Question paper, Set 5

$49 - 64n^2 = 7^2 - (8n)^2$

$= (7 + 8n)(7 - 8n)$

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

Answer: $(7 + 8n)(7 - 8n)$

$25y^2 + 20y + 4 = (5y)^2 + 2^2 + 2(5y)(2)$

$= (5y + 2)^2$

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

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

Answer: $(5y + 2)^2$

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

$= (7y - 9z)^2$

$= (7y - 9z)(7y - 9z)$

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

Answer: $(7y - 9z)^2$

$3 \times 14 = 42$: $6 \times 7 = 42$ and $6 + 7 = 13$

$= 3t^2 + 6t + 7t + 14$

$= 3t(t + 2) + 7(t + 2)$

$= (t + 2)(3t + 7)$

Answer: $(t + 2)(3t + 7)$

$4 \times 40 = 160$

$(-5) \times (-32) = 160$ and $-5 - 32 = -37$

So the numbers are $-5$ and $-32$.

Answer: $-5$ and $-32$

$12 \times (-25) = -300$: $15 \times (-20) = -300$ and $15 - 20 = -5$

$= 12z^2 + 15z - 20z - 25$

$= 3z(4z + 5) - 5(4z + 5)$

$= (4z + 5)(3z - 5)$

Answer: $(4z + 5)(3z - 5)$

$29 \times (-36) = -1044$

$1044 = 2 \times 2 \times 3 \times 3 \times 29 = 12 \times 87$, and $12 - 87 = -75$

$= 29p^2 + 12p - 87p - 36$

$= p(29p + 12) - 3(29p + 12)$

$= (29p + 12)(p - 3)$

Answer: $(29p + 12)(p - 3)$

Writing the terms in order: $-b^2 - 2b + 48$

$(-1) \times 48 = -48$: $(-8) \times 6 = -48$ and $-8 + 6 = -2$

$= -b^2 - 8b + 6b + 48$

$= -b(b + 8) + 6(b + 8)$

$= (b + 8)(6 - b)$

Answer: $(b + 8)(6 - b)$

Let $p + q = t$

$(p + q)^2 + 2(p + q) - 24 = t^2 + 2t - 24$

$1 \times (-24) = -24$: $(-4) \times 6 = -24$ and $-4 + 6 = 2$

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

$= t(t - 4) + 6(t - 4)$

$= (t - 4)(t + 6)$

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

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

Answer: $(p + q - 4)(p + q + 6)$

$2 \times 2 = 4$

$1 \times 4 = 4$ and $1 + 4 = 5$

So the numbers are $1$ and $4$.

Answer: $1$ and $4$

$\displaystyle 6t^2 - \frac{1}{5}t - \frac{2}{15} = \frac{90t^2 - 3t - 2}{15}$

$\displaystyle = \frac{1}{15}(90t^2 - 3t - 2)$

$90 \times (-2) = -180$: $12 \times (-15) = -180$ and $12 - 15 = -3$

$\displaystyle = \frac{1}{15}[90t^2 + 12t - 15t - 2]$

$\displaystyle = \frac{1}{15}[6t(15t + 2) - 1(15t + 2)]$

$\displaystyle = \frac{1}{15}(15t + 2)(6t - 1)$

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

Answer: $\displaystyle \frac{1}{15}(15t + 2)(6t - 1)$, i.e. $\displaystyle \left(3t + \frac{2}{5}\right)\left(2t - \frac{1}{3}\right)$

$9p^2 + 9q^2 + 4r^2 + 18pq - 12qr - 12rp = (3p)^2 + (3q)^2 + (2r)^2 + 2(3p)(3q) - 2(3q)(2r) - 2(2r)(3p)$

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

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

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

$8z^3 + 12z^2 + 6z + 1 = (2z)^3 + 1^3 + 3(2z)^2(1) + 3(2z)(1)^2$

$= (2z + 1)^3$

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

Answer: $(2z + 1)^3$

$216x^6 - z^6 = (6x^2)^3 - (z^2)^3$

$= (6x^2 - z^2)[(6x^2)^2 + (6x^2)(z^2) + (z^2)^2]$

$= (6x^2 - z^2)(36x^4 + 6x^2z^2 + z^4)$

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

Answer: $(6x^2 - z^2)(36x^4 + 6x^2z^2 + z^4)$

$125y^6 + 64 = (5y^2)^3 + 4^3$

$= (5y^2 + 4)[(5y^2)^2 - (5y^2)(4) + 4^2]$

$= (5y^2 + 4)(25y^4 - 20y^2 + 16)$

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

Answer: $(5y^2 + 4)(25y^4 - 20y^2 + 16)$

$x^3 - 19x + 30 = x^3 - 4x - 15x + 30$

$= x(x^2 - 4) - 15(x - 2)$

$= x(x - 2)(x + 2) - 15(x - 2)$

$= (x - 2)[x(x + 2) - 15]$

$= (x - 2)(x^2 + 2x - 15)$

Now $x^2 + 2x - 15$: $(-3) \times 5 = -15$ and $-3 + 5 = 2$

$x^2 + 2x - 15 = x^2 - 3x + 5x - 15$

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

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

So $x^3 - 19x + 30 = (x - 2)(x - 3)(x + 5)$

Answer: $(x - 2)(x - 3)(x + 5)$

$1024a^4 + 1 = 1^2 + (32a^2)^2 + 2(1)(32a^2) - 64a^2$

$= (32a^2 + 1)^2 - (8a)^2$

$= (32a^2 + 1 + 8a)(32a^2 + 1 - 8a)$

$= (32a^2 + 8a + 1)(32a^2 - 8a + 1)$

Answer: $(32a^2 + 8a + 1)(32a^2 - 8a + 1)$

Here $(a - 3b) + (3b - 5) + (5 - a) = 0$

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

$\therefore (a - 3b)^3 + (3b - 5)^3 + (5 - a)^3 = 3(a - 3b)(3b - 5)(5 - a)$

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

$256t^9 - t = t(256t^8 - 1)$

$= t[(16t^4)^2 - 1^2]$

$= t(16t^4 + 1)(16t^4 - 1)$

$= t(16t^4 + 1)[(4t^2)^2 - 1^2]$

$= t(16t^4 + 1)(4t^2 + 1)(4t^2 - 1)$

$= t(16t^4 + 1)(4t^2 + 1)[(2t)^2 - 1^2]$

$= t(16t^4 + 1)(4t^2 + 1)(2t + 1)(2t - 1)$

Answer: $t(16t^4 + 1)(4t^2 + 1)(2t + 1)(2t - 1)$

$p(p - 2q)^3 + 6p^2q(p - 2q) = p(p - 2q)[(p - 2q)^2 + 6pq]$

$= p(p - 2q)[p^2 - 4pq + 4q^2 + 6pq]$

$= p(p - 2q)(p^2 + 2pq + 4q^2)$

Answer: $p(p - 2q)(p^2 + 2pq + 4q^2)$