MCQ test, Set 1

Answer: (D) $(11x + 4z)(11x - 4z)$

$121x^2 - 16z^2 = (11x)^2 - (4z)^2$

$= (11x + 4z)(11x - 4z)$

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

Answer: (A) $(2m + 1)^2$

$4m^2 + 4m + 1 = (2m)^2 + 1^2 + 2(2m)(1)$

$= (2m + 1)^2$

$= (2m + 1)(2m + 1)$

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

Answer: (D) $(7b - 1)^2$

$49b^2 - 14b + 1 = (7b)^2 + 1^2 - 2(7b)(1)$

$= (7b - 1)^2$

$= (7b - 1)(7b - 1)$

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

Answer: (D) $(2m + 7 + 5n)(2m + 7 - 5n)$

$4m^2 + 28m + 49 - 25n^2 = (2m + 7)^2 - (5n)^2$

$= (2m + 7 + 5n)(2m + 7 - 5n)$

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

Answer: (C) $2$ and $20$

$8 \times 5 = 40$

$2 \times 20 = 40$ and $2 + 20 = 22$

So the numbers are $2$ and $20$.

Answer: (D) $(2a + 1)(5a + 6)$

$10 \times 6 = 60$: $5 \times 12 = 60$ and $5 + 12 = 17$

$= 10a^2 + 5a + 12a + 6$

$= 5a(2a + 1) + 6(2a + 1)$

$= (2a + 1)(5a + 6)$

Answer: (B) $(4y - 5)(y - 4)$

$4 \times 20 = 80$: $(-5) \times (-16) = 80$ and $-5 - 16 = -21$

$= 4y^2 - 5y - 16y + 20$

$= y(4y - 5) - 4(4y - 5)$

$= (4y - 5)(y - 4)$

Answer: (A) $(2p + 3)(5p - 2)$

$10 \times (-6) = -60$: $(-4) \times 15 = -60$ and $-4 + 15 = 11$

$= 10p^2 - 4p + 15p - 6$

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

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

Answer: (A) $-4$ and $12$

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

$(-4) \times 12 = -48$ and $-4 + 12 = 8$

So the numbers are $-4$ and $12$.

Answer: (D) $(19t - 39)(t + 7)$

$19 \times (-273) = -5187$

$5187 = 3 \times 7 \times 13 \times 19 = 39 \times 133$, and $-39 + 133 = 94$

$= 19t^2 - 39t + 133t - 273$

$= t(19t - 39) + 7(19t - 39)$

$= (19t - 39)(t + 7)$

Answer: (C) $(2x - 7)(5 - x)$

Writing the terms in order: $-2x^2 + 17x - 35$

$(-2) \times (-35) = 70$: $7 \times 10 = 70$ and $7 + 10 = 17$

$= -2x^2 + 7x + 10x - 35$

$= -x(2x - 7) + 5(2x - 7)$

$= (2x - 7)(5 - x)$

Answer: (D) $(m - p + 2)(m - p + 5)$

Let $m - p = t$

$(m - p)^2 + 7(m - p) + 10 = t^2 + 7t + 10$

$1 \times 10 = 10$: $2 \times 5 = 10$ and $2 + 5 = 7$

$= t^2 + 2t + 5t + 10$

$= t(t + 2) + 5(t + 2)$

$= (t + 2)(t + 5)$

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

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

Answer: (D) $(3x + z)(4x + 3z)$

$12 \times 3 = 36$: $4 \times 9 = 36$ and $4 + 9 = 13$

$= 12x^2 + 4xz + 9xz + 3z^2$

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

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

Answer: (C) $5$ and $-7$

$5 \times (-7) = -35$

$5 \times (-7) = -35$ and $5 - 7 = -2$

So the numbers are $5$ and $-7$.

Answer: (B) $(a - 6\sqrt{5})(a - \sqrt{5})$

Product $= 30$, sum $= -7\sqrt{5}$: $(-\sqrt{5}) \times (-6\sqrt{5}) = 30$ and $-\sqrt{5} + (-6\sqrt{5}) = -7\sqrt{5}$

$= a^2 - \sqrt{5}a - 6\sqrt{5}a + 30$

$= a(a - \sqrt{5}) - 6\sqrt{5}(a - \sqrt{5})$

$= (a - \sqrt{5})(a - 6\sqrt{5})$

Answer: (B) $\displaystyle \left(3y - \frac{1}{4}\right)\left(y + \frac{1}{3}\right)$

$\displaystyle 3y^2 + \frac{3}{4}y - \frac{1}{12} = \frac{36y^2 + 9y - 1}{12}$

$\displaystyle = \frac{1}{12}(36y^2 + 9y - 1)$

$36 \times (-1) = -36$: $(-3) \times 12 = -36$ and $-3 + 12 = 9$

$\displaystyle = \frac{1}{12}[36y^2 - 3y + 12y - 1]$

$\displaystyle = \frac{1}{12}[3y(12y - 1) + 1(12y - 1)]$

$\displaystyle = \frac{1}{12}(12y - 1)(3y + 1)$

$\displaystyle = \left(y + \frac{1}{3}\right)\left(3y - \frac{1}{4}\right)$

Answer: (A) $(2p + q + 4)(3p - 4q + 28)$

Let $p + 4 = x$ and $q - 4 = y$

$6(p + 4)^2 - 5(p + 4)(q - 4) - 4(q - 4)^2 = 6x^2 - 5xy - 4y^2$

$6 \times (-4) = -24$: $3 \times (-8) = -24$ and $3 - 8 = -5$

$= 6x^2 + 3xy - 8xy - 4y^2$

$= 3x(2x + y) - 4y(2x + y)$

$= (2x + y)(3x - 4y)$

$= [2(p + 4) + (q - 4)][3(p + 4) - 4(q - 4)]$

$= (2p + 8 + q - 4)(3p + 12 - 4q + 16)$

$= (2p + q + 4)(3p - 4q + 28)$

Answer: (C) $(2p + 2q - 3r)^2$

$4p^2 + 4q^2 + 9r^2 + 8pq - 12qr - 12rp = (2p)^2 + (2q)^2 + (3r)^2 + 2(2p)(2q) - 2(2q)(3r) - 2(3r)(2p)$

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

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

Answer: (B) $(p + 6)^3$

$p^3 + 18p^2 + 108p + 216 = p^3 + 6^3 + 3(p)^2(6) + 3(p)(6)^2$

$= (p + 6)^3$

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

Answer: (C) $(x - 1)(x^2 + 10x + 37)$

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

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

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

$= (x - 1)[x^2 + 6x + 9 + 4x + 12 + 16]$

$= (x - 1)(x^2 + 10x + 37)$

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

Answer: (B) $(25m^2n^2 - 20mn + 16)(5mn + 4)$

$64 + 125m^3n^3 = 4^3 + (5mn)^3$

$= (4 + 5mn)[4^2 - (4)(5mn) + (5mn)^2]$

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

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

Answer: (B) $(4x^2 + 6xy + 9y^2)(4x^2 - 6xy + 9y^2)$

$16x^4 + 36x^2y^2 + 81y^4 = (4x^2)^2 + 2(4x^2)(9y^2) + (9y^2)^2 - 36x^2y^2$

$= (4x^2 + 9y^2)^2 - (6xy)^2$

$= (4x^2 + 9y^2 + 6xy)(4x^2 + 9y^2 - 6xy)$

$= (4x^2 + 6xy + 9y^2)(4x^2 - 6xy + 9y^2)$

Answer: (A) $3(2x - 3y)(3y - 5)(5 - 2x)$

Here $(2x - 3y) + (3y - 5) + (5 - 2x) = 0$

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

$\therefore (2x - 3y)^3 + (3y - 5)^3 + (5 - 2x)^3 = 3(2x - 3y)(3y - 5)(5 - 2x)$

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

$16t^5 - t = t(16t^4 - 1)$

$= t[(4t^2)^2 - 1^2]$

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

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

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

Answer: (A) $(2y^2 + 5)(y + 5)$

$10y^2 + (2y^2 + 5)y + 25 = 10y^2 + 2y^3 + 5y + 25$

$= 2y^3 + 10y^2 + 5y + 25$

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

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