MCQ test, Set 8

Answer: (C) $(1 + 2p)(1 + 4p^2)(1 - 2p)$

$1 - 16p^4 = 1^2 - (4p^2)^2$

$= (1 + 4p^2)(1 - 4p^2)$

$= (1 + 4p^2)[1^2 - (2p)^2]$

$= (1 + 4p^2)(1 + 2p)(1 - 2p)$

Answer: (D) $(2p + 9q)^2$

$4p^2 + 36pq + 81q^2 = (2p)^2 + (9q)^2 + 2(2p)(9q)$

$= (2p + 9q)^2$

$= (2p + 9q)(2p + 9q)$

using $a^2 + b^2 + 2ab = (a + b)^2$

Answer: (B) $(7t - 2)^2$

$49t^2 - 28t + 4 = (7t)^2 + 2^2 - 2(7t)(2)$

$= (7t - 2)^2$

$= (7t - 2)(7t - 2)$

using $a^2 + b^2 - 2ab = (a - b)^2$

Answer: (A) $(a - 6 + b)(a - 6 - b)$

$a^2 - 12a + 36 - b^2 = (a - 6)^2 - b^2$

$= (a - 6 + b)(a - 6 - b)$

using $a^2 - 2ab + b^2 = (a - b)^2$ and $a^2 - b^2 = (a + b)(a - b)$

Answer: (A) $5$ and $21$

$15 \times 7 = 105$

$5 \times 21 = 105$ and $5 + 21 = 26$

So the numbers are $5$ and $21$.

Answer: (C) $(3t + 1)(t + 8)$

$3 \times 8 = 24$: $1 \times 24 = 24$ and $1 + 24 = 25$

$= 3t^2 + t + 24t + 8$

$= t(3t + 1) + 8(3t + 1)$

$= (3t + 1)(t + 8)$

Answer: (D) $-3$ and $-16$

$12 \times 4 = 48$

$(-3) \times (-16) = 48$ and $-3 - 16 = -19$

So the numbers are $-3$ and $-16$.

Answer: (A) $(5x - 3)(x + 5)$

$5 \times (-15) = -75$: $(-3) \times 25 = -75$ and $-3 + 25 = 22$

$= 5x^2 - 3x + 25x - 15$

$= x(5x - 3) + 5(5x - 3)$

$= (5x - 3)(x + 5)$

Answer: (B) $(2n - 3)(3n + 5)$

$6 \times (-15) = -90$: $(-9) \times 10 = -90$ and $-9 + 10 = 1$

$= 6n^2 - 9n + 10n - 15$

$= 3n(2n - 3) + 5(2n - 3)$

$= (2n - 3)(3n + 5)$

Answer: (D) $(17t + 21)(t - 7)$

$17 \times (-147) = -2499$

$2499 = 3 \times 7 \times 7 \times 17 = 21 \times 119$, and $21 - 119 = -98$

$= 17t^2 + 21t - 119t - 147$

$= t(17t + 21) - 7(17t + 21)$

$= (17t + 21)(t - 7)$

Answer: (C) $(7 - 3n)(n + 6)$

Writing the terms in order: $-3n^2 - 11n + 42$

$(-3) \times 42 = -126$: $(-18) \times 7 = -126$ and $-18 + 7 = -11$

$= -3n^2 - 18n + 7n + 42$

$= -3n(n + 6) + 7(n + 6)$

$= (n + 6)(7 - 3n)$

Answer: (B) $(2a + 3b + 7)(6a + 9b + 7)$

Let $2a + 3b = p$

$3(2a + 3b)^2 + 28(2a + 3b) + 49 = 3p^2 + 28p + 49$

$3 \times 49 = 147$: $7 \times 21 = 147$ and $7 + 21 = 28$

$= 3p^2 + 7p + 21p + 49$

$= p(3p + 7) + 7(3p + 7)$

$= (3p + 7)(p + 7)$

$= [3(2a + 3b) + 7][(2a + 3b) + 7]$

$= (6a + 9b + 7)(2a + 3b + 7)$

Answer: (C) $(3x - 4y)(4x + y)$

$12 \times (-4) = -48$: $3 \times (-16) = -48$ and $3 - 16 = -13$

$= 12x^2 + 3xy - 16xy - 4y^2$

$= 3x(4x + y) - 4y(4x + y)$

$= (4x + y)(3x - 4y)$

Answer: (B) $6$ and $7$

$21 \times 2 = 42$

$6 \times 7 = 42$ and $6 + 7 = 13$

So the numbers are $6$ and $7$.

Answer: (B) $(3t + 2\sqrt{2})(\sqrt{2}t + 5)$

$3\sqrt{2} \times 10\sqrt{2} = 3 \times 10 \times 2 = 60$: $15 \times 4 = 60$ and $15 + 4 = 19$

$= 3\sqrt{2}t^2 + 15t + 4t + 10\sqrt{2}$

$= 3t(\sqrt{2}t + 5) + 2\sqrt{2}(\sqrt{2}t + 5)$

$= (\sqrt{2}t + 5)(3t + 2\sqrt{2})$

Answer: (A) $\displaystyle (4t - 3)\left(t - \frac{3}{4}\right)$

$\displaystyle 4t^2 - 6t + \frac{9}{4} = \frac{16t^2 - 24t + 9}{4}$

$\displaystyle = \frac{1}{4}(16t^2 - 24t + 9)$

$\displaystyle = \frac{1}{4}[(4t)^2 + 3^2 - 2(4t)(3)]$

$\displaystyle = \frac{1}{4}(4t - 3)^2$

$\displaystyle = \left(t - \frac{3}{4}\right)(4t - 3)$

Answer: (C) $(2a - c + 13)(5a + 3c + 16)$

Let $a + 5 = x$ and $c - 3 = y$

$10(a + 5)^2 + (a + 5)(c - 3) - 3(c - 3)^2 = 10x^2 + xy - 3y^2$

$10 \times (-3) = -30$: $(-5) \times 6 = -30$ and $-5 + 6 = 1$

$= 10x^2 - 5xy + 6xy - 3y^2$

$= 5x(2x - y) + 3y(2x - y)$

$= (2x - y)(5x + 3y)$

$= [2(a + 5) - (c - 3)][5(a + 5) + 3(c - 3)]$

$= (2a + 10 - c + 3)(5a + 25 + 3c - 9)$

$= (2a - c + 13)(5a + 3c + 16)$

Answer: (A) $(a + 2b - 4c)^2$

$a^2 + 4b^2 + 16c^2 + 4ab - 16bc - 8ca = a^2 + (2b)^2 + (4c)^2 + 2(a)(2b) - 2(2b)(4c) - 2(4c)(a)$

$= (a + 2b - 4c)^2$

using $a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2$

Answer: (D) $(y - 3z)^3$

$y^3 - 9y^2z + 27yz^2 - 27z^3 = y^3 - (3z)^3 - 3(y)^2(3z) + 3(y)(3z)^2$

$= (y - 3z)^3$

using $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$

Answer: (D) $(2m - 3)(4m^2 + 3)$

$8m^3 - 12m^2 + 6m - 9 = (2m)^3 - 3(2m)^2(1) + 3(2m)(1)^2 - 1^3 - 8$

$= (2m - 1)^3 - 2^3$

$= (2m - 1 - 2)[(2m - 1)^2 + (2m - 1)2 + 2^2]$

$= (2m - 3)[4m^2 - 4m + 1 + 4m - 2 + 4]$

$= (2m - 3)(4m^2 + 3)$

using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Answer: (C) $(3ab + 4)(9a^2b^2 - 12ab + 16)$

$27a^3b^3 + 64 = (3ab)^3 + 4^3$

$= (3ab + 4)[(3ab)^2 - (3ab)(4) + 4^2]$

$= (3ab + 4)(9a^2b^2 - 12ab + 16)$

using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Answer: (D) $(y^2 + 4y + 16)(y^2 - 4y + 16)$

$y^4 + 16y^2 + 256 = (y^2)^2 + 2(y^2)(16) + 16^2 - 16y^2$

$= (y^2 + 16)^2 - (4y)^2$

$= (y^2 + 16 + 4y)(y^2 + 16 - 4y)$

$= (y^2 + 4y + 16)(y^2 - 4y + 16)$

Answer: (A) $3(m - 3n)(3n - 2)(2 - m)$

Here $(m - 3n) + (3n - 2) + (2 - m) = 0$

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$

$\therefore (m - 3n)^3 + (3n - 2)^3 + (2 - m)^3 = 3(m - 3n)(3n - 2)(2 - m)$

Answer: (B) $y(2 - 3y)(3y + 2)(9y^2 + 4)$

$16y - 81y^5 = y(16 - 81y^4)$

$= y[4^2 - (9y^2)^2]$

$= y(9y^2 + 4)(4 - 9y^2)$

$= y(9y^2 + 4)[2^2 - (3y)^2]$

$= y(9y^2 + 4)(3y + 2)(2 - 3y)$

Answer: (D) $(5m^2 + 4)(m + 5)$

$25m^2 + (5m^2 + 4)m + 20 = 25m^2 + 5m^3 + 4m + 20$

$= 5m^3 + 25m^2 + 4m + 20$

$= 5m^2(m + 5) + 4(m + 5)$

$= (m + 5)(5m^2 + 4)$