Question paper, Set 1

$p^2 + 10p + 25 = p^2 + 5^2 + 2(p)(5)$

$= (p + 5)^2$

$= (p + 5)(p + 5)$

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

Answer: $(p + 5)^2$

$16a^2 - 56a + 49 = (4a)^2 + 7^2 - 2(4a)(7)$

$= (4a - 7)^2$

$= (4a - 7)(4a - 7)$

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

Answer: $(4a - 7)^2$

$p^2 - 4p + 4 - 9q^2 = (p - 2)^2 - (3q)^2$

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

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

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

$5 \times 12 = 60$

$3 \times 20 = 60$ and $3 + 20 = 23$

So the numbers are $3$ and $20$.

Answer: $3$ and $20$

$3 \times 6 = 18$

$(-1) \times (-18) = 18$ and $-1 - 18 = -19$

So the numbers are $-1$ and $-18$.

Answer: $-1$ and $-18$

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

$3 \times (-24) = -72$ and $3 - 24 = -21$

So the numbers are $3$ and $-24$.

Answer: $3$ and $-24$

$19 \times (-176) = -3344$

$3344 = 2 \times 2 \times 2 \times 2 \times 11 \times 19 = 22 \times 152$, and $-22 + 152 = 130$

$= 19m^2 - 22m + 152m - 176$

$= m(19m - 22) + 8(19m - 22)$

$= (19m - 22)(m + 8)$

Answer: $(19m - 22)(m + 8)$

Writing the terms in order: $-5m^2 - 41m + 36$

$(-5) \times 36 = -180$: $(-45) \times 4 = -180$ and $-45 + 4 = -41$

$= -5m^2 - 45m + 4m + 36$

$= -5m(m + 9) + 4(m + 9)$

$= (m + 9)(4 - 5m)$

Answer: $(m + 9)(4 - 5m)$

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

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

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

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

Answer: $(2x - y)(x - y)$

$4\sqrt{2} \times 3\sqrt{2} = 4 \times 3 \times 2 = 24$: $(-2) \times (-12) = 24$ and $-2 - 12 = -14$

$= 4\sqrt{2}m^2 - 2m - 12m + 3\sqrt{2}$

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

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

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

$\displaystyle a^2 - \frac{3}{20}a - \frac{9}{20} = \frac{20a^2 - 3a - 9}{20}$

$\displaystyle = \frac{1}{20}(20a^2 - 3a - 9)$

Answer: $\displaystyle \frac{1}{20}(20a^2 - 3a - 9)$

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

$6(a - 3)^2 + (a - 3)(c - 5) - 12(c - 5)^2 = 6x^2 + xy - 12y^2$

$6 \times (-12) = -72$: $(-8) \times 9 = -72$ and $-8 + 9 = 1$

$= 6x^2 - 8xy + 9xy - 12y^2$

$= 2x(3x - 4y) + 3y(3x - 4y)$

$= (3x - 4y)(2x + 3y)$

$= [3(a - 3) - 4(c - 5)][2(a - 3) + 3(c - 5)]$

$= (3a - 9 - 4c + 20)(2a - 6 + 3c - 15)$

$= (3a - 4c + 11)(2a + 3c - 21)$

Answer: $(3a - 4c + 11)(2a + 3c - 21)$

$4p^2 + 9q^2 + 16r^2 - 12pq - 24qr + 16rp = (2p)^2 + (3q)^2 + (4r)^2 - 2(2p)(3q) - 2(3q)(4r) + 2(4r)(2p)$

$= (2p - 3q + 4r)^2$

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

Answer: $(2p - 3q + 4r)^2$

$t^3 + 12t^2 + 48t + 64 = t^3 + 4^3 + 3(t)^2(4) + 3(t)(4)^2$

$= (t + 4)^3$

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

Answer: $(t + 4)^3$

$64p^6 - 125 = (4p^2)^3 - 5^3$

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

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

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

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

$y^6z^6 + 216 = (y^2z^2)^3 + 6^3$

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

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

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

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

$x^3 - 43x + 42 = x^3 - x - 42x + 42$

$= x(x^2 - 1) - 42(x - 1)$

$= x(x - 1)(x + 1) - 42(x - 1)$

$= (x - 1)[x(x + 1) - 42]$

$= (x - 1)(x^2 + x - 42)$

Now $x^2 + x - 42$: $(-6) \times 7 = -42$ and $-6 + 7 = 1$

$x^2 + x - 42 = x^2 - 6x + 7x - 42$

$= x(x - 6) + 7(x - 6)$

$= (x - 6)(x + 7)$

So $x^3 - 43x + 42 = (x - 1)(x - 6)(x + 7)$

Answer: $(x - 1)(x - 6)(x + 7)$

$p^4 + 16p^2q^2 + 256q^4 = (p^2)^2 + 2(p^2)(16q^2) + (16q^2)^2 - 16p^2q^2$

$= (p^2 + 16q^2)^2 - (4pq)^2$

$= (p^2 + 16q^2 + 4pq)(p^2 + 16q^2 - 4pq)$

$= (p^2 + 4pq + 16q^2)(p^2 - 4pq + 16q^2)$

Answer: $(p^2 + 4pq + 16q^2)(p^2 - 4pq + 16q^2)$

Here $(3a - b) + (b - 4c) + (4c - 3a) = 0$

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

$\therefore (3a - b)^3 + (b - 4c)^3 + (4c - 3a)^3 = 3(3a - b)(b - 4c)(4c - 3a)$

Answer: $3(3a - b)(b - 4c)(4c - 3a)$

$48x^5 - 243xz^4 = 3x(16x^4 - 81z^4)$

$= 3x[(4x^2)^2 - (9z^2)^2]$

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

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

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

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