Factorization: MCQ test, Set 6 | Expresswala Academy

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Factorize $256 - m^4$ .
A $(-m + 4)(-m^2 + 16)(m + 4)$ B $(-m + 4)^2(m^2 + 16)$ C $(16 + m^2)(4 + m)(4 - m)$ D $(m + 4)^2(m^2 + 16)$

Answer and solution

Answer: (C) $(16 + m^2)(4 + m)(4 - m)$

$256 - m^4 = 16^2 - (m^2)^2$

$= (16 + m^2)(16 - m^2)$

$= (16 + m^2)[4^2 - m^2]$

$= (16 + m^2)(4 + m)(4 - m)$
The factors of $y^2 + 4y + 4$ are
A $(y + 2)(y - 2)$ B $(2y + 1)^2$ C $(y - 2)^2$ D $(y + 2)^2$

Answer and solution

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

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

$= (y + 2)^2$

$= (y + 2)(y + 2)$

using $a^2 + b^2 + 2ab = (a + b)^2$
The factors of $36z^2 + 84z + 49$ are
A $(6z + 7)(6z - 7)$ B $(7z + 6)^2$ C $(6z + 7)^2$ D $(6z - 7)^2$

Answer and solution

Answer: (C) $(6z + 7)^2$

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

$= (6z + 7)^2$

$= (6z + 7)(6z + 7)$

using $a^2 + b^2 + 2ab = (a + b)^2$
The factors of $25m^2 - 80m + 64$ are
A $(5m - 8)^2$ B $(5m + 8)^2$ C $(5m + 8)(5m - 8)$ D $(8m - 5)^2$

Answer and solution

Answer: (A) $(5m - 8)^2$

$25m^2 - 80m + 64 = (5m)^2 + 8^2 - 2(5m)(8)$

$= (5m - 8)^2$

$= (5m - 8)(5m - 8)$

using $a^2 + b^2 - 2ab = (a - b)^2$
$25n^2 - 40n + 16 =$
A $(5n + 4)^2$ B $(5n - 4)^2$ C $(5n + 4)(5n - 4)$ D $(4n - 5)^2$

Answer and solution

Answer: (B) $(5n - 4)^2$

$25n^2 - 40n + 16 = (5n)^2 + 4^2 - 2(5n)(4)$

$= (5n - 4)^2$

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

using $a^2 + b^2 - 2ab = (a - b)^2$
Factorize $p^2 + 2p + 1 - 36q^2$ .
A $(p + 1 + 6q)(p - 1 - 6q)$ B $(p - 1 + 6q)(p - 1 - 6q)$ C $(p + 1 + 6q)(p + 1 - 6q)$ D $(p + 1 - 6q)^2$

Answer and solution

Answer: (C) $(p + 1 + 6q)(p + 1 - 6q)$

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

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

using $a^2 + 2ab + b^2 = (a + b)^2$ and $a^2 - b^2 = (a + b)(a - b)$
The factors of $5b^2 + 19b + 18$ are
A $(5b + 2)(b + 9)$ B $(5b - 9)(b - 2)$ C $(5b - 9)(b + 2)$ D $(5b + 9)(b + 2)$

Answer and solution

Answer: (D) $(5b + 9)(b + 2)$

$5 \times 18 = 90$ : $9 \times 10 = 90$ and $9 + 10 = 19$

$= 5b^2 + 9b + 10b + 18$

$= b(5b + 9) + 2(5b + 9)$

$= (5b + 9)(b + 2)$
To factorize $3p^2 - 22p + 24$ by splitting the middle term, we need two numbers whose product is $3 \times 24$ and whose sum is $-22$ . They are
A $4$ and $18$ B $-4$ and $-18$ C $-12$ and $-6$ D $-9$ and $-8$

Answer and solution

Answer: (B) $-4$ and $-18$

$3 \times 24 = 72$

$(-4) \times (-18) = 72$ and $-4 - 18 = -22$

So the numbers are $-4$ and $-18$ .
To factorize $5t^2 - 26t - 24$ by splitting the middle term, we need two numbers whose product is $5 \times (-24)$ and whose sum is $-26$ . They are
A $4$ and $-30$ B $-40$ and $3$ C $-4$ and $30$ D $-24$ and $5$

Answer and solution

Answer: (A) $4$ and $-30$

$5 \times (-24) = -120$

$4 \times (-30) = -120$ and $4 - 30 = -26$

So the numbers are $4$ and $-30$ .
Factorize $29x^2 + 28x - 60$ .
A $(29x + 2)(x - 30)$ B $(29x + 30)(x - 2)$ C $(29x - 30)(x + 2)$ D $(29x + 30)(x + 2)$

Answer and solution

Answer: (C) $(29x - 30)(x + 2)$

$29 \times (-60) = -1740$

$1740 = 2 \times 2 \times 3 \times 5 \times 29 = 30 \times 58$ , and $-30 + 58 = 28$

$= 29x^2 - 30x + 58x - 60$

$= x(29x - 30) + 2(29x - 30)$

$= (29x - 30)(x + 2)$
$11n - 9 - 2n^2 =$
A $(2n + 9)(n + 1)$ B $(2n - 9)(n - 1)$ C $(-2n - 9)(1 - n)$ D $(9 - 2n)(n - 1)$

Answer and solution

Answer: (D) $(9 - 2n)(n - 1)$

Writing the terms in order: $-2n^2 + 11n - 9$

$(-2) \times (-9) = 18$ : $2 \times 9 = 18$ and $2 + 9 = 11$

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

$= -2n(n - 1) + 9(n - 1)$

$= (n - 1)(9 - 2n)$
Factorize $13t + 30 - 3t^2$ .
A $(3t - 5)(t + 6)$ B $(3t + 5)(6 - t)$ C $(5 - 3t)(6 - t)$ D $(3t + 5)(t - 6)$

Answer and solution

Answer: (B) $(3t + 5)(6 - t)$

Writing the terms in order: $-3t^2 + 13t + 30$

$(-3) \times 30 = -90$ : $(-5) \times 18 = -90$ and $-5 + 18 = 13$

$= -3t^2 - 5t + 18t + 30$

$= -t(3t + 5) + 6(3t + 5)$

$= (3t + 5)(6 - t)$
Factorize $(m - 2n)^2 - 5(m - 2n) - 14$ .
A $(m - 2n + 2)(m - 2n + 7)$ B $(m - 2n + 7)(m - 2n - 2)$ C $(m - 2n + 2)(m - 2n - 7)$ D $(m - 2n - 2)(m - 2n - 7)$

Answer and solution

Answer: (C) $(m - 2n + 2)(m - 2n - 7)$

Let $m - 2n = p$

$(m - 2n)^2 - 5(m - 2n) - 14 = p^2 - 5p - 14$

$1 \times (-14) = -14$ : $2 \times (-7) = -14$ and $2 - 7 = -5$

$= p^2 + 2p - 7p - 14$

$= p(p + 2) - 7(p + 2)$

$= (p + 2)(p - 7)$

$= [(m - 2n) + 2][(m - 2n) - 7]$

$= (m - 2n + 2)(m - 2n - 7)$
On factorizing, $2m^2 - 11mn - 6n^2 =$
A $(2m - n)(m + 6n)$ B $(2m + n)(m - 6n)$ C $(2m - 6n)(m + n)$ D $(2m + n)(m + 6n)$

Answer and solution

Answer: (B) $(2m + n)(m - 6n)$

$2 \times (-6) = -12$ : $1 \times (-12) = -12$ and $1 - 12 = -11$

$= 2m^2 + mn - 12mn - 6n^2$

$= m(2m + n) - 6n(2m + n)$

$= (2m + n)(m - 6n)$
On factorizing, $a^2 - \sqrt{5}a - 30 =$
A $(a + 2\sqrt{5})(a - 3\sqrt{5})$ B $(a + 3\sqrt{5})(a - 2\sqrt{5})$ C $(a + 2\sqrt{5})(a + 3\sqrt{5})$ D $(a - 2\sqrt{5})(a - 3\sqrt{5})$

Answer and solution

Answer: (A) $(a + 2\sqrt{5})(a - 3\sqrt{5})$

Product $= -30$ , sum $= -\sqrt{5}$ : $2\sqrt{5} \times (-3\sqrt{5}) = -30$ and $2\sqrt{5} + (-3\sqrt{5}) = -\sqrt{5}$

$= a^2 + 2\sqrt{5}a - 3\sqrt{5}a - 30$

$= a(a + 2\sqrt{5}) - 3\sqrt{5}(a + 2\sqrt{5})$

$= (a + 2\sqrt{5})(a - 3\sqrt{5})$
The factors of $\displaystyle 11m^2 + 8m + \frac{16}{11}$ are
A $\displaystyle (11m - 4)\left(m + \frac{4}{11}\right)$ B $(11m + 4)(m + 4)$ C $\displaystyle (11m + 4)\left(m + \frac{4}{11}\right)$ D $\displaystyle (11m - 4)\left(m - \frac{4}{11}\right)$

Answer and solution

Answer: (C) $\displaystyle (11m + 4)\left(m + \frac{4}{11}\right)$

$\displaystyle 11m^2 + 8m + \frac{16}{11} = \frac{121m^2 + 88m + 16}{11}$

$\displaystyle = \frac{1}{11}(121m^2 + 88m + 16)$

$\displaystyle = \frac{1}{11}[(11m)^2 + 4^2 + 2(11m)(4)]$

$\displaystyle = \frac{1}{11}(11m + 4)^2$

$\displaystyle = \left(m + \frac{4}{11}\right)(11m + 4)$
On factorizing, $5(a + 1)^2 + 16(a + 1)(b - 3) + 3(b - 3)^2 =$
A $(5a + b + 2)(a + 3b + 8)$ B $(5a + b - 2)(a + 3b - 2)$ C $(5a + b - 2)(a + 3b - 8)$ D $(5a + b + 2)(a + 3b - 8)$

Answer and solution

Answer: (D) $(5a + b + 2)(a + 3b - 8)$

Let $a + 1 = x$ and $b - 3 = y$

$5(a + 1)^2 + 16(a + 1)(b - 3) + 3(b - 3)^2 = 5x^2 + 16xy + 3y^2$

$5 \times 3 = 15$ : $1 \times 15 = 15$ and $1 + 15 = 16$

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

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

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

$= [5(a + 1) + (b - 3)][(a + 1) + 3(b - 3)]$

$= (5a + 5 + b - 3)(a + 1 + 3b - 9)$

$= (5a + b + 2)(a + 3b - 8)$
The factors of $4p^2 + 4q^2 + 9r^2 - 8pq - 12qr + 12rp$ are
A $(2p + 2q - 3r)^2$ B $(2p - 2q + 3r)^2$ C $(4p - 4q + 9r)^2$ D $(2p + 2q + 3r)^2$

Answer and solution

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

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

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

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

Answer and solution

Answer: (A) $(2x + y)^3$

$8x^3 + 12x^2y + 6xy^2 + y^3 = (2x)^3 + y^3 + 3(2x)^2(y) + 3(2x)(y)^2$

$= (2x + y)^3$

using $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
The factors of $8a^3 + 36a^2c + 54ac^2 - 37c^3$ are
A $(2a - c)(4a^2 + 12ac + 25c^2)$ B $(2a + 7c)(4a^2 + 4ac + 13c^2)$ C $(2a - c)(4a^2 + 4ac + 13c^2)$ D $(2a - c)(4a^2 + 20ac + 37c^2)$

Answer and solution

Answer: (D) $(2a - c)(4a^2 + 20ac + 37c^2)$

$8a^3 + 36a^2c + 54ac^2 - 37c^3 = (2a)^3 + 3(2a)^2(3c) + 3(2a)(3c)^2 + (3c)^3 - 64c^3$

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

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

$= (2a - c)[4a^2 + 12ac + 9c^2 + 8ac + 12c^2 + 16c^2]$

$= (2a - c)(4a^2 + 20ac + 37c^2)$

using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$8 + 27y^3z^3 =$
A $(3yz + 2)(9y^2z^2 - 6yz + 4)$ B $(3yz + 2)(9y^2z^2 + 6yz + 4)$ C $(3yz + 2)(9y^2z^2 - 12yz + 4)$ D $(2 - 3yz)(9y^2z^2 + 6yz + 4)$

Answer and solution

Answer: (A) $(3yz + 2)(9y^2z^2 - 6yz + 4)$

$8 + 27y^3z^3 = 2^3 + (3yz)^3$

$= (2 + 3yz)[2^2 - (2)(3yz) + (3yz)^2]$

$= (3yz + 2)(9y^2z^2 - 6yz + 4)$

using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
Which of the following is a factor of $x^3 - 13x - 12$ ?
A $x + 2$ B $x - 1$ C $x - 2$ D $x + 3$

Answer and solution

Answer: (D) $x + 3$

$x^3 - 13x - 12 = x^3 - x - 12x - 12$

$= x(x^2 - 1) - 12(x + 1)$

$= (x + 1)(x^2 - x - 12)$

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

So $x + 3$ is a factor.
On factorizing, $(5a - b)^3 + (b - 2)^3 + (2 - 5a)^3 =$
A $-3(5a - b)(b - 2)(2 - 5a)$ B $0$ C $3(5a - b)(b - 2)(2 - 5a)$ D $(5a - b)(b - 2)(2 - 5a)$

Answer and solution

Answer: (C) $3(5a - b)(b - 2)(2 - 5a)$

Here $(5a - b) + (b - 2) + (2 - 5a) = 0$

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

$\therefore (5a - b)^3 + (b - 2)^3 + (2 - 5a)^3 = 3(5a - b)(b - 2)(2 - 5a)$
$5a^5b - 5ab^5 =$
A $5ab(a - b)^2(a^2 + b^2)$ B $5ab(a + b)(a - b)(a^2 + b^2)$ C $5ab(a + b)^2(a^2 + b^2)$ D $5ab(a + b)^2(a - b)^2$

Answer and solution

Answer: (B) $5ab(a + b)(a - b)(a^2 + b^2)$

$5a^5b - 5ab^5 = 5ab(a^4 - b^4)$

$= 5ab[(a^2)^2 - (b^2)^2]$

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

$= 5ab(a^2 + b^2)(a + b)(a - b)$
The factors of $5a^2 + (a^2 + 3)a + 15$ are
A $(a + 5)(a^2 + 3)$ B $(a + 5)(a^2 - 3)$ C $(a - 5)(a^2 + 3)$ D $(a + 3)(a^2 + 5)$

Answer and solution

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

$5a^2 + (a^2 + 3)a + 15 = 5a^2 + a^3 + 3a + 15$

$= a^3 + 5a^2 + 3a + 15$

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

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

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MCQ test, Set 6