Question paper, Set 4

$144x^2 - 169z^2 = (12x)^2 - (13z)^2$

$= (12x + 13z)(12x - 13z)$

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

Answer: $(12x + 13z)(12x - 13z)$

$25x^2 + 90x + 81 = (5x)^2 + 9^2 + 2(5x)(9)$

$= (5x + 9)^2$

$= (5x + 9)(5x + 9)$

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

Answer: $(5x + 9)^2$

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

$= (7t - 3)^2$

$= (7t - 3)(7t - 3)$

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

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

$x^2 + 6x + 9 - 25y^2 = (x + 3)^2 - (5y)^2$

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

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

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

$15 \times 7 = 105$

$3 \times 35 = 105$ and $3 + 35 = 38$

So the numbers are $3$ and $35$.

Answer: $3$ and $35$

$10 \times 21 = 210$: $(-14) \times (-15) = 210$ and $-14 - 15 = -29$

$= 10a^2 - 14a - 15a + 21$

$= 2a(5a - 7) - 3(5a - 7)$

$= (5a - 7)(2a - 3)$

Answer: $(5a - 7)(2a - 3)$

$12 \times (-5) = -60$

$4 \times (-15) = -60$ and $4 - 15 = -11$

So the numbers are $4$ and $-15$.

Answer: $4$ and $-15$

Writing the terms in order: $-2t^2 + 13t - 20$

$(-2) \times (-20) = 40$: $5 \times 8 = 40$ and $5 + 8 = 13$

$= -2t^2 + 5t + 8t - 20$

$= -t(2t - 5) + 4(2t - 5)$

$= (2t - 5)(4 - t)$

Answer: $(2t - 5)(4 - t)$

Let $3m - 2p = t$

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

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

$= t^2 - t + 2t - 2$

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

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

$= [(3m - 2p) - 1][(3m - 2p) + 2]$

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

Answer: $(3m - 2p - 1)(3m - 2p + 2)$

$12 \times (-5) = -60$: $6 \times (-10) = -60$ and $6 - 10 = -4$

$= 12x^2 + 6xz - 10xz - 5z^2$

$= 6x(2x + z) - 5z(2x + z)$

$= (2x + z)(6x - 5z)$

Answer: $(2x + z)(6x - 5z)$

Product $= 84$, sum $= -7\sqrt{7}$: $(-3\sqrt{7}) \times (-4\sqrt{7}) = 84$ and $-3\sqrt{7} + (-4\sqrt{7}) = -7\sqrt{7}$

$= t^2 - 3\sqrt{7}t - 4\sqrt{7}t + 84$

$= t(t - 3\sqrt{7}) - 4\sqrt{7}(t - 3\sqrt{7})$

$= (t - 3\sqrt{7})(t - 4\sqrt{7})$

Answer: $(t - 3\sqrt{7})(t - 4\sqrt{7})$

$\displaystyle 2y^2 + \frac{7}{30}y - \frac{1}{30} = \frac{60y^2 + 7y - 1}{30}$

$\displaystyle = \frac{1}{30}(60y^2 + 7y - 1)$

Answer: $\displaystyle \frac{1}{30}(60y^2 + 7y - 1)$

Let $m - 4 = x$ and $n - 3 = y$

$3(m - 4)^2 - 19(m - 4)(n - 3) + 6(n - 3)^2 = 3x^2 - 19xy + 6y^2$

$3 \times 6 = 18$: $(-1) \times (-18) = 18$ and $-1 - 18 = -19$

$= 3x^2 - xy - 18xy + 6y^2$

$= x(3x - y) - 6y(3x - y)$

$= (3x - y)(x - 6y)$

$= [3(m - 4) - (n - 3)][(m - 4) - 6(n - 3)]$

$= (3m - 12 - n + 3)(m - 4 - 6n + 18)$

$= (3m - n - 9)(m - 6n + 14)$

Answer: $(3m - n - 9)(m - 6n + 14)$

$27a^3 + 54a^2 + 36a + 7 = (3a)^3 + 3(3a)^2(2) + 3(3a)(2)^2 + 2^3 - 1$

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

$= (3a + 2 - 1)[(3a + 2)^2 + (3a + 2)1 + 1^2]$

$= (3a + 1)[9a^2 + 12a + 4 + 3a + 2 + 1]$

$= (3a + 1)(9a^2 + 15a + 7)$

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

Answer: $(3a + 1)(9a^2 + 15a + 7)$

$125m^3 + 8 = (5m)^3 + 2^3$

$= (5m + 2)[(5m)^2 - (5m)(2) + 2^2]$

$= (5m + 2)(25m^2 - 10m + 4)$

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

Answer: $(5m + 2)(25m^2 - 10m + 4)$

$x^3 - 52x - 96 = x^3 - 4x - 48x - 96$

$= x(x^2 - 4) - 48(x + 2)$

$= x(x + 2)(x - 2) - 48(x + 2)$

$= (x + 2)[x(x - 2) - 48]$

$= (x + 2)(x^2 - 2x - 48)$

Now $x^2 - 2x - 48$: $6 \times (-8) = -48$ and $6 - 8 = -2$

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

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

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

So $x^3 - 52x - 96 = (x + 2)(x + 6)(x - 8)$

Answer: $(x + 2)(x + 6)(x - 8)$

$81a^4 + 64c^4 = (9a^2)^2 + (8c^2)^2 + 2(9a^2)(8c^2) - 144a^2c^2$

$= (9a^2 + 8c^2)^2 - (12ac)^2$

$= (9a^2 + 8c^2 + 12ac)(9a^2 + 8c^2 - 12ac)$

$= (9a^2 + 12ac + 8c^2)(9a^2 - 12ac + 8c^2)$

Answer: $(9a^2 + 12ac + 8c^2)(9a^2 - 12ac + 8c^2)$

Here $(m - 5n) + (5n - 4p) + (4p - m) = 0$

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

$\therefore (m - 5n)^3 + (5n - 4p)^3 + (4p - m)^3 = 3(m - 5n)(5n - 4p)(4p - m)$

Answer: $3(m - 5n)(5n - 4p)(4p - m)$

$81ac^4 - a^5 = a(81c^4 - a^4)$

$= a[(9c^2)^2 - (a^2)^2]$

$= a(a^2 + 9c^2)(9c^2 - a^2)$

$= a(a^2 + 9c^2)[(3c)^2 - a^2]$

$= a(a^2 + 9c^2)(a + 3c)(3c - a)$

Answer: $a(a^2 + 9c^2)(a + 3c)(3c - a)$

$20y^2 + (4y^2 + 1)y + 5 = 20y^2 + 4y^3 + y + 5$

$= 4y^3 + 20y^2 + y + 5$

$= 4y^2(y + 5) + 1(y + 5)$

$= (y + 5)(4y^2 + 1)$

Answer: $(y + 5)(4y^2 + 1)$