MCQ test, Set 3

Answer: (A) $(2 + 3m)(2 - 3m)(4 + 9m^2)$

$16 - 81m^4 = 4^2 - (9m^2)^2$

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

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

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

Answer: (C) $(5m + 7p)^2$

$25m^2 + 70mp + 49p^2 = (5m)^2 + (7p)^2 + 2(5m)(7p)$

$= (5m + 7p)^2$

$= (5m + 7p)(5m + 7p)$

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

Answer: (D) $(6p - 1)^2$

$36p^2 - 12p + 1 = (6p)^2 + 1^2 - 2(6p)(1)$

$= (6p - 1)^2$

$= (6p - 1)(6p - 1)$

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

Answer: (B) $(m - 7p)^2$

$m^2 - 14mp + 49p^2 = m^2 + (7p)^2 - 2(m)(7p)$

$= (m - 7p)^2$

$= (m - 7p)(m - 7p)$

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

Answer: (B) $(2m - 5 + 4n)(2m - 5 - 4n)$

$4m^2 - 20m + 25 - 16n^2 = (2m - 5)^2 - (4n)^2$

$= (2m - 5 + 4n)(2m - 5 - 4n)$

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

Answer: (C) $5$ and $9$

$5 \times 9 = 45$

$5 \times 9 = 45$ and $5 + 9 = 14$

So the numbers are $5$ and $9$.

Answer: (B) $(4n - 5)(n - 6)$

$4 \times 30 = 120$: $(-5) \times (-24) = 120$ and $-5 - 24 = -29$

$= 4n^2 - 5n - 24n + 30$

$= n(4n - 5) - 6(4n - 5)$

$= (4n - 5)(n - 6)$

Answer: (C) $(2x + 5)(5x - 3)$

$10 \times (-15) = -150$: $(-6) \times 25 = -150$ and $-6 + 25 = 19$

$= 10x^2 - 6x + 25x - 15$

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

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

Answer: (A) $(11x + 19)(x + 6)$

$11 \times 114 = 1254$

$1254 = 2 \times 3 \times 11 \times 19 = 19 \times 66$, and $19 + 66 = 85$

$= 11x^2 + 19x + 66x + 114$

$= x(11x + 19) + 6(11x + 19)$

$= (11x + 19)(x + 6)$

Answer: (A) $(11n + 29)(n + 4)$

$11 \times 116 = 1276$

$1276 = 2 \times 2 \times 11 \times 29 = 29 \times 44$, and $29 + 44 = 73$

$= 11n^2 + 29n + 44n + 116$

$= n(11n + 29) + 4(11n + 29)$

$= (11n + 29)(n + 4)$

Answer: (B) $(9 - 2n)(n + 8)$

Writing the terms in order: $-2n^2 - 7n + 72$

$(-2) \times 72 = -144$: $(-16) \times 9 = -144$ and $-16 + 9 = -7$

$= -2n^2 - 16n + 9n + 72$

$= -2n(n + 8) + 9(n + 8)$

$= (n + 8)(9 - 2n)$

Answer: (D) $(2x + 6z - 5)(x + 3z + 2)$

Let $x + 3z = p$

$2(x + 3z)^2 - (x + 3z) - 10 = 2p^2 - p - 10$

$2 \times (-10) = -20$: $4 \times (-5) = -20$ and $4 - 5 = -1$

$= 2p^2 + 4p - 5p - 10$

$= 2p(p + 2) - 5(p + 2)$

$= (p + 2)(2p - 5)$

$= [(x + 3z) + 2][2(x + 3z) - 5]$

$= (x + 3z + 2)(2x + 6z - 5)$

Answer: (C) $1$ and $-30$

$5 \times (-6) = -30$

$1 \times (-30) = -30$ and $1 - 30 = -29$

So the numbers are $1$ and $-30$.

Answer: (A) $(3\sqrt{7}t + 4)(t - 6\sqrt{7})$

$3\sqrt{7} \times (-24\sqrt{7}) = 3 \times (-24) \times 7 = -504$: $4 \times (-126) = -504$ and $4 - 126 = -122$

$= 3\sqrt{7}t^2 + 4t - 126t - 24\sqrt{7}$

$= t(3\sqrt{7}t + 4) - 6\sqrt{7}(3\sqrt{7}t + 4)$

$= (3\sqrt{7}t + 4)(t - 6\sqrt{7})$

Answer: (D) $\displaystyle (15t + 2)\left(t + \frac{2}{15}\right)$

$\displaystyle 15t^2 + 4t + \frac{4}{15} = \frac{225t^2 + 60t + 4}{15}$

$\displaystyle = \frac{1}{15}(225t^2 + 60t + 4)$

$\displaystyle = \frac{1}{15}[(15t)^2 + 2^2 + 2(15t)(2)]$

$\displaystyle = \frac{1}{15}(15t + 2)^2$

$\displaystyle = \left(t + \frac{2}{15}\right)(15t + 2)$

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

Let $a + 2 = x$ and $c + 1 = y$

$6(a + 2)^2 + (a + 2)(c + 1) - (c + 1)^2 = 6x^2 + xy - y^2$

$6 \times (-1) = -6$: $(-2) \times 3 = -6$ and $-2 + 3 = 1$

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

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

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

$= [3(a + 2) - (c + 1)][2(a + 2) + (c + 1)]$

$= (3a + 6 - c - 1)(2a + 4 + c + 1)$

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

Answer: (A) $(3a - 4b + 4c)^2$

$9a^2 + 16b^2 + 16c^2 - 24ab - 32bc + 24ca = (3a)^2 + (4b)^2 + (4c)^2 - 2(3a)(4b) - 2(4b)(4c) + 2(4c)(3a)$

$= (3a - 4b + 4c)^2$

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

Answer: (B) $(3y - 4z)^3$

$27y^3 - 108y^2z + 144yz^2 - 64z^3 = (3y)^3 - (4z)^3 - 3(3y)^2(4z) + 3(3y)(4z)^2$

$= (3y - 4z)^3$

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

Answer: (C) $(x - z)(x^2 + 10xz + 37z^2)$

$x^3 + 9x^2z + 27xz^2 - 37z^3 = x^3 + 3(x)^2(3z) + 3(x)(3z)^2 + (3z)^3 - 64z^3$

$= (x + 3z)^3 - (4z)^3$

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

$= (x - z)[x^2 + 6xz + 9z^2 + 4xz + 12z^2 + 16z^2]$

$= (x - z)(x^2 + 10xz + 37z^2)$

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

Answer: (D) $(16m^2 - 12mn + 9n^2)(4m + 3n)$

$64m^3 + 27n^3 = (4m)^3 + (3n)^3$

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

$= (4m + 3n)(16m^2 - 12mn + 9n^2)$

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

Answer: (D) $(x + 1)(x + 7)(x - 8)$

$x^3 - 57x - 56 = x^3 - x - 56x - 56$

$= x(x^2 - 1) - 56(x + 1)$

$= x(x + 1)(x - 1) - 56(x + 1)$

$= (x + 1)[x(x - 1) - 56]$

$= (x + 1)(x^2 - x - 56)$

Now $x^2 - x - 56$: $7 \times (-8) = -56$ and $7 - 8 = -1$

$x^2 - x - 56 = x^2 + 7x - 8x - 56$

$= x(x + 7) - 8(x + 7)$

$= (x + 7)(x - 8)$

So $x^3 - 57x - 56 = (x + 1)(x + 7)(x - 8)$

Answer: (C) $(m^2 + 5m + 25)(m^2 - 5m + 25)$

$m^4 + 25m^2 + 625 = (m^2)^2 + 2(m^2)(25) + 25^2 - 25m^2$

$= (m^2 + 25)^2 - (5m)^2$

$= (m^2 + 25 + 5m)(m^2 + 25 - 5m)$

$= (m^2 + 5m + 25)(m^2 - 5m + 25)$

Answer: (A) $3(2p - q)(q - 3r)(3r - 2p)$

Here $(2p - q) + (q - 3r) + (3r - 2p) = 0$

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

$\therefore (2p - q)^3 + (q - 3r)^3 + (3r - 2p)^3 = 3(2p - q)(q - 3r)(3r - 2p)$

Answer: (D) $m(3 - m)(m + 3)(m^2 + 9)$

$81m - m^5 = m(81 - m^4)$

$= m[9^2 - (m^2)^2]$

$= m(m^2 + 9)(9 - m^2)$

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

$= m(m^2 + 9)(m + 3)(3 - m)$

Answer: (B) $(4y^2 + 9)(y + 5)$

$20y^2 + (4y^2 + 9)y + 45 = 20y^2 + 4y^3 + 9y + 45$

$= 4y^3 + 20y^2 + 9y + 45$

$= 4y^2(y + 5) + 9(y + 5)$

$= (y + 5)(4y^2 + 9)$