Exercise 2.4 Practice

(i) $(x+5)(x+9)$:
Using identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:
Here $a=5, b=9$.
$= x^2 + (5+9)x + (5)(9) = x^2 + 14x + 45$.

(ii) $(x-4)(x-8)$:
Using identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:
Here $a=-4, b=-8$.
$= x^2 + (-4-8)x + (-4)(-8) = x^2 - 12x + 32$.

(iii) $(2x+3)(2x-7)$:
Using identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x$ replaced by $2x$:
$= (2x)^2 + (3-7)(2x) + (3)(-7) = 4x^2 - 8x - 21$.

(iv) $(m^2 + \frac{5}{2})(m^2 - \frac{5}{2})$:
Using identity $(a+b)(a-b) = a^2 - b^2$:
$= (m^2)^2 - (\frac{5}{2})^2 = m^4 - \frac{25}{4}$.

(v) $(5-3y)(5+3y)$:
Using identity $(a-b)(a+b) = a^2 - b^2$:
$= (5)^2 - (3y)^2 = 25 - 9y^2$.

(i) $102 \times 106$:
Write as $(100+2)(100+6)$. Use $(x+a)(x+b) = x^2 + (a+b)x + ab$.
$= 100^2 + (2+6)100 + (2)(6) = 10000 + 800 + 12 = 10812$.

(ii) $94 \times 97$:
Write as $(100-6)(100-3)$.
$= 100^2 + (-6-3)100 + (-6)(-3) = 10000 - 900 + 18 = 9118$.

(iii) $105 \times 95$:
Write as $(100+5)(100-5)$. Use $(a+b)(a-b) = a^2 - b^2$.
$= 100^2 - 5^2 = 10000 - 25 = 9975$.

(i) $16x^2 + 24xy + 9y^2$:
Rewrite as $(4x)^2 + 2(4x)(3y) + (3y)^2$.
Using $a^2 + 2ab + b^2 = (a+b)^2$, where $a=4x, b=3y$.
$= (4x + 3y)^2$.

(ii) $25y^2 - 10y + 1$:
Rewrite as $(5y)^2 - 2(5y)(1) + (1)^2$.
Using $a^2 - 2ab + b^2 = (a-b)^2$, where $a=5y, b=1$.
$= (5y - 1)^2$.

(iii) $x^2 - \frac{y^2}{81}$:
Rewrite as $(x)^2 - (\frac{y}{9})^2$.
Using $a^2 - b^2 = (a+b)(a-b)$.
$= (x + \frac{y}{9})(x - \frac{y}{9})$.

(i) $(2x + 3y + z)^2$:
$= (2x)^2 + (3y)^2 + z^2 + 2(2x)(3y) + 2(3y)(z) + 2(z)(2x)$
$= 4x^2 + 9y^2 + z^2 + 12xy + 6yz + 4zx$.

(ii) $(x - 2y + 3z)^2$:
$= x^2 + (-2y)^2 + (3z)^2 + 2(x)(-2y) + 2(-2y)(3z) + 2(3z)(x)$
$= x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6zx$.

(iii) $(-3x + y + 2z)^2$:
$= (-3x)^2 + y^2 + (2z)^2 + 2(-3x)(y) + 2(y)(2z) + 2(2z)(-3x)$
$= 9x^2 + y^2 + 4z^2 - 6xy + 4yz - 12zx$.

(iv) $(2a - 5b - 3c)^2$:
$= (2a)^2 + (-5b)^2 + (-3c)^2 + 2(2a)(-5b) + 2(-5b)(-3c) + 2(-3c)(2a)$
$= 4a^2 + 25b^2 + 9c^2 - 20ab + 30bc - 12ac$.

(v) $(-x + 4y - 2z)^2$:
$= (-x)^2 + (4y)^2 + (-2z)^2 + 2(-x)(4y) + 2(4y)(-2z) + 2(-2z)(-x)$
$= x^2 + 16y^2 + 4z^2 - 8xy - 16yz + 4xz$.

(vi) $[\frac{1}{3}a - \frac{1}{2}b + 2]^2$:
$= (\frac{a}{3})^2 + (-\frac{b}{2})^2 + 2^2 + 2(\frac{a}{3})(-\frac{b}{2}) + 2(-\frac{b}{2})(2) + 2(2)(\frac{a}{3})$
$= \frac{a^2}{9} + \frac{b^2}{4} + 4 - \frac{ab}{3} - 2b + \frac{4a}{3}$.

(i) $9x^2 + 16y^2 + 4z^2 + 24xy - 16yz - 12xz$:
Since $yz$ and $xz$ terms are negative, $z$ must have the opposite sign to $x$ and $y$. Let $z$ be negative.
$= (3x)^2 + (4y)^2 + (-2z)^2 + 2(3x)(4y) + 2(4y)(-2z) + 2(-2z)(3x)$
$= (3x + 4y - 2z)^2$.

(ii) $3x^2 + y^2 + z^2 - 2\sqrt{3}xy - 2\sqrt{3}xz + 2yz$:
Since $xy$ and $xz$ are negative, $x$ has the opposite sign. Let $x$ be negative.
$= (-\sqrt{3}x)^2 + y^2 + z^2 + 2(-\sqrt{3}x)(y) + 2(y)(z) + 2(z)(-\sqrt{3}x)$
$= (-\sqrt{3}x + y + z)^2$.

Identity: $(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$.

(i) $(3x+2)^3$:
$= (3x)^3 + 2^3 + 3(3x)(2)(3x+2)$
$= 27x^3 + 8 + 18x(3x+2) = 27x^3 + 54x^2 + 36x + 8$.

(ii) $(3a-2b)^3$:
$= (3a)^3 - (2b)^3 - 3(3a)(2b)(3a-2b)$
$= 27a^3 - 8b^3 - 18ab(3a-2b) = 27a^3 - 8b^3 - 54a^2b + 36ab^2$.

(iii) $[\frac{2}{3}x + 1]^3$:
$= (\frac{2}{3}x)^3 + 1^3 + 3(\frac{2}{3}x)(1)(\frac{2}{3}x+1)$
$= \frac{8}{27}x^3 + 1 + 2x(\frac{2}{3}x+1) = \frac{8}{27}x^3 + \frac{4}{3}x^2 + 2x + 1$.

(iv) $[x - \frac{3}{4}y]^3$:
$= x^3 - (\frac{3}{4}y)^3 - 3(x)(\frac{3}{4}y)(x-\frac{3}{4}y)$
$= x^3 - \frac{27}{64}y^3 - \frac{9}{4}xy(x-\frac{3}{4}y) = x^3 - \frac{27}{64}y^3 - \frac{9}{4}x^2y + \frac{27}{16}xy^2$.

(i) $(98)^3$: Write as $(100-2)^3$.
$= 100^3 - 2^3 - 3(100)(2)(100-2) = 1000000 - 8 - 600(98) = 1000000 - 8 - 58800 = 941192$.

(ii) $(103)^3$: Write as $(100+3)^3$.
$= 100^3 + 3^3 + 3(100)(3)(100+3) = 1000000 + 27 + 900(103) = 1000000 + 27 + 92700 = 1092727$.

(iii) $(999)^3$: Write as $(1000-1)^3$.
$= 1000^3 - 1^3 - 3(1000)(1)(1000-1) = 1000000000 - 1 - 3000(999) = 1000000000 - 1 - 2997000 = 997002999$.

(i) $27a^3 + 8b^3 + 54a^2b + 36ab^2$:
$= (3a)^3 + (2b)^3 + 3(3a)^2(2b) + 3(3a)(2b)^2$. This matches $(x+y)^3$.
$= (3a+2b)^3$.

(ii) $64a^3 - 27b^3 - 144a^2b + 108ab^2$:
$= (4a)^3 - (3b)^3 - 3(4a)^2(3b) + 3(4a)(3b)^2$. This matches $(x-y)^3$.
$= (4a-3b)^3$.

(iii) $125 - 27x^3 - 225x + 135x^2$:
$= 5^3 - (3x)^3 - 3(5)^2(3x) + 3(5)(3x)^2 = (5-3x)^3$.

(iv) $8p^3 - \frac{1}{27} - 4p^2 + \frac{2}{3}p$:
$= (2p)^3 - (\frac{1}{3})^3 - 3(2p)^2(\frac{1}{3}) + 3(2p)(\frac{1}{3})^2 = (2p - \frac{1}{3})^3$.

(i) Verify $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$:
RHS $= a(a^2-ab+b^2) + b(a^2-ab+b^2)$
$= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3 = \text{LHS}$.

(ii) Verify $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:
RHS $= a(a^2+ab+b^2) - b(a^2+ab+b^2)$
$= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3 = \text{LHS}$.

(i) $64a^3 + 27b^3$: Write as $(4a)^3 + (3b)^3$.
Using $a^3+b^3 = (a+b)(a^2-ab+b^2)$:
$= (4a + 3b)((4a)^2 - (4a)(3b) + (3b)^2) = (4a + 3b)(16a^2 - 12ab + 9b^2)$.

(ii) $125x^3 - 343y^3$: Write as $(5x)^3 - (7y)^3$.
Using $a^3-b^3 = (a-b)(a^2+ab+b^2)$:
$= (5x - 7y)((5x)^2 + (5x)(7y) + (7y)^2) = (5x - 7y)(25x^2 + 35xy + 49y^2)$.

Identity: $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Rewrite expression as: $(2x)^3 + y^3 + (3z)^3 - 3(2x)(y)(3z)$.
Here $a=2x, b=y, c=3z$.

Apply identity:
$= (2x+y+3z)((2x)^2 + y^2 + (3z)^2 - (2x)y - y(3z) - (3z)(2x))$
$= (2x+y+3z)(4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6zx)$.

Start with RHS:
$\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]$

Expand the squares:
$\frac{1}{2}(a+b+c)[(a^2-2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2)]$ $= \frac{1}{2}(a+b+c)[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]$

Factor out 2 from the bracket: $= \frac{1}{2}(a+b+c) \cdot 2 [a^2 + b^2 + c^2 - ab - bc - ca]$

$= (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$
This matches the identity for $a^3 + b^3 + c^3 - 3abc$. Hence Verified.

Identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Given that $a+b+c = 0$. Substitute this into the identity:

$a^3 + b^3 + c^3 - 3abc = (0) \times (a^2 + b^2 + c^2 - ab - bc - ca) = 0$.

$\Rightarrow a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$.

(i) $(-15)^3 + (9)^3 + (6)^3$:
Let $x=-15, y=9, z=6$.
$x+y+z = -15+9+6 = 0$.
Since sum is 0, $x^3+y^3+z^3 = 3xyz$.
$= 3(-15)(9)(6) = -2430$.

(ii) $(30)^3 + (-18)^3 + (-12)^3$:
Let $x=30, y=-18, z=-12$.
$x+y+z = 30-18-12 = 0$.
Since sum is 0, $x^3+y^3+z^3 = 3xyz$.
$= 3(30)(-18)(-12) = 19440$.

(i) $25a^2 - 15a + 2$:
We need to factorise. Product = $25 \times 2 = 50$. Sum = -15. Factors are -10 and -5.
$25a^2 - 10a - 5a + 2 = 5a(5a-2) - 1(5a-2) = (5a-2)(5a-1)$.
Possible length/breadth: $(5a-2)$ and $(5a-1)$.

(ii) $16x^2 - 24x + 5$:
Product = $16 \times 5 = 80$. Sum = -24. Factors are -20 and -4.
$16x^2 - 20x - 4x + 5 = 4x(4x-5) - 1(4x-5) = (4x-5)(4x-1)$.
Possible length/breadth: $(4x-5)$ and $(4x-1)$.

(i) $2x^2 - 8x$:
Factor out common term $2x$.
$= 2x(x - 4)$. This can be written as $2 \times x \times (x-4)$.
Dimensions: $2, x, (x-4)$.

(ii) $6ky^2 + 3ky - 45k$:
Factor out common term $3k$.
$= 3k(2y^2 + y - 15)$.
Factorise quadratic $2y^2 + y - 15$: Split $y$ into $6y$ and $-5y$.
$2y^2 + 6y - 5y - 15 = 2y(y+3) - 5(y+3) = (2y-5)(y+3)$.
Dimensions: $3k, (2y-5), (y+3)$.