22 questions · timed · auto-graded
(2a + b + 2)2
Solution:
4a2 + b2 + 4ab + 8a + 4b + 4
= 4a2 + b2 + 4 + 4ab + 4b + 8a
= (2a)2 + b2 + 22 + 2 × 2a × b + 2 × b × 2 + 2 × 2a × 2
= (2a + b +2)2
0
Solution:
$\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
⇒ a2 + b2 = -ab
⇒ a2 + b2 + ab = 0
Thus, we have:
(a3 - b3) = (a - b)(a2 + b2 + ab)
= (a - b) × 0
= 0
108
Solution:
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
= (x + y + z)[(x + y + z)2 - 3(xy + yz + zx)]
= 9 × (81 - 3 × 23)
= 9 × 12
= 108
(3x + 2)(x2 + 1)
Solution:
3x3 + 2x2 + 3x + 2
= x2(3x + 2) + 1(3x + 2)
= (3x + 2)(x2 + 1)
(2x + 5)(3x + 1)
Solution:
6x2 + 17x + 5
= 6x2 + 15x + 2x + 5
= 3x(2x + 5) + 1(2x + 5)
= (2x + 5)(3x + 1)
(2x + 3)(2x - 1)
Solution:
4x2 + 4x - 3
= 4x2 + 6x - 2x - 3
= 2x(2x + 3) - 1(2x + 3)
= (2x + 3)(2x - 1)
2
Solution:
(x + 1) is a factor of 2x2 + kx
So, -1 is a zero of 2x2 + kx
Thus, we have:
2 × (-1)2 + k × (-1) = 0
⇒ 2 - k = 0
⇒ k = 2
497
Solution:
(249)2 - (248)2
We know
a2 - b2 = (a + b)(a - b)
So,
(249)2 - (248)2
(249 - 248)(249 + 248)
= 497
27
Solution:
(x + 3)3
= x3 + 33 + 9x(x + 3)
= x3 + 27 + 9x2 + 27x
So, the coefficient of x in (x + 3)3 is 27.
39951
Solution:
207 × 193
= (200 + 7)(200 - 7)
= (200)2 - (7)2
= 40000 - 49
= 39951
3xy
Solution:
(x + y)3 - (x3 + y3)
= x3 + y3 + 3xy(x + y) - (x3 + y3)
= 3xy(x + y)
Thus, the factors of (x + y)3 - (x3 + y3) are 3xy and (x + y)
(x - 7)(x + 3)
Solution:
x2 - 4x - 21
x2 - 7x + 3x - 21
= x(x - 7) + 3(x - 7)
= (x - 7)(x + 3)
x3 + 2x2 - x - 2
Solution:
Let:
f(x) = x3 - 2x2 + x + 2
By the factor theorem, (x + 1) will be a factor of f(x) if f(-1) = 0.
We have:
f(-1) = (-1)3 - 2 × (-1)2 + (-1) + 2
= -1 - 2 - 1 + 2
$=-2\neq0$
Hence, (x + 1) is not a factor of f(x) = x3 - 2x2 + x + 2.
Now,
Let:
f(x) = x3 + 2x2 + x - 2
By the factor theorem, (x + 1) will be a factor of f(x) if f(-1) = 0.
We have:
f(-1) = (-1)3 + 2 × (-1)2 + (-1) - 2
= -1 + 2 - 1 - 2
$=-2\neq0$
Hence, (x + 1) is not factor of f(x) = x3 + 2x2 + x - 2.
Now,
Let:
f(x) = x3 + 2x2 - x - 2
By the factor theorem, (x + 1) will be a factor of f(x) if f(-1) = 0.
We have:
f(-1) = (-1)3 + 2 × (-1)2 - (-1) - 2
= -1 + 2 + 1 - 2
= 0
Hence, (x + 1) is a factor of f(x) = x3 + 2x2 - x - 2.
3abc
Solution:
a + b + c = 0
⇒ a + b = -c
⇒ (a + b)3 = (-c)3
⇒ a3 + b3 + 3ab(a + b) = -c3
⇒ a3 + b3 + 3ab(-c) = -c3
⇒ a3 + b3 + c3 = 3abc
0
Solution:
$\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
⇒ x2 + y2 = -xy
⇒ x2 + y2 + xy = 0
Thus, we have:
(x3 - y3) = (x - y)(x2 + y2 + xy)
= (x - y) × 0
= 0
10x
Solution:
(25x2 - 1) + (1 + 5x)2
= (5x - 1)(5x + 1) + (1 + 5x)2
= (5x + 1)[(5x - 1) + (1 + 5x)]
= (5x + 1)(10x)
So, the factors of (25x2 - 1) + (1 + 5x)2 are (5x + 1) and 10x
3
Solution:
a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$
93940
Solution:
305 × 308 = (300 + 5)(300 + 8)
= (300)2 + 300 × (5 + 8) + 5 × 8
= 90000 + 3900 + 40
= 93940
9984
Solution:
104 × 96 = (100 + 4)(100 - 4)
= 1002 - 42
= (10000 - 16)
= 9984
m = 7, n = -18
Solution:
Let:
p(x) = x3 + 10x2 + mx + n
Now,
x + 2 = 0 ⇒ x = -2
(x + 2) is a factor of p(x).
So, we have p(-2)2 + m × (-2) + n = 0
⇒ (-2)3 + 10 × (-2)2 + m × (-2) + n = 0
⇒ -8 + 40 - 2m + n = 0
⇒ 32 - 2m + n = 0
⇒ 2m - n = 32 ...(i)
Now,
x - 1 = 0 ⇒ x = 1
Also,
(x - 1) is a factor of p(x)
We have:
p(1) = 0
⇒ 13 + 10 × 12 + m × 1 + n = 0
⇒ 1 + 10 + m + n = 0
⇒ 11 + m + n = 0
⇒ m + n = -11 ...(ii)
From (i) and (ii),
We get:
3m = 21 ⇒ m = 7
By substituting the value of m in (i), we get n = -18
$\therefore\ $m = 7 and n = -18
5
Solution:
(x + 5) is a factor of p(x) = x3 - 20x + 5k
$\therefore\ $p(-5) = 0
⇒ (-5)3 - 20 × (-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ 5k = 25
⇒ k = 5
$\frac{1}{4}$
Solution:
$\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$
$9\text{x}^2-\frac{1}{4}$ $\Big(\therefore\ \big(\text{a}^2-\text{b}^2\big)=(\text{a}+\text{b})(\text{a}-\text{b})\Big)$
$=9\text{x}^2-\text{p}$
$\Rightarrow\text{p}=\frac{1}{4}$