MCQ test, Set 10

Answer: (A) $(12n + 1)(12n - 1)$

$144n^2 - 1 = (12n)^2 - 1^2$

$= (12n + 1)(12n - 1)$

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

Answer: (D) $(7y + 9)^2$

$49y^2 + 126y + 81 = (7y)^2 + 9^2 + 2(7y)(9)$

$= (7y + 9)^2$

$= (7y + 9)(7y + 9)$

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

Answer: (A) $(7x + 6)^2$

$49x^2 + 84x + 36 = (7x)^2 + 6^2 + 2(7x)(6)$

$= (7x + 6)^2$

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

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

Answer: (B) $(5z - 7)^2$

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

$= (5z - 7)^2$

$= (5z - 7)(5z - 7)$

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

Answer: (B) $(2p - 5)^2$

$4p^2 - 20p + 25 = (2p)^2 + 5^2 - 2(2p)(5)$

$= (2p - 5)^2$

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

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

Answer: (A) $(2x - 3y + 6)(2x - 3y - 6)$

$4x^2 - 12xy + 9y^2 - 36 = (2x - 3y)^2 - 6^2$

$= (2x - 3y + 6)(2x - 3y - 6)$

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

Answer: (C) $14$ and $20$

$8 \times 35 = 280$

$14 \times 20 = 280$ and $14 + 20 = 34$

So the numbers are $14$ and $20$.

Answer: (B) $(2n - 1)(4n - 3)$

$8 \times 3 = 24$: $(-4) \times (-6) = 24$ and $-4 - 6 = -10$

$= 8n^2 - 4n - 6n + 3$

$= 4n(2n - 1) - 3(2n - 1)$

$= (2n - 1)(4n - 3)$

Answer: (C) $(2a - 9)(3a + 2)$

$6 \times (-18) = -108$: $4 \times (-27) = -108$ and $4 - 27 = -23$

$= 6a^2 + 4a - 27a - 18$

$= 2a(3a + 2) - 9(3a + 2)$

$= (3a + 2)(2a - 9)$

Answer: (B) $(23x - 25)(x + 8)$

$23 \times (-200) = -4600$

$4600 = 2 \times 2 \times 2 \times 5 \times 5 \times 23 = 25 \times 184$, and $-25 + 184 = 159$

$= 23x^2 - 25x + 184x - 200$

$= x(23x - 25) + 8(23x - 25)$

$= (23x - 25)(x + 8)$

Answer: (A) $(11n - 15)(n + 3)$

$11 \times (-45) = -495$

$495 = 3 \times 3 \times 5 \times 11 = 15 \times 33$, and $-15 + 33 = 18$

$= 11n^2 - 15n + 33n - 45$

$= n(11n - 15) + 3(11n - 15)$

$= (11n - 15)(n + 3)$

Answer: (D) $(3z - 1)(4 - z)$

Writing the terms in order: $-3z^2 + 13z - 4$

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

$= -3z^2 + z + 12z - 4$

$= -z(3z - 1) + 4(3z - 1)$

$= (3z - 1)(4 - z)$

Answer: (D) $(9 - 2t)(t - 3)$

Writing the terms in order: $-2t^2 + 15t - 27$

$(-2) \times (-27) = 54$: $6 \times 9 = 54$ and $6 + 9 = 15$

$= -2t^2 + 6t + 9t - 27$

$= -2t(t - 3) + 9(t - 3)$

$= (t - 3)(9 - 2t)$

Answer: (C) $(x + 4z + 4)(x + 4z - 5)$

Let $x + 4z = p$

$(x + 4z)^2 - (x + 4z) - 20 = p^2 - p - 20$

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

$= p^2 + 4p - 5p - 20$

$= p(p + 4) - 5(p + 4)$

$= (p + 4)(p - 5)$

$= [(x + 4z) + 4][(x + 4z) - 5]$

$= (x + 4z + 4)(x + 4z - 5)$

Answer: (A) $4$ and $-5$

$10 \times (-2) = -20$

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

So the numbers are $4$ and $-5$.

Answer: (C) $(x - 3\sqrt{7})(x - 6\sqrt{7})$

Product $= 126$, sum $= -9\sqrt{7}$: $(-3\sqrt{7}) \times (-6\sqrt{7}) = 126$ and $-3\sqrt{7} + (-6\sqrt{7}) = -9\sqrt{7}$

$= x^2 - 3\sqrt{7}x - 6\sqrt{7}x + 126$

$= x(x - 3\sqrt{7}) - 6\sqrt{7}(x - 3\sqrt{7})$

$= (x - 3\sqrt{7})(x - 6\sqrt{7})$

Answer: (D) $\displaystyle \left(2y + \frac{3}{7}\right)\left(y + \frac{1}{2}\right)$

$\displaystyle 2y^2 + \frac{10}{7}y + \frac{3}{14} = \frac{28y^2 + 20y + 3}{14}$

$\displaystyle = \frac{1}{14}(28y^2 + 20y + 3)$

$28 \times 3 = 84$: $6 \times 14 = 84$ and $6 + 14 = 20$

$\displaystyle = \frac{1}{14}[28y^2 + 6y + 14y + 3]$

$\displaystyle = \frac{1}{14}[2y(14y + 3) + 1(14y + 3)]$

$\displaystyle = \frac{1}{14}(14y + 3)(2y + 1)$

$\displaystyle = \left(2y + \frac{3}{7}\right)\left(y + \frac{1}{2}\right)$

Answer: (B) $(5a + 2b - 14)(a + 5b + 11)$

Let $a - 4 = x$ and $b + 3 = y$

$5(a - 4)^2 + 27(a - 4)(b + 3) + 10(b + 3)^2 = 5x^2 + 27xy + 10y^2$

$5 \times 10 = 50$: $2 \times 25 = 50$ and $2 + 25 = 27$

$= 5x^2 + 2xy + 25xy + 10y^2$

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

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

$= [5(a - 4) + 2(b + 3)][(a - 4) + 5(b + 3)]$

$= (5a - 20 + 2b + 6)(a - 4 + 5b + 15)$

$= (5a + 2b - 14)(a + 5b + 11)$

Answer: (B) $(x - 3y + 2z)^2$

$x^2 + 9y^2 + 4z^2 - 6xy - 12yz + 4zx = x^2 + (3y)^2 + (2z)^2 - 2(x)(3y) - 2(3y)(2z) + 2(2z)(x)$

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

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

Answer: (D) $(t - 6)^3$

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

$= (t - 6)^3$

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

Answer: (B) $(3t - 5)(9t^2 + 6t + 13)$

$27t^3 - 27t^2 + 9t - 65 = (3t)^3 - 3(3t)^2(1) + 3(3t)(1)^2 - 1^3 - 64$

$= (3t - 1)^3 - 4^3$

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

$= (3t - 5)[9t^2 - 6t + 1 + 12t - 4 + 16]$

$= (3t - 5)(9t^2 + 6t + 13)$

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

Answer: (A) $(25a^2b^2 - 10ab + 4)(5ab + 2)$

$125a^3b^3 + 8 = (5ab)^3 + 2^3$

$= (5ab + 2)[(5ab)^2 - (5ab)(2) + 2^2]$

$= (5ab + 2)(25a^2b^2 - 10ab + 4)$

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

Answer: (D) $3(4x - 3y)(3y - 5)(5 - 4x)$

Here $(4x - 3y) + (3y - 5) + (5 - 4x) = 0$

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

$\therefore (4x - 3y)^3 + (3y - 5)^3 + (5 - 4x)^3 = 3(4x - 3y)(3y - 5)(5 - 4x)$

Answer: (C) $2x(x + 2z)(x - 2z)(x^2 + 4z^2)$

$2x^5 - 32xz^4 = 2x(x^4 - 16z^4)$

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

$= 2x(x^2 + 4z^2)(x^2 - 4z^2)$

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

$= 2x(x^2 + 4z^2)(x + 2z)(x - 2z)$

Answer: (C) $(4m^2 + 3)(m + 4)$

$16m^2 + (4m^2 + 3)m + 12 = 16m^2 + 4m^3 + 3m + 12$

$= 4m^3 + 16m^2 + 3m + 12$

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

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