Question paper, Set 9

$25 - 144t^2 = 5^2 - (12t)^2$

$= (5 + 12t)(5 - 12t)$

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

Answer: $(5 + 12t)(5 - 12t)$

$4m^2 - 12mn + 9n^2 = (2m)^2 + (3n)^2 - 2(2m)(3n)$

$= (2m - 3n)^2$

$= (2m - 3n)(2m - 3n)$

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

Answer: $(2m - 3n)^2$

$4p^2 - 24pq + 36q^2 - 9 = (2p - 6q)^2 - 3^2$

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

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

Answer: $(2p - 6q + 3)(2p - 6q - 3)$

$5 \times 24 = 120$

$6 \times 20 = 120$ and $6 + 20 = 26$

So the numbers are $6$ and $20$.

Answer: $6$ and $20$

$15 \times 8 = 120$

$(-5) \times (-24) = 120$ and $-5 - 24 = -29$

So the numbers are $-5$ and $-24$.

Answer: $-5$ and $-24$

$2 \times (-3) = -6$

$2 \times (-3) = -6$ and $2 - 3 = -1$

So the numbers are $2$ and $-3$.

Answer: $2$ and $-3$

$11 \times (-78) = -858$

$858 = 2 \times 3 \times 11 \times 13 = 22 \times 39$, and $-22 + 39 = 17$

$= 11t^2 - 22t + 39t - 78$

$= 11t(t - 2) + 39(t - 2)$

$= (t - 2)(11t + 39)$

Answer: $(t - 2)(11t + 39)$

Writing the terms in order: $-z^2 + 3z + 28$

$(-1) \times 28 = -28$: $(-4) \times 7 = -28$ and $-4 + 7 = 3$

$= -z^2 - 4z + 7z + 28$

$= -z(z + 4) + 7(z + 4)$

$= (z + 4)(7 - z)$

Answer: $(z + 4)(7 - z)$

Let $m + p = t$

$(m + p)^2 + 3(m + p) - 28 = t^2 + 3t - 28$

$1 \times (-28) = -28$: $(-4) \times 7 = -28$ and $-4 + 7 = 3$

$= t^2 - 4t + 7t - 28$

$= t(t - 4) + 7(t - 4)$

$= (t - 4)(t + 7)$

$= [(m + p) - 4][(m + p) + 7]$

$= (m + p - 4)(m + p + 7)$

Answer: $(m + p - 4)(m + p + 7)$

$4 \times (-6) = -24$: $1 \times (-24) = -24$ and $1 - 24 = -23$

$= 4a^2 + ab - 24ab - 6b^2$

$= a(4a + b) - 6b(4a + b)$

$= (4a + b)(a - 6b)$

Answer: $(4a + b)(a - 6b)$

$6\sqrt{5} \times \sqrt{5} = 6 \times 1 \times 5 = 30$: $3 \times 10 = 30$ and $3 + 10 = 13$

$= 6\sqrt{5}m^2 + 3m + 10m + \sqrt{5}$

$= 3m(2\sqrt{5}m + 1) + \sqrt{5}(2\sqrt{5}m + 1)$

$= (2\sqrt{5}m + 1)(3m + \sqrt{5})$

Answer: $(2\sqrt{5}m + 1)(3m + \sqrt{5})$

$\displaystyle y^2 - \frac{7}{20}y - \frac{3}{20} = \frac{20y^2 - 7y - 3}{20}$

$\displaystyle = \frac{1}{20}(20y^2 - 7y - 3)$

Answer: $\displaystyle \frac{1}{20}(20y^2 - 7y - 3)$

Let $a - 5 = x$ and $c - 3 = y$

$6(a - 5)^2 - 19(a - 5)(c - 3) + 15(c - 3)^2 = 6x^2 - 19xy + 15y^2$

$6 \times 15 = 90$: $(-9) \times (-10) = 90$ and $-9 - 10 = -19$

$= 6x^2 - 9xy - 10xy + 15y^2$

$= 3x(2x - 3y) - 5y(2x - 3y)$

$= (2x - 3y)(3x - 5y)$

$= [2(a - 5) - 3(c - 3)][3(a - 5) - 5(c - 3)]$

$= (2a - 10 - 3c + 9)(3a - 15 - 5c + 15)$

$= (2a - 3c - 1)(3a - 5c)$

Answer: $(2a - 3c - 1)(3a - 5c)$

$16a^2 + 18b^2 + 2c^2 + 24\sqrt{2}ab - 12bc - 8\sqrt{2}ca = (4a)^2 + (3\sqrt{2}b)^2 + (\sqrt{2}c)^2 + 2(4a)(3\sqrt{2}b) - 2(3\sqrt{2}b)(\sqrt{2}c) - 2(\sqrt{2}c)(4a)$

$= (4a + 3\sqrt{2}b - \sqrt{2}c)^2$

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

Answer: $(4a + 3\sqrt{2}b - \sqrt{2}c)^2$

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

$= (x - 3)^3$

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

Answer: $(x - 3)^3$

$27 + 125p^3 = 3^3 + (5p)^3$

$= (3 + 5p)[3^2 - (3)(5p) + (5p)^2]$

$= (5p + 3)(25p^2 - 15p + 9)$

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

Answer: $(5p + 3)(25p^2 - 15p + 9)$

$625a^4 + 4c^4 = (25a^2)^2 + (2c^2)^2 + 2(25a^2)(2c^2) - 100a^2c^2$

$= (25a^2 + 2c^2)^2 - (10ac)^2$

$= (25a^2 + 2c^2 + 10ac)(25a^2 + 2c^2 - 10ac)$

$= (25a^2 + 10ac + 2c^2)(25a^2 - 10ac + 2c^2)$

Answer: $(25a^2 + 10ac + 2c^2)(25a^2 - 10ac + 2c^2)$

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

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

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

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

$16x^5y - 81xy^5 = xy(16x^4 - 81y^4)$

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

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

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

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

Answer: $xy(4x^2 + 9y^2)(2x + 3y)(2x - 3y)$

$(y - 1)(y^2 + 16) - 8y^2 + 8y = (y - 1)(y^2 + 16) - 8y(y - 1)$

$= (y - 1)(y^2 + 16 - 8y)$

$= (y - 1)(y^2 - 8y + 16)$

$= (y - 1)(y - 4)^2$

Answer: $(y - 1)(y - 4)^2$