(a)

\begin{align} \text{By } & \text{Factor theorem,} \\ P(2) & = 2(2)^3 - 4(2)^2 + (2) - 2 \\ & = 0 \\ \\ \therefore D(x) & \text{ is a factor of } P(x). \end{align}

(b)

\begin{align} \text{By } & \text{Factor theorem,} \\ P\left(-{1 \over 2}\right) & = 2\left(-{1 \over 2}\right)^3 - 4\left(-{1 \over 2}\right)^2 + \left(-{1 \over 2}\right) - 2 \\ \\ & = -{15 \over 4} \\ \\ \therefore D(x) & \text{ is not a factor of } P(x). \end{align}

(c)

\begin{align} \text{By } & \text{Factor theorem,} \\ P(-3) & = (-3)^4 - 2(-3)[(-3)^2 - 8] - 12(-3)^2 + 11 \\ & = -10 \\ \\ \therefore D(x) & \text{ is not a factor of } P(x). \end{align}

\begin{align} \text{By } & \text{Factor theorem,} \\ P(-2) & = 3(-2)^2 - 4a(-2) - 4a^2 \\ 0 & = 12 + 8a - 4a^2 \\ 0 & = -4a^2 + 8a + 12 \\ 0 & = 4a^2 - 8a - 12 \\ 0 & = a^2 - 2a - 3 \\ 0 & = (a + 1)(a - 3) \\ \\ a + 1 = 0 \phantom{000}&\text{or}\phantom{000} a - 3 = 0 \\ a = - 1 \phantom{0.} &\phantom{or000-3} a = 3 \end{align}

(i)

\begin{align} \text{By } & \text{Factor theorem,} \\ g(2) & = 2(2)^3 + (2)^2 + p(2) - 4 \\ 0 & = 16 + 4 + 2p - 4 \\ 0 & = 16 + 2p \\ -2p & = 16 \\ p & = {16 \over -2} \\ & = -8 \end{align}

(ii)

\begin{align} g(x) & = 2x^3 + x^2 + px - 4 \\ & = 2x^3 + x^2 + (-8)x - 4 \\ & = 2x^3 + x^2 - 8x - 4 \end{align} $$ \require{enclose} \begin{array}{rll} 2x^2 + 5x + 2 \phantom{00000}\\ x - 2 \enclose{longdiv}{ 2x^3 + x^2 - 8x - 4 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 - 4x^2 ){\phantom{0000000}}} \\ 5x^2 - 8x - 4 \phantom{0} \\ -\underline{( 5x^2 - 10x ){\phantom{00.}}} \\ 2x - 4\phantom{0} \\ -\underline{( 2x - 4 )} \\ 0\phantom{0} \end{array} $$ \begin{align} g(x) & = 2x^3 + x^2 - 8x - 4 \\ & = (x - 2)(2x^2 + 5x + 2) \\ & = (x - 2)(2x + 1)(x + 2) \end{align}

Use the identities: \begin{align} a^3 + b^3 & = (a + b)(a^2 - ab + b^2) \\ \\ a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \end{align}

(a)

\begin{align} x^3 - 64 & = x^3 - )^3 \\ & = (x - 4)[x^2 + (x)(4) + 4^2] \\ & = (x - 4)(x^2 + 4x + 16) \end{align}

(b)

\begin{align} 27a^3 + 125x^3 & = (3a)^3 + (5x)^3 \\ & = (3a + 5x)[(3a)^2 - (3a)(5x) + (5x)^2] \\ & = (3a + 5x)(9a^2 - 15ax + 25x^2) \end{align}

(c)

\begin{align} 8 + 27x^3 & = 2^3 + (3x)^3 \\ & = (2 + 3x)[2^2 - (2)(3x) + (3x)^2] \\ & = (2 + 3x)(4 - 6x + 9x^2) \end{align}

(d)

\begin{align} 432x^3 - 2y^3 & = 2(216x^3 - y^3) \\ & = 2[ (6x)^3 - y^3 ] \\ & = 2(6x - y)[ (6x)^2 + (6x)(y) + y^2 ] \\ & = 2(6x - y)(36x^2 + 6xy + y^2) \end{align}

(i)

\begin{align} x^2 - 3x - 4 & = (x + 1)(x - 4) \\ \\ \\ \text{By Factor} & \text{ theorem,} \\ f(-1) & = a(-1)^3 + b(-1)^2 - 5(-1) + 2a \\ 0 & = -a + b + 5 + 2a \\ 0 & = a + b + 5 \\ -b - 5 & = a \phantom{000} \text{ --- (1)} \\ \\ \text{By Factor} & \text{ theorem,} \\ f(4) & = a(4)^3 + b(4)^2 - 5(4) + 2a \\ 0 & = 64a + 16b - 20 + 2a \\ 20 & = 66a + 16b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 20 & = 66(-b - 5) + 16b \\ 20 & = -66b - 330 + 16b \\ 20 + 330 & = -66b + 16b \\ 350 & = -50b \\ {350 \over -50} & = b \\ -7 & = b \\ \\ \text{Substitute } b & = -7 \text{ into (1),} \\ -(-7) - 5 & = a \\ 7 - 5 & = a \\ 2 & = a \\ \\ \therefore a & = 2, b = -7 \end{align}

(ii)

\begin{align} f(x) & = ax^3 + bx^2 - 5x + 2a \\ & = (2)x^3 + (-7)x^2 - 5x + 2(2) \\ & = 2x^3 - 7x^2 - 5x + 4 \end{align} $$ \require{enclose} \begin{array}{rll} 2x - 1 \phantom{0000000000.}\\ x^2 - 3x - 4 \enclose{longdiv}{ 2x^3 - 7x^2 - 5x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 - 6x^2 - 8x ){\phantom{000.}}} \\ - x^2 + 3x + 4 \phantom{0} \\ -\underline{( -x^2 + 3x + 4 ){\phantom{.}}} \\ 0\phantom{0} \end{array} $$ $$ \therefore \text{Third factor is } 2x - 1 $$

(i)

\begin{align} f(1) & = (2p + 1)(1)^2 + p(1) + 2p^2 \\ 0 & = (2p + 1)(1) + p + 2p^2 \\ 0 & = 2p + 1 + p + 2p^2 \\ 0 & = 2p^2 + 3p + 1 \\ 0 & = (2p + 1)(p + 1) \\ \\ 2p + 1 = 0 \phantom{000}&\text{or}\phantom{000} p + 1 = 0 \\ 2p = -1 \phantom{0.} & \phantom{or000+1} p = - 1 \\ p = -{1 \over 2} \phantom{0} & \end{align}

(ii)

\begin{align} f(-2) & = (2p + 1)(-2)^2 + p(-2) + 2p^2 \\ 0 & = (2p + 1)(4) - 2p + 2p^2 \\ 0 & = 8p + 4 - 2p + 2p^2 \\ 0 & = 2p^2 + 6p + 4 \\ 0 & = p^2 + 3p + 2 \\ 0 & = (p + 2)(p + 1) \\ \\ p + 2 = 0 \phantom{000}&\text{or}\phantom{000} p + 1 = 0 \\ p = - 2 \phantom{0.}&\phantom{or000+1} p = -1 \end{align}

(iii)

\begin{align} \text{When } p = -1 \text{ or } -{1 \over 2}, & \phantom{0} f(x) \text{ has a factor of } x - 1. \\ \\ \text{When } p = -2 \text{ or } -1, & \phantom{0} f(x) \text{ has a factor of } x + 2. \\ \\ \\ \text{For } f(x) \text{ to have a factor of } & x + 2 \text{ but not } x - 1, \\ p & = -2 \end{align}

\begin{align} f(a) & = 2(a)^2 + 3p(a) - 2q \\ 0 & = 2a^2 + 3ap - 2q \phantom{000} \text{ --- (1)} \\ \\ g(a) & = (a)^2 + q \\ 0 & = a^2 + q \\ -q & = a^2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 0 & = 2(-q) + 3ap - 2q \\ 0 & = -2q + 3ap - 2q \\ 0 & = 3ap - 4q \\ 4q & = 3ap \\ {4q \over 3p} & = a \\ \\ \text{Substitute } a & = {4q \over 3p} \text{ into (1),} \\ 0 & = 2\left(4q \over 3p\right)^2 + 3\left(4q \over 3p\right)p - 2q \\ 0 & = 2\left(16q^2 \over 9p^2\right) + \left(4q \over p\right)p - 2q \\ 0 & = {32q^2 \over 9p^2} + 4q - 2q \\ 0 & = {32q^2 \over 9p^2} + 2q \\ 0 & = 32q^2 + 18p^2q \\ 0 & = 16q^2 + 9p^2q \\ 0 & = q(16q + 9p^2) \\ \\ q = 0 \text{ (Reject)} \phantom{000}&\text{or}\phantom{000} 16q + 9p^2 = 0 \text{ (Shown)} \end{align}

\begin{align} \text{By Remainder} & \text{ theorem,} \\ f(-2a) & = 3(-2a)^3 - 5a(-2a)^2 + ka^2 (-2a) + 4a^3 \\ -32a^3 & = -24a^3 - 20a^3 - 2ka^3 + 4a^3 \\ -32a^3 & = -40a^3 - 2ka^3 \\ -32a^3 + 40a^3 & = -2ka^3 \\ 8a^3 & = -2ka^3 \\ 8 & = -2k \\ {8 \over -2} & = k \\ -4 & = k \end{align} \begin{align} f(x) & = 3x^3 - 5ax^2 + ka^2x + 4a^3 \\ & = 3x^3 - 5ax^2 + (-4)a^2x + 4a^3 \\ & = 3x^3 - 5ax^2 - 4a^2x + 4a^3 \\ \\ x^2 - ax - 2a^2 & = (x + a)(x - 2a) \\ \\ \text{By Factor} & \text{ theorem,} \\ f(-a) & = 3(-a)^3 - 5a(-a)^2 - 4a^2 (-a) + 4a^3 \\ & = -3a^3 - 5a^3 + 4a^3 + 4a^3 \\ & = 0 \\ \\ \therefore x + a & \text{ is a factor of } f(x). \\ \\ \text{By Factor} & \text{ theorem,} \\ f(2a) & = 3(2a)^3 - 5a(2a)^2 - 4a^2(2a) + 4a^3 \\ & = 24a^3 - 20a^3 - 8a^3 + 4a^3 \\ & = 0 \\ \\ \therefore x - 2a & \text{ is a factor of } f(x). \\ \\ \therefore x^2 - ax - 2a^2 & \text{ is a factor of } f(x). \end{align}

\begin{align} \text{By Factor} & \text{ theorem,} \\ f(2b) & = (2b)^3 - b(2b)^2 - 4b^2(2b) + 4b^3 \\ & = 8b^3 - 4b^3 - 8b^3 + 4b^3 \\ & = 0 \\ \\ \therefore a - 2b & \text{ is a factor of } f(a). \\ \\ \text{By Remainder} & \text{ theorem,} \\ f(-b) & = (-b)^3 - b(-b)^2 - 4b^2(-b) + 4b^3 \\ & = -b^3 - b^3 + 4b^3 + 4b^3 \\ & = 6b^3 \\ \\ \therefore \text{Remainder} & = 6b^3 \end{align}

Use the identities: \begin{align} a^3 + b^3 & = (a + b)(a^2 - ab + b^2) \\ \\ a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \end{align}

(a)

\begin{align} 2x^4 + 54x & = 2x(x^3 + 27) \\ & = 2x(x^3 + 3^3) \\ & = 2x(x + 3)[x^2 - (x)(3) + 3^2] \\ & = 2x(x + 3)(x^2 - 3x + 9) \end{align}

(b)

\begin{align} (1 + x)^3 + 64 & = (1 + x)^3 + 4^3 \\ & = [(1 + x) + 4][(1 + x)^2 - (1 + x)(4) + 4^2] \\ & = (1 + x + 4)(1 + 2x + x^2 - 4 - 4x + 16) \\ & = (x + 1 + 4)(x^2 + 2x - 4x + 1 - 4 + 16) \\ & = (x + 5)(x^2 - 2x + 13) \end{align}

(c)

\begin{align} (x + y)^3 - (x - y)^3 & = [(x + y) - (x - y)][(x + y)^2 + (x + y)(x - y) + (x - y)^2] \\ & = (x + y - x + y)(x^2 + 2xy + y^2 + x^2 - y^2 + x^2 - 2xy + y^2) \\ & = (x - x + y + y)(x^2 + x^2 + x^2 + 2xy - 2xy + y^2 - y^2 + y^2) \\ & = 2y(3x^2 + y^2) \end{align}

(d)

\begin{align} 1000x^3 - y^6 & = (10x)^3 - (y^2)^3 \\ & = (10x - y^2)[(10x)^2 + (10x)(y^2) + (y^2)^2] \\ & = (10x - y^2)(100x^2 + 10xy^2 + y^4) \end{align}

Use the identity: \begin{align} a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \end{align}

(i)

\begin{align} x^6 - 64 & = (x^3)^2 - 8^2 \\ & = (x^3 + 8)(x^3 - 8) \\ & = (x^3 + 2^3)(x^3 - 2^3) \\ & = (x + 2)[x^2 - (x)(2) + 2^2] (x - 2)[x^2 + (x)(2) + 2^2] \\ & = (x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4) \\ & = (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) \end{align}

(ii)

\begin{align} x^6 - 64 & = (x^2 + 4)^2 - 4x^2 \\ (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) & = (x^2 + 4)^2 - 4x^2 \\ (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) & = (x^2 + 4)^2 - (2x)^2 \\ (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) & = [(x^2 + 4) + 2x][(x^2 + 4) - 2x] \\ (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) & = (x^2 + 2x + 4)(x^2 - 2x + 4) \\ (x + 2)(x - 2)(x^2 + 2x + 4)(x^2 - 2x + 4) - (x^2 + 2x + 4)(x^2 - 2x + 4) & = 0 \\ (x^2 + 2x + 4)(x^2 - 2x + 4)[(x + 2)(x - 2) - 1] & = 0 \end{align} \begin{align} x^2 + 2x + 4 = 0 \text{ (No solutions)} \phantom{00} \text{ or } \phantom{00} x^2 - 2x + 4 = 0 \text{ (No solutions) } \phantom{00} \text{ or } \phantom{00} (x + 2)(x - 2) - 1 & = 0 \\ x^2 - 4 - 1 & = 0 \\ x^2 - 5 & = 0 \\ x^2 & = 5 \\ x & = \pm \sqrt{5} \end{align}

(i)

\begin{align} x^{2 \over 3} - 2x^{1 \over 3} + 3 & = 0 \phantom{00000} [\text{Roots } \alpha, \beta] \\ (x^{1 \over 3})^2 - 2x^{1 \over 3} + 3 & = 0 \\ \\ \text{Let } u & = x^{1 \over 3}, \\ u^2 - 2u + 3 & = 0 \phantom{00000} \left[ \text{Roots } \alpha^{1 \over 3}, \beta^{1 \over 3} \right] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha^{1 \over 3} + \beta^{1 \over 3} & = -{-2 \over 1} \\ & = 2 \\ \\ \text{Product of roots} & = {c \over a} \\ \alpha^{1 \over 3} \beta^{1 \over 3} & = {3 \over 1} \\ & = 3 \end{align}

(ii)

\begin{align} a^3 + b^3 & = (a + b)(a^2 - ab + b^2) \end{align} \begin{align} \alpha + \beta & = (\alpha^{1 \over 3})^3 + (\beta^{1 \over 3})^3 \\ \\ & = (\alpha^{1 \over 3} + \beta^{1 \over 3})\left[ (\alpha^{1 \over 3})^2 - (\alpha^{1 \over 3})(\beta^{1 \over 3}) + (\beta^{1 \over 3})^2 \right] \\ \\ & = (2)\left[ (\alpha^{1 \over 3})^2 + (\beta^{1 \over 3})^2 - \alpha^{1 \over 3} \beta^{1 \over 3} \right] \\ \\ & = (2) \left[ (\alpha^{1 \over 3} + \beta^{1 \over 3})^2 - 2 \alpha^{1 \over 3} \beta^{1 \over 3} - \alpha^{1 \over 3} \beta^{1 \over 3} \right] \\ \\ & = (2) [ (2)^2 - 2(3) - 3] \\ \\ & = -10 \end{align}

\begin{align} a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \end{align}
\begin{align} (2n + 1)^3 - (2n - 1)^3 & = [(2n + 1) - (2n - 1)][ (2n + 1)^2 + (2n + 1)(2n - 1) + (2n - 1)^2] \\ \\ & = (2n + 1 - 2n + 1)( 4n^2 + 4n + 1 + 4n^2 - 1 + 4n^2 - 4n + 1 ) \\ \\ & = (2n - 2n + 1 + 1)( 4n^2 + 4n^2 + 4n^2 + 4n - 4n + 1 - 1 + 1) \\ \\ & = 2(12n^2 + 1) \end{align}
\begin{align} \text{Difference between the cube of two consecutive positive odd numbers} & = (2n + 1)^3 - (2n - 1)^3 \\ \\ & = 2(12n^2 + 1) \\ \\ {2(12n^2 + 1) \over 4} & = {12n^2 + 1 \over 2} \\ \\ \text{Since } 12n^2 + 1 \text{ is an odd number,} & \phantom{0} {(12n^2 + 1) \over 2} \text{ is not an integer.} \\ \\ \therefore \text{The difference between the cube of two consecutive} &\text{ positive odd numbers can never be divisible by 4}. \end{align}

\begin{align} \text{Volume of cube (before removal)} & = a \times a \times a \\ & = a^3 \\ \\ \text{Volume of cube removed} & = b \times b \times b \\ & = b^3 \\ \\ \\ \text{Volume of bottom half of remaining solid} & = a \times a \times (a - b) \\ & = a^2(a - b) \\ \\ \text{Volume of top half of remaining solid} & = (a - b) \times a \times b + b \times (a - b) \times b \\ & = ab(a - b) + b^2 (a - b) \\ \\ \\ \therefore a^3 - b^3 & = a^2 (a - b) + ab(a - b) + b^2 (a - b) \\ & = (a - b)(a^2 + ab + b^2) \end{align}