Question paper, Set 6

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

$= (4 + m^2p^2)(4 - m^2p^2)$

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

$= (4 + m^2p^2)(2 + mp)(2 - mp)$

Answer: $(4 + m^2p^2)(2 + mp)(2 - mp)$

$4t^2 + 36t + 81 = (2t)^2 + 9^2 + 2(2t)(9)$

$= (2t + 9)^2$

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

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

Answer: $(2t + 9)^2$

$9m^2 - 42m + 49 = (3m)^2 + 7^2 - 2(3m)(7)$

$= (3m - 7)^2$

$= (3m - 7)(3m - 7)$

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

Answer: $(3m - 7)^2$

$9a^2 - 18ab + 9b^2 - 25 = (3a - 3b)^2 - 5^2$

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

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

Answer: $(3a - 3b + 5)(3a - 3b - 5)$

$5 \times 5 = 25$

$1 \times 25 = 25$ and $1 + 25 = 26$

So the numbers are $1$ and $25$.

Answer: $1$ and $25$

$15 \times 18 = 270$

$(-10) \times (-27) = 270$ and $-10 - 27 = -37$

So the numbers are $-10$ and $-27$.

Answer: $-10$ and $-27$

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

$6 \times (-20) = -120$ and $6 - 20 = -14$

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

Answer: $6$ and $-20$

$29 \times (-243) = -7047$

$7047 = 3 \times 3 \times 3 \times 3 \times 3 \times 29 = 27 \times 261$, and $-27 + 261 = 234$

$= 29z^2 - 27z + 261z - 243$

$= z(29z - 27) + 9(29z - 27)$

$= (29z - 27)(z + 9)$

Answer: $(29z - 27)(z + 9)$

Writing the terms in order: $-12t^2 - 4t + 1$

$(-12) \times 1 = -12$: $(-6) \times 2 = -12$ and $-6 + 2 = -4$

$= -12t^2 - 6t + 2t + 1$

$= -6t(2t + 1) + 1(2t + 1)$

$= (2t + 1)(1 - 6t)$

Answer: $(2t + 1)(1 - 6t)$

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

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

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

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

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

$7\sqrt{5} \times (-20\sqrt{5}) = 7 \times (-20) \times 5 = -700$: $5 \times (-140) = -700$ and $5 - 140 = -135$

$= 7\sqrt{5}y^2 + 5y - 140y - 20\sqrt{5}$

$= y(7\sqrt{5}y + 5) - 4\sqrt{5}(7\sqrt{5}y + 5)$

$= (7\sqrt{5}y + 5)(y - 4\sqrt{5})$

Answer: $(7\sqrt{5}y + 5)(y - 4\sqrt{5})$

$\displaystyle 6t^2 + \frac{73}{35}t + \frac{6}{35} = \frac{210t^2 + 73t + 6}{35}$

$\displaystyle = \frac{1}{35}(210t^2 + 73t + 6)$

Answer: $\displaystyle \frac{1}{35}(210t^2 + 73t + 6)$

Let $p - 1 = x$ and $q - 1 = y$

$4(p - 1)^2 + 29(p - 1)(q - 1) + 30(q - 1)^2 = 4x^2 + 29xy + 30y^2$

$4 \times 30 = 120$: $5 \times 24 = 120$ and $5 + 24 = 29$

$= 4x^2 + 5xy + 24xy + 30y^2$

$= x(4x + 5y) + 6y(4x + 5y)$

$= (4x + 5y)(x + 6y)$

$= [4(p - 1) + 5(q - 1)][(p - 1) + 6(q - 1)]$

$= (4p - 4 + 5q - 5)(p - 1 + 6q - 6)$

$= (4p + 5q - 9)(p + 6q - 7)$

Answer: $(4p + 5q - 9)(p + 6q - 7)$

$8x^2 + 8y^2 + z^2 - 16xy + 4\sqrt{2}yz - 4\sqrt{2}zx = (2\sqrt{2}x)^2 + (2\sqrt{2}y)^2 + z^2 - 2(2\sqrt{2}x)(2\sqrt{2}y) + 2(2\sqrt{2}y)(z) - 2(z)(2\sqrt{2}x)$

$= (-2\sqrt{2}x + 2\sqrt{2}y + z)^2$

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

Answer: $(-2\sqrt{2}x + 2\sqrt{2}y + z)^2$

$p^3 - 3p^2 + 3p - 1 = p^3 - 1^3 - 3(p)^2(1) + 3(p)(1)^2$

$= (p - 1)^3$

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

Answer: $(p - 1)^3$

$a^6 - 64b^3 = (a^2)^3 - (4b)^3$

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

$= (a^2 - 4b)(a^4 + 4a^2b + 16b^2)$

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

Answer: $(a^2 - 4b)(a^4 + 4a^2b + 16b^2)$

$216a^3 + c^3 = (6a)^3 + c^3$

$= (6a + c)[(6a)^2 - (6a)(c) + c^2]$

$= (6a + c)(36a^2 - 6ac + c^2)$

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

Answer: $(6a + c)(36a^2 - 6ac + c^2)$

$16m^4 + 100m^2 + 625 = (4m^2)^2 + 2(4m^2)(25) + 25^2 - 100m^2$

$= (4m^2 + 25)^2 - (10m)^2$

$= (4m^2 + 25 + 10m)(4m^2 + 25 - 10m)$

$= (4m^2 + 10m + 25)(4m^2 - 10m + 25)$

Answer: $(4m^2 + 10m + 25)(4m^2 - 10m + 25)$

Here $(5p - 3q) + (3q - r) + (r - 5p) = 0$

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

$\therefore (5p - 3q)^3 + (3q - r)^3 + (r - 5p)^3 = 3(5p - 3q)(3q - r)(r - 5p)$

Answer: $3(5p - 3q)(3q - r)(r - 5p)$

$2m(m + 3n)^3 - 18m^2n(m + 3n) = 2m(m + 3n)[(m + 3n)^2 - 9mn]$

$= 2m(m + 3n)[m^2 + 6mn + 9n^2 - 9mn]$

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

Answer: $2m(m + 3n)(m^2 - 3mn + 9n^2)$