MCQ test, Set 2

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

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

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

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

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

Answer: (A) $(9t + 8)(9t - 8)$

$81t^2 - 64 = (9t)^2 - 8^2$

$= (9t + 8)(9t - 8)$

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

Answer: (C) $(3a + 5b)^2$

$9a^2 + 30ab + 25b^2 = (3a)^2 + (5b)^2 + 2(3a)(5b)$

$= (3a + 5b)^2$

$= (3a + 5b)(3a + 5b)$

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

Answer: (B) $(2x + 5y)^2$

$4x^2 + 20xy + 25y^2 = (2x)^2 + (5y)^2 + 2(2x)(5y)$

$= (2x + 5y)^2$

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

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

Answer: (A) $(7m - 5)^2$

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

$= (7m - 5)^2$

$= (7m - 5)(7m - 5)$

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

Answer: (C) $(x - 7 + 6y)(x - 7 - 6y)$

$x^2 - 14x + 49 - 36y^2 = (x - 7)^2 - (6y)^2$

$= (x - 7 + 6y)(x - 7 - 6y)$

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

Answer: (A) $(3a + 2)(5a + 4)$

$15 \times 8 = 120$: $10 \times 12 = 120$ and $10 + 12 = 22$

$= 15a^2 + 10a + 12a + 8$

$= 5a(3a + 2) + 4(3a + 2)$

$= (3a + 2)(5a + 4)$

Answer: (C) $-3$ and $-4$

$4 \times 3 = 12$

$(-3) \times (-4) = 12$ and $-3 - 4 = -7$

So the numbers are $-3$ and $-4$.

Answer: (C) $-3$ and $-10$

$5 \times 6 = 30$

$(-3) \times (-10) = 30$ and $-3 - 10 = -13$

So the numbers are $-3$ and $-10$.

Answer: (D) $-4$ and $9$

$12 \times (-3) = -36$

$(-4) \times 9 = -36$ and $-4 + 9 = 5$

So the numbers are $-4$ and $9$.

Answer: (C) $(19b + 10)(b + 9)$

$19 \times 90 = 1710$

$1710 = 2 \times 3 \times 3 \times 5 \times 19 = 10 \times 171$, and $10 + 171 = 181$

$= 19b^2 + 10b + 171b + 90$

$= b(19b + 10) + 9(19b + 10)$

$= (19b + 10)(b + 9)$

Answer: (B) $(1 - t)(t + 3)$

Writing the terms in order: $-t^2 - 2t + 3$

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

$= -t^2 - 3t + t + 3$

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

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

Answer: (B) $(3t - 5)(8 - t)$

Writing the terms in order: $-3t^2 + 29t - 40$

$(-3) \times (-40) = 120$: $5 \times 24 = 120$ and $5 + 24 = 29$

$= -3t^2 + 5t + 24t - 40$

$= -t(3t - 5) + 8(3t - 5)$

$= (3t - 5)(8 - t)$

Answer: (D) $(3a - 4c + 6)(9a - 12c - 2)$

Let $3a - 4c = p$

$3(3a - 4c)^2 + 16(3a - 4c) - 12 = 3p^2 + 16p - 12$

$3 \times (-12) = -36$: $(-2) \times 18 = -36$ and $-2 + 18 = 16$

$= 3p^2 - 2p + 18p - 12$

$= p(3p - 2) + 6(3p - 2)$

$= (3p - 2)(p + 6)$

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

$= (9a - 12c - 2)(3a - 4c + 6)$

Answer: (B) $(5x - 3y)(x - 3y)$

$5 \times 9 = 45$: $(-3) \times (-15) = 45$ and $-3 - 15 = -18$

$= 5x^2 - 3xy - 15xy + 9y^2$

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

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

Answer: (D) $(6\sqrt{2}t + 1)(t - 3\sqrt{2})$

$6\sqrt{2} \times (-3\sqrt{2}) = 6 \times (-3) \times 2 = -36$: $1 \times (-36) = -36$ and $1 - 36 = -35$

$= 6\sqrt{2}t^2 + t - 36t - 3\sqrt{2}$

$= t(6\sqrt{2}t + 1) - 3\sqrt{2}(6\sqrt{2}t + 1)$

$= (6\sqrt{2}t + 1)(t - 3\sqrt{2})$

Answer: (A) $\displaystyle (11a + 3)\left(a + \frac{3}{11}\right)$

$\displaystyle 11a^2 + 6a + \frac{9}{11} = \frac{121a^2 + 66a + 9}{11}$

$\displaystyle = \frac{1}{11}(121a^2 + 66a + 9)$

$\displaystyle = \frac{1}{11}[(11a)^2 + 3^2 + 2(11a)(3)]$

$\displaystyle = \frac{1}{11}(11a + 3)^2$

$\displaystyle = \left(a + \frac{3}{11}\right)(11a + 3)$

Answer: (B) $(3m + n + 13)(m + 4n + 19)$

Let $m + 3 = x$ and $n + 4 = y$

$3(m + 3)^2 + 13(m + 3)(n + 4) + 4(n + 4)^2 = 3x^2 + 13xy + 4y^2$

$3 \times 4 = 12$: $1 \times 12 = 12$ and $1 + 12 = 13$

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

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

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

$= [3(m + 3) + (n + 4)][(m + 3) + 4(n + 4)]$

$= (3m + 9 + n + 4)(m + 3 + 4n + 16)$

$= (3m + n + 13)(m + 4n + 19)$

Answer: (D) $(2x + y - 2z)^2$

$4x^2 + y^2 + 4z^2 + 4xy - 4yz - 8zx = (2x)^2 + y^2 + (2z)^2 + 2(2x)(y) - 2(y)(2z) - 2(2z)(2x)$

$= (2x + y - 2z)^2$

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

Answer: (A) $(b - 2)^3$

$b^3 - 6b^2 + 12b - 8 = b^3 - 2^3 - 3(b)^2(2) + 3(b)(2)^2$

$= (b - 2)^3$

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

Answer: (A) $(m - 5p)(m^2 + 5mp + 25p^2)$

$m^3 - 125p^3 = m^3 - (5p)^3$

$= (m - 5p)[m^2 + (m)(5p) + (5p)^2]$

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

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

Answer: (D) $(16x^2z^2 - 12xz + 9)(4xz + 3)$

$27 + 64x^3z^3 = 3^3 + (4xz)^3$

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

$= (4xz + 3)(16x^2z^2 - 12xz + 9)$

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

Answer: (B) $3(5m - 3n)(3n - 2p)(2p - 5m)$

Here $(5m - 3n) + (3n - 2p) + (2p - 5m) = 0$

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

$\therefore (5m - 3n)^3 + (3n - 2p)^3 + (2p - 5m)^3 = 3(5m - 3n)(3n - 2p)(2p - 5m)$

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

$m^5n - 16mn^5 = mn(m^4 - 16n^4)$

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

$= mn(m^2 + 4n^2)(m^2 - 4n^2)$

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

$= mn(m^2 + 4n^2)(m + 2n)(m - 2n)$

Answer: (D) $(5x^2 + 3)(x + 5)$

$25x^2 + (5x^2 + 3)x + 15 = 25x^2 + 5x^3 + 3x + 15$

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

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

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