = (100 + 3)(100 + 7)
Using (x + a)(x + b) = x² + (a + b)x + ab
= 100² + (3 + 7)(100) + (3)(7) = 10000 + 1000 + 21 = 11021
(ii) 95 × 96
95 × 96 = (100 - 5)(100 - 4)
Using (x + a)(x + b) = x² + (a + b)x + ab
= 100² + (-5 - 4)(100) + (-5)(-4) = 10000 - 900 + 20 = 9120
(iii) 104 × 96
104 × 96 = (100 + 4)(100 - 4)
Using (a + b)(a - b) = a² - b²
= 100² - 4² = 10000 - 16 = 9984
(i) 9x² + 6xy + y²
Using a² + 2ab + b² = (a + b)²
= (3x)² + 2(3x)(y) + y² = (3x + y)²
(ii) 4y² - 4y + 1
Using a² - 2ab + b² = (a - b)²
= (2y)² - 2(2y)(1) + 1² = (2y - 1)²
(iii) x² - y²/100
Using a² - b² = (a + b)(a - b)
= (x)² - (y/10)² = (x + y/10)(x - y/10)
(i) (x + 2y + 4z)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= 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
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (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
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (-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
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (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
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (-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
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (1/4 a)² + (-1/2 b)² + 1² + 2(1/4 a)(-1/2 b) + 2(-1/2 b)(1) + 2(1)(1/4 a)
= 1/16 a² + 1/4 b² + 1 - 1/4 ab - b + 1/2 a
(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
This is of the form a² + b² + c² + 2ab + 2bc + 2ca
Where a = 2x, b = 3y, c = -4z
= (2x + 3y - 4z)²
(ii) 2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8xz
This is of the form a² + b² + c² + 2ab + 2bc + 2ca
Where a = -√2 x, b = y, c = 2√2 z
= (-√2 x + y + 2√2 z)²
(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
Using (a + b)³ = a³ + b³ + 3ab(a + b)
= (3/2 x)³ + 1³ + 3(3/2 x)(1)(3/2 x + 1)
= 27/8 x³ + 1 + 9/2 x(3/2 x + 1)
= 27/8 x³ + 1 + 27/4 x² + 9/2 x
(iv) (x - 2/3 y)³
Using (a - b)³ = a³ - b³ - 3ab(a - b)
= x³ - (2/3 y)³ - 3(x)(2/3 y)(x - 2/3 y)
= x³ - 8/27 y³ - 2xy(x - 2/3 y)
= x³ - 8/27 y³ - 2x²y + 4/3 xy²
(i) 99³
99³ = (100 - 1)³
Using (a - b)³ = a³ - b³ - 3ab(a - b)
= 100³ - 1³ - 3(100)(1)(100 - 1)
= 1000000 - 1 - 300(99)
= 1000000 - 1 - 29700
= 970299
(ii) 102³
102³ = (100 + 2)³
Using (a + b)³ = a³ + b³ + 3ab(a + b)
= 100³ + 2³ + 3(100)(2)(100 + 2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208
(iii) 998³
998³ = (1000 - 2)³
Using (a - b)³ = a³ - b³ - 3ab(a - b)
= 1000³ - 2³ - 3(1000)(2)(1000 - 2)
= 1000000000 - 8 - 6000(998)
= 1000000000 - 8 - 5988000
= 994011992
(i) 8a³ + b³ + 12a²b + 6ab²
This is of the form a³ + b³ + 3a²b + 3ab² = (a + b)³
Where a = 2a, b = b
= (2a + b)³
(ii) 8a³ - b³ - 12a²b + 6ab²
This is of the form a³ - b³ - 3a²b + 3ab² = (a - b)³
Where a = 2a, b = b
= (2a - b)³
(iii) 27 - 125a³ - 135a + 225a²
Rearranging: 27 - 135a + 225a² - 125a³
This is of the form a³ - b³ - 3a²b + 3ab² = (a - b)³
Where a = 3, b = 5a
= (3 - 5a)³
(iv) 64a³ - 27b³ - 144a²b + 108ab²
This is of the form a³ - b³ - 3a²b + 3ab² = (a - b)³
Where a = 4a, b = 3b
= (4a - 3b)³
(v) 27p³ - 1/216 - 9/2 p² + 1/4 p
Rearranging: 27p³ - 9/2 p² + 1/4 p - 1/216
This is of the form a³ - b³ - 3a²b + 3ab² = (a - b)³
Where a = 3p, b = 1/6
= (3p - 1/6)³
(i) x³ + y³ = (x + y)(x² - xy + y²)
RHS = (x + y)(x² - xy + y²)
= x(x² - xy + y²) + y(x² - xy + y²)
= x³ - x²y + xy² + x²y - xy² + y³
= x³ + y³ = LHS
Hence verified.
(ii) x³ - y³ = (x - y)(x² + xy + y²)
RHS = (x - y)(x² + xy + y²)
= x(x² + xy + y²) - y(x² + xy + y²)
= x³ + x²y + xy² - x²y - xy² - y³
= x³ - y³ = LHS
Hence verified.
(i) 27y³ + 125z³
Using a³ + b³ = (a + b)(a² - ab + b²)
= (3y)³ + (5z)³
= (3y + 5z)[(3y)² - (3y)(5z) + (5z)²]
= (3y + 5z)(9y² - 15yz + 25z²)
(ii) 64m³ - 343n³
Using a³ - b³ = (a - b)(a² + ab + b²)
= (4m)³ - (7n)³
= (4m - 7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m - 7n)(16m² + 28mn + 49n²)
Using identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Here, a = 3x, b = y, c = z
= (3x + y + z)[(3x)² + y² + z² - (3x)(y) - (y)(z) - (z)(3x)]
= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)
We know that: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
Now, x² + y² + z² - xy - yz - zx = 1/2[2x² + 2y² + 2z² - 2xy - 2yz - 2zx]
= 1/2[(x² - 2xy + y²) + (y² - 2yz + z²) + (z² - 2zx + x²)]
= 1/2[(x - y)² + (y - z)² + (z - x)²]
So, x³ + y³ + z³ - 3xyz = (x + y + z) × 1/2[(x - y)² + (y - z)² + (z - x)²]
= 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
Hence verified.
We know that: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
If x + y + z = 0, then:
x³ + y³ + z³ - 3xyz = 0 × (x² + y² + z² - xy - yz - zx) = 0
Therefore, x³ + y³ + z³ = 3xyz
Hence proved.
(i) (-12)³ + (7)³ + (5)³
Let x = -12, y = 7, z = 5
x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
So, (-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260
(ii) (28)³ + (-15)³ + (-13)³
Let x = 28, y = -15, z = -13
x + y + z = 28 - 15 - 13 = 0
We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
So, (28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13) = 16380
(i) Area: 25a² - 35a + 12
Factorising: 25a² - 35a + 12
= 25a² - 15a - 20a + 12
= 5a(5a - 3) - 4(5a - 3)
= (5a - 3)(5a - 4)
So, possible length and breadth are (5a - 3) and (5a - 4)
(ii) Area: 35y² + 13y - 12
Factorising: 35y² + 13y - 12
= 35y² + 28y - 15y - 12
= 7y(5y + 4) - 3(5y + 4)
= (5y + 4)(7y - 3)
So, possible length and breadth are (5y + 4) and (7y - 3)
(i) Volume: 3x² - 12x
Factorising: 3x² - 12x = 3x(x - 4)
So, possible dimensions are 3, x, and (x - 4)
(ii) Volume: 12ky² + 8ky - 20k
Factorising: 12ky² + 8ky - 20k = 4k(3y² + 2y - 5)
Now, 3y² + 2y - 5 = 3y² + 5y - 3y - 5 = y(3y + 5) - 1(3y + 5) = (3y + 5)(y - 1)
So, 12ky² + 8ky - 20k = 4k(3y + 5)(y - 1)
So, possible dimensions are 4k, (3y + 5), and (y - 1)
The highest power of x in the polynomial is 4.
Therefore, the degree of the polynomial is 4.
To find the zero, set p(x) = 0:
3x - 2 = 0
3x = 2
x = 2/3
Therefore, the zero of the polynomial is 2/3.
p(2) = (2)² - 4 = 4 - 4 = 0
Since p(2) = 0, x = 2 is a zero of the polynomial.
We need to find two numbers whose sum is 5 and product is 6.
The numbers are 2 and 3.
So, x² + 5x + 6 = (x + 2)(x + 3)
Using (a + b)² = a² + 2ab + b²
(2x + 3y)² = (2x)² + 2(2x)(3y) + (3y)²
= 4x² + 12xy + 9y²
Using a³ + b³ = (a + b)(a² - ab + b²)
8x³ + 27y³ = (2x)³ + (3y)³
= (2x + 3y)[(2x)² - (2x)(3y) + (3y)²]
= (2x + 3y)(4x² - 6xy + 9y²)
p(1) = (1)³ - 1 = 1 - 1 = 0
Since p(1) = 0, x = 1 is a zero of the polynomial.
If (x - 2) is a factor, then p(2) = 0
p(2) = 2(2)² + k(2) - 14 = 8 + 2k - 14 = 2k - 6
Set 2k - 6 = 0
2k = 6
k = 3