Factorization: MCQ test, Set 5 | Expresswala Academy

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

The factors of $a^4b^4 - 1$ are
A $(a^2b^2 + 1)(ab + 1)^2$ B $(a^2b^2 + 1)(ab + 1)(ab - 1)$ C $(a^2b^2 - 1)(ab + 1)(ab - 1)$ D $(a^2b^2 + 1)(ab - 1)^2$

Answer and solution

Answer: (B) $(a^2b^2 + 1)(ab + 1)(ab - 1)$

$a^4b^4 - 1 = (a^2b^2)^2 - 1^2$

$= (a^2b^2 + 1)(a^2b^2 - 1)$

$= (a^2b^2 + 1)[(ab)^2 - 1^2]$

$= (a^2b^2 + 1)(ab + 1)(ab - 1)$
$64m^2 - 121n^2 =$
A $(8m + 11n)^2$ B $(8m + 11n)(8m - 11n)$ C $(64m + 11n)(64m - 11n)$ D $(8m - 11n)^2$

Answer and solution

Answer: (B) $(8m + 11n)(8m - 11n)$

$64m^2 - 121n^2 = (8m)^2 - (11n)^2$

$= (8m + 11n)(8m - 11n)$

using $a^2 - b^2 = (a + b)(a - b)$
Factorize $x^2 + 12xz + 36z^2$ .
A $(6x + z)^2$ B $(x - 6z)^2$ C $(x + 6z)^2$ D $(x + 6z)(x - 6z)$

Answer and solution

Answer: (C) $(x + 6z)^2$

$x^2 + 12xz + 36z^2 = x^2 + (6z)^2 + 2(x)(6z)$

$= (x + 6z)^2$

$= (x + 6z)(x + 6z)$

using $a^2 + b^2 + 2ab = (a + b)^2$
On factorizing, $4m^2 - 28m + 49 =$
A $(2m + 7)(2m - 7)$ B $(7m - 2)^2$ C $(2m + 7)^2$ D $(2m - 7)^2$

Answer and solution

Answer: (D) $(2m - 7)^2$

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

$= (2m - 7)^2$

$= (2m - 7)(2m - 7)$

using $a^2 + b^2 - 2ab = (a - b)^2$
On factorizing, $4x^2 + 4x + 1 - 16y^2 =$
A $(2x + 1 - 4y)^2$ B $(2x + 1 + 4y)(2x + 1 - 4y)$ C $(2x - 1 + 4y)(2x - 1 - 4y)$ D $(2x + 1 + 4y)(2x - 1 - 4y)$

Answer and solution

Answer: (B) $(2x + 1 + 4y)(2x + 1 - 4y)$

$4x^2 + 4x + 1 - 16y^2 = (2x + 1)^2 - (4y)^2$

$= (2x + 1 + 4y)(2x + 1 - 4y)$

using $a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - b^2 = (a + b)(a - b)$
To factorize $2x^2 + 21x + 40$ by splitting the middle term, we need two numbers whose product is $2 \times 40$ and whose sum is $21$ . They are
A $5$ and $16$ B $-5$ and $-16$ C $8$ and $10$ D $4$ and $20$

Answer and solution

Answer: (A) $5$ and $16$

$2 \times 40 = 80$

$5 \times 16 = 80$ and $5 + 16 = 21$

So the numbers are $5$ and $16$ .
$4z^2 - 25z + 6 =$
A $(4z - 1)(z + 6)$ B $(4z - 6)(z - 1)$ C $(4z - 1)(z - 6)$ D $(4z + 1)(z + 6)$

Answer and solution

Answer: (C) $(4z - 1)(z - 6)$

$4 \times 6 = 24$ : $(-1) \times (-24) = 24$ and $-1 - 24 = -25$

$= 4z^2 - z - 24z + 6$

$= z(4z - 1) - 6(4z - 1)$

$= (4z - 1)(z - 6)$
The factors of $5z^2 - 17z + 14$ are
A $(5z - 7)(z - 2)$ B $(5z + 7)(z + 2)$ C $(5z - 2)(z - 7)$ D $(5z + 7)(z - 2)$

Answer and solution

Answer: (A) $(5z - 7)(z - 2)$

$5 \times 14 = 70$ : $(-7) \times (-10) = 70$ and $-7 - 10 = -17$

$= 5z^2 - 7z - 10z + 14$

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

$= (5z - 7)(z - 2)$
The factors of $10z^2 - 37z - 36$ are
A $(2z + 9)(5z - 4)$ B $(2z - 9)(5z + 4)$ C $(2z + 4)(5z - 9)$ D $(2z - 9)(5z - 4)$

Answer and solution

Answer: (B) $(2z - 9)(5z + 4)$

$10 \times (-36) = -360$ : $8 \times (-45) = -360$ and $8 - 45 = -37$

$= 10z^2 + 8z - 45z - 36$

$= 2z(5z + 4) - 9(5z + 4)$

$= (5z + 4)(2z - 9)$
$11t^2 - 50t + 51 =$
A $(11t - 3)(t - 17)$ B $(11t - 17)(t - 3)$ C $(11t + 17)(t + 3)$ D $(11t + 17)(t - 3)$

Answer and solution

Answer: (B) $(11t - 17)(t - 3)$

$11 \times 51 = 561$

$561 = 3 \times 11 \times 17 = 17 \times 33$ , and $-17 - 33 = -50$

$= 11t^2 - 17t - 33t + 51$

$= t(11t - 17) - 3(11t - 17)$

$= (11t - 17)(t - 3)$
$a + 30 - a^2 =$
A $(6 - a)(a + 5)$ B $(a + 5)(a - 6)$ C $(a + 6)(a - 5)$ D $(5 - a)(6 - a)$

Answer and solution

Answer: (A) $(6 - a)(a + 5)$

Writing the terms in order: $-a^2 + a + 30$

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

$= -a^2 - 5a + 6a + 30$

$= -a(a + 5) + 6(a + 5)$

$= (a + 5)(6 - a)$
Factorize $(p - 3q)^2 - 3(p - 3q) - 18$ .
A $(p - 3q + 3)(p - 3q + 6)$ B $(p - 3q + 6)(p - 3q - 3)$ C $(p - 3q + 3)(p - 3q - 6)$ D $(p - 3q - 3)(p - 3q - 6)$

Answer and solution

Answer: (C) $(p - 3q + 3)(p - 3q - 6)$

Let $p - 3q = t$

$(p - 3q)^2 - 3(p - 3q) - 18 = t^2 - 3t - 18$

$1 \times (-18) = -18$ : $3 \times (-6) = -18$ and $3 - 6 = -3$

$= t^2 + 3t - 6t - 18$

$= t(t + 3) - 6(t + 3)$

$= (t + 3)(t - 6)$

$= [(p - 3q) + 3][(p - 3q) - 6]$

$= (p - 3q + 3)(p - 3q - 6)$
To factorize $7x^2 + 5xy - 2y^2$ by splitting the middle term, we need two numbers whose product is $7 \times (-2)$ and whose sum is $5$ . They are
A $-14$ and $1$ B $-1$ and $14$ C $-2$ and $7$ D $2$ and $-7$

Answer and solution

Answer: (C) $-2$ and $7$

$7 \times (-2) = -14$

$(-2) \times 7 = -14$ and $-2 + 7 = 5$

So the numbers are $-2$ and $7$ .
To factorize $16x^2 + 8xz - 3z^2$ by splitting the middle term, we need two numbers whose product is $16 \times (-3)$ and whose sum is $8$ . They are
A $-4$ and $12$ B $-6$ and $8$ C $-3$ and $16$ D $4$ and $-12$

Answer and solution

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

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

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

So the numbers are $-4$ and $12$ .
On factorizing, $4\sqrt{2}y^2 - 31y - 4\sqrt{2} =$
A $(4\sqrt{2}y + 1)(y - 4\sqrt{2})$ B $(4\sqrt{2}y - 1)(y + 4\sqrt{2})$ C $(4\sqrt{2}y + 1)(y + 4\sqrt{2})$ D $(4\sqrt{2}y - 1)(y - 4\sqrt{2})$

Answer and solution

Answer: (A) $(4\sqrt{2}y + 1)(y - 4\sqrt{2})$

$4\sqrt{2} \times (-4\sqrt{2}) = 4 \times (-4) \times 2 = -32$ : $1 \times (-32) = -32$ and $1 - 32 = -31$

$= 4\sqrt{2}y^2 + y - 32y - 4\sqrt{2}$

$= y(4\sqrt{2}y + 1) - 4\sqrt{2}(4\sqrt{2}y + 1)$

$= (4\sqrt{2}y + 1)(y - 4\sqrt{2})$
Factorize $\displaystyle 3t^2 + \frac{43}{10}t - \frac{3}{10}$ .
A $\displaystyle \left(3t - \frac{1}{2}\right)\left(t + \frac{3}{5}\right)$ B $\displaystyle \left(3t + \frac{1}{5}\right)\left(t - \frac{3}{2}\right)$ C $\displaystyle \left(3t + \frac{1}{5}\right)\left(t + \frac{3}{2}\right)$ D $\displaystyle \left(3t - \frac{1}{5}\right)\left(t + \frac{3}{2}\right)$

Answer and solution

Answer: (D) $\displaystyle \left(3t - \frac{1}{5}\right)\left(t + \frac{3}{2}\right)$

$\displaystyle 3t^2 + \frac{43}{10}t - \frac{3}{10} = \frac{30t^2 + 43t - 3}{10}$

$\displaystyle = \frac{1}{10}(30t^2 + 43t - 3)$

$30 \times (-3) = -90$ : $(-2) \times 45 = -90$ and $-2 + 45 = 43$

$\displaystyle = \frac{1}{10}[30t^2 - 2t + 45t - 3]$

$\displaystyle = \frac{1}{10}[2t(15t - 1) + 3(15t - 1)]$

$\displaystyle = \frac{1}{10}(15t - 1)(2t + 3)$

$\displaystyle = \left(t + \frac{3}{2}\right)\left(3t - \frac{1}{5}\right)$
The factors of $4(p + 4)^2 - 5(p + 4)(q - 4) - 6(q - 4)^2$ are
A $(4p + 3q - 4)(p - 2q + 12)$ B $(4p + 3q)(p - 2q)$ C $(4p + 3q + 4)(p - 2q - 12)$ D $(4p + 3q + 4)(p - 2q + 12)$

Answer and solution

Answer: (D) $(4p + 3q + 4)(p - 2q + 12)$

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

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

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

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

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

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

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

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

$= (4p + 3q + 4)(p - 2q + 12)$
Factorize $x^2 + 9y^2 + 9z^2 + 6xy - 18yz - 6zx$ .
A $(-x + 3y + 3z)^2$ B $(x + 3y + 3z)^2$ C $(x + 3y - 3z)^2$ D $(x + 9y - 9z)^2$

Answer and solution

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

$x^2 + 9y^2 + 9z^2 + 6xy - 18yz - 6zx = x^2 + (3y)^2 + (3z)^2 + 2(x)(3y) - 2(3y)(3z) - 2(3z)(x)$

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

using $a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = (a + b - c)^2$
$b^3 + 6b^2 + 12b + 8 =$
A $(b - 2)^3$ B $(b + 2)(b^2 - 2b + 4)$ C $(b + 2)^2(b - 2)$ D $(b + 2)^3$

Answer and solution

Answer: (D) $(b + 2)^3$

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

$= (b + 2)^3$

using $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$216x^3y^3 - 1 =$
A $(36x^2y^2 + 6xy + 1)(6xy - 1)$ B $(36x^2y^2 + 12xy + 1)(6xy - 1)$ C $(36x^2y^2 - 6xy + 1)(6xy + 1)$ D $(36x^2y^2 - 6xy + 1)(6xy - 1)$

Answer and solution

Answer: (A) $(36x^2y^2 + 6xy + 1)(6xy - 1)$

$216x^3y^3 - 1 = (6xy)^3 - 1^3$

$= (6xy - 1)[(6xy)^2 + (6xy)(1) + 1^2]$

$= (6xy - 1)(36x^2y^2 + 6xy + 1)$

using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Factorize $216 + p^3$ .
A $(p + 6)(p^2 + 6p + 36)$ B $(6 - p)(p^2 + 6p + 36)$ C $(p + 6)(p^2 - 12p + 36)$ D $(p + 6)(p^2 - 6p + 36)$

Answer and solution

Answer: (D) $(p + 6)(p^2 - 6p + 36)$

$216 + p^3 = 6^3 + p^3$

$= (6 + p)[6^2 - (6)(p) + p^2]$

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

using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
The factors of $4x^4 + 81$ are
A $(2x^2 + 9)^2$ B $(6x + 9 - 2x^2)(9 - 2x^2 - 6x)$ C $(2x^2 + 3x + 9)(2x^2 - 3x + 9)$ D $(2x^2 + 6x + 9)(2x^2 - 6x + 9)$

Answer and solution

Answer: (D) $(2x^2 + 6x + 9)(2x^2 - 6x + 9)$

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

$= (2x^2 + 9)^2 - (6x)^2$

$= (2x^2 + 9 + 6x)(2x^2 + 9 - 6x)$

$= (2x^2 + 6x + 9)(2x^2 - 6x + 9)$
$(2m - n)^3 + (n - 5)^3 + (5 - 2m)^3 =$
A $(2m - n)(n - 5)(5 - 2m)$ B $3(2m - n)(n - 5)(5 - 2m)$ C $0$ D $-3(2m - n)(n - 5)(5 - 2m)$

Answer and solution

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

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

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

$\therefore (2m - n)^3 + (n - 5)^3 + (5 - 2m)^3 = 3(2m - n)(n - 5)(5 - 2m)$
$405x - 80x^5 =$
A $5x(3 - 2x)^2(4x^2 + 9)$ B $5x(2x + 3)^2(3 - 2x)^2$ C $5x(2x + 3)(3 - 2x)(4x^2 + 9)$ D $5x(2x + 3)^2(4x^2 + 9)$

Answer and solution

Answer: (C) $5x(2x + 3)(3 - 2x)(4x^2 + 9)$

$405x - 80x^5 = 5x(81 - 16x^4)$

$= 5x[9^2 - (4x^2)^2]$

$= 5x(4x^2 + 9)(9 - 4x^2)$

$= 5x(4x^2 + 9)[3^2 - (2x)^2]$

$= 5x(4x^2 + 9)(2x + 3)(3 - 2x)$
Factorize $9x^2 + (3x^2 + 4)x + 12$ .
A $(3x^2 + 3)(x + 4)$ B $(3x^2 - 4)(x + 3)$ C $(3x^2 + 4)(x - 3)$ D $(3x^2 + 4)(x + 3)$

Answer and solution

Answer: (D) $(3x^2 + 4)(x + 3)$

$9x^2 + (3x^2 + 4)x + 12 = 9x^2 + 3x^3 + 4x + 12$

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

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

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

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MCQ test, Set 5