Question 13 Marks
Simplify:
$(9 x-7)(2 x-5)-(3 x-8)(5 x-3)$
Answer$(9 x-7)(2 x-5)$
$=9 x \times(2 x-5)-7(2 x-5)$
$=18 x^{(1+1)}-45 x-14 x+35$
$=18 x^2-59 x+35$
$(3 x-8)(5 x-3)$
$=3 x(5 x-3)-8(5 x-3)$
$=15 x^2-9 x-40 x+24$
$=15 x^2-49 x+24$
$\therefore(2 x-5)-(3 x-8)(5 x-3)$
$=18 x^2-59 x+35-\left(15 x^2-49 x+24\right)$
$=18 x^2-15 x^2-59 x+49 x+35-24$
$=3 x 2-10 x+11$
View full question & answer→Question 23 Marks
Simplify:
$(5 x-3)(x+4)-(2 x+5)(3 x-4)$
Answer$(5 x-3)(x+4)$
$=5 x(x+4)-3(x+4)$
$=5 x^{(1+1)}+20 x-3 x-12$
$=5 x^2+17 x-12$
$(2 x+5)(3 x-4)$
$=2 x(3 x-4)+5(3 x-4)$
$=6 x^{(1+1)}-8 x+15 x-20$
$=6 x^2+7 x-20$
$\therefore(5 x-3)(x+4)-(2 x+5)(3 x-4)$
$=5 x^2+17 x-12-\left(6 x^2+7 x-20\right)^{\prime}$
$=5 x^2-6 x^2+17 x-7 x-12+20$
$=-x^2+10 x+8$
View full question & answer→Question 33 Marks
Add the following expression:
$\frac{8}{5}\text{x}+\frac{11}{7}\text{y},+\frac{9}{4}\text{xy},-\frac{3}{2}\text{x}-\frac{5}{3}\text{y}-\frac{9}{5}\text{xy}$
Answer$\Big(\frac{8}{5}\text{x}-\frac{11}{7}\text{y}+\frac{9}{4}\text{xy}\Big)+\Big(\frac{-3}{2}\text{x}\frac{-5}{3}\text{y}-\frac{9}{5}\text{xy}\Big)$
$=\frac{8}{5}\text{x}-\frac{-3}{2}\text{x}+\frac{11}{7}\text{y}-\frac{5}{3}\text{y}+\frac{9}{4}\text{xy}-\frac{9}{5}\text{xy}$
$=\Big(\frac{8}{5}-\frac{-3}{2}\Big)\text{x}+\Big(\frac{11}{7}-\frac{5}{3}\Big)\text{y}+\Big(\frac{9}{4}-\frac{9}{5}\Big)\text{xy}$
$=\frac{16-15}{10}\text{x}+\frac{33-35}{21}\text{y}+\frac{45-36}{20}\text{xy}$
$=\frac{1}{10}\text{x}+\Big(\frac{-2}{21}\Big)\text{y}+\frac{9}{20}\text{xy}$
$=\frac{1}{10}\text{x}-\frac{2}{21}\text{y}+\frac{9}{20}\text{xy}$
View full question & answer→Question 43 Marks
Find the product $s\left(s^2-s t\right)$ and evaluate it fore $s=2$ and $t=3$.
Answer$s\left(s^2-s t\right)$
$= s \times s^2-s \times s t$
$= s^{(1+2)}-s^{(1+1)} \times t$
$= s^3-s^2 t$
When $s=2$ and $t=3$, we get,
$\text { L.H.S. }=s\left(s^2-st\right)$
$=2\left(2^2-2 \times 3\right)$
$=2 \times(4-6)$
$=-4$
$\text { R.H.S. }=s^3-s^2 t$
$=2^3-2^2 \times 3$
$=8-12$
$=-4$
$\text { L.H.S. }=\text { R.H.S. }$
$\therefore s\left(s^2-st\right)=s^3-s^2 t$
View full question & answer→Question 53 Marks
Subtract the sum of $\left(8 a-6 a^2+9\right)$ and $\left(-10 a-8+8 a^2\right)$ from -3 .
Answersum of $\left(8 a-6 a^2+9\right)$ and $\left(-10 a-8+8 a^2\right)$
$=8 a-6 a^2+9+(-10 a)-8+8 a^2$
$=8 a-10 a-6 a^2+8 a^2+9-8$
$=-2 a+2 a^2+1$
Now $-3-\left(-2 a+2 a^2+1\right)$
$=\left(-3+2 a-2 a^2-1\right)$
$=-4+2 a-2 a^2$
View full question & answer→Question 63 Marks
Subtract the sum of $\left(8 m-7 n+6 p^2\right)$ and $\left(-3 m-4 n-p^2\right)$ from the sum of $\left(2 m+4 n-3 p^2\right)$ and $\left(-m-n-p^2\right)$.
AnswerSum of $\left(8 m-7 n+6 p^2\right)$ and $\left(-3 m-4 n-p^2\right)$
$=\left(8 m-7 n+6 p^2\right)+\left(-3 m-4 n-p^2\right)$
$=8 m-7 n+6 p^2-3 m-4 n-p^2$
$=8 m-3 m-7 n-4 n+6 p^2-p^2$
$=5 m-11 n+5 p^2$
Sum of $\left(2 m+4 n-3 p^2\right)$ and $\left(-m-n-p^2\right)$
$=\left(2 m+4 n-3 p^2\right)+\left(-m-n-p^2\right)$
$=2 m+4 n-3 p^2-m-n-p^2$
$=2 m-m+4 n-n-3 p^2-p^2$
$=m+3 n-4 p^2$
$\text { Now }\left(m+3 n-4 p^2\right)-\left(5 m-11 n+5 p^2\right)$
$=m+3 n-4 p^2-5 m+11 n-5 p^2$
$=m-5 m+3 n+11 n-4 p^2-5 p^2$
$=-4 m+14 n-9 p^2$
View full question & answer→Question 73 Marks
Add the following expression:
$\frac{2}{3}\text{a}-\frac{4}{5}\text{b}+\frac{3}{5}\text{c},-\frac{3}{4}\text{a}-\frac{5}{2}\text{b}+\frac{2}{3}\text{c}, \frac{5}{2}\text{a}+\frac{7}{4}\text{b}-\frac{5}{6}\text{c}$
Answer$\Big(\frac{2}{3}\text{a}-\frac{4}{5}\text{b}+\frac{3}{5}\text{c}\Big)+\Big(\frac{-3}{4}\text{a}-\frac{5}{2}\text{b}+\frac{2}{3}\text{c}\Big)\\+\Big(\frac{5}{2}\text{a}+\frac{7}{4}\text{b}-\frac{5}{6}\text{c}\Big)$
$=\Big(\frac{2}{3}\text{a}-\frac{3}{4}\text{a}+\frac{5}{2}\text{a}\Big)+\Big(\frac{-4}{5}\text{b}-\frac{5}{2}\text{b}+\frac{7}{4}\text{b}\Big)\\+\Big(\frac{3}{5}\text{c}+\frac{2}{3}\text{c}-\frac{5}{6}\text{c}\Big)$
$=\Big(\frac{2}{3}-\frac{3}{4}+\frac{5}{2}\Big)\text{a}+\Big(\frac{-4}{5}-\frac{5}{2}+\frac{7}{4}\Big)\text{b}+\Big(\frac{3}{5}+\frac{2}{3}-\frac{5}{6}\Big)\text{c}$
$=\frac{8-9+30}{12}\text{a}+\frac{-16-50+35}{20}\text{b}+\frac{18+20-25}{30}\text{c}$
$=\frac{29}{12}\text{a}+\Big(\frac{-31}{20}\Big)\text{b}+\frac{13}{30}\text{c}$
$=\frac{29}{12}\text{a}-\frac{-31}{20}\text{b}+\frac{13}{30}\text{c}$
View full question & answer→Question 83 Marks
Find the following product:
$\frac{-4}{27}\text{xyz}\Big(\frac{9}{2}\text{x}^2\text{yz}-\frac{3}{4}\text{xyz}^2\Big)$
Answer$=\frac{-4}{27}\text{xyz}\times \frac{9}{2}\text{x}^2\text{yz}-\Big(\frac{-4}{27}\text{xyz}\times \frac{3}{4}\text{xyz}^2\Big)$
$=\frac{-4}{27}\times \frac{9}{2}\times \text{x}\times \text{x}^2\times \text{y}\times \text{y}\times \text{z}\times \text{z}\\+\frac{4}{27}\times \frac{3}{4}\times \text{x}\times \text{x}\times \text{y}\times \text{y}\times\text{z}\times \text{z}^2$
$=\frac{-2}{3}\times \text{x}^{1+2}\times \text{y}^{1+1}\times \text{z}^{1+1}\\+\frac{1}{9}\times\text{x}^{1+1}\times \text{y}^{1+1}\times \text{z}^{1+2}$
$=\frac{-2}{3}\text{x}^3\text{y}^2\text{z}^2+\frac{1}{9}\text{x}^2\text{y}^2\text{z}^3$
View full question & answer→Question 93 Marks
Find the product $-3 y\left(x y+y^2\right)$ and evaluate it fore $x=4$ and $y =5$.
Answer$-3 y\left(x y+y^2\right)$
$=-3 y \times x z y-3 y \times y^2$
$=-3 \times x \times y \times y-3 \times y \times y^2$
$=-3 \times x \times y^{(1+1)}-3 \times y^{(1+2)}$
$=-3 x y^2-3 y^3$
When $x=4$ and $y=5$, we get,
L.H.S. $=-3 y\left(x y+y^2\right)$
$=-3 \times 5\left(4 \times 5+5^2\right)$
$=-15 \times(20+25)$
$=-675$
R.H.S. $=-3 x y^2-3 y^3$
$=-3 \times 4 \times 5^2-3 \times 5^3$
$=-300-375$
$=-675$
L.H.S. = R.H.S.
$\therefore-3 y\left(x y+y^2\right)$
$=-3 x y^2-3 y^3$
View full question & answer→Question 103 Marks
Find the following product:
$\frac{2}{3}\text{abc}(\text{a}^2+\text{b}^2-3\text{c}^2)$
Answer$=\frac{2}{3}\text{abc}\times \text{a}^2+\frac{2}{3}\text{abc}\times \text{b}^2-\frac{2}{3}\text{abc}\times 3\text{c}^2$
$=\frac{2}{3}\text{a}\times \text{a}^2\times \text{b}\times \text{c}+\frac{2}{3}\text{a}\times \text{b}\times \text{b}^2\times \text{c}\\-\frac{2}{3}\times 3\times \text{a}\times \text{b}\times \text{c}\times \text{c}^2$
$=\frac{2}{3}\times \text{a}^{1+2}\times \text{b}\times \text{c}+\frac{2}{3}\times \text{a}\times \text{b}^{1+2}\times \text{c}\\-2\times \text{a}\times \text{b}\times \text{c}^{1+2}$
$=\frac{2}{3}\text{a}^3\text{bc}+\frac{2}{3}\text{ab}^3\text{c}-2\text{abc}^3$
View full question & answer→Question 113 Marks
Find the products:
Find the value of $\left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right)$, when $a=1$ and $b=0.5$.
Answer$2.3 a ^5 b^2 \times 1.2 a ^2 b^2$
$=2.3 \times 1.2 a^{5+2}, b^{2+2}$
$=2.76 a^7 b^4$
$=2.76 \times(1)^7 \times\left(\frac{1}{2}\right)^4$
$=2.76 \times 1 \times \frac{1}{16}=\frac{2.76}{16}$
$=0.1735$
View full question & answer→Question 123 Marks
Find the products:
$\Big(\frac{4}{3}\text{x}^2\text{yz}\Big)\times \Big(\frac{1}{3}\text{y}^2\text{zx}\Big)\times (-6\text{xyz}^2)$
Answer$\Big(\frac{4}{3}\text{x}^2\text{yz}\Big)\times \Big(\frac{1}{3}\text{y}^2\text{zx}\Big)\times (-6\text{xyz}^2)$
$=\frac{4}{3}\times \frac{1}{3}\times (-6)\\\times \text{x}^2\times \text{x}\times \text{x}\times \text{y}\times \text{y}^2\times \text{y}\times \text{z}\times \text{z}\times \text{z}^2$
$=\frac{-8}{3}\times \text{x}^{2+1+1}\times \text{y}^{1+2+1}\times \text{z}^{1+1+2}$
$=\frac{-8}{3}\text{x}^4\text{y}^4\text{z}^4$
View full question & answer→Question 133 Marks
Find the product $24 x^2(1-2 x)$ and evaluate it fore $x=2$.
Answer$24 X^2(1-2 X)$
$= 24 X^2 \times 1-24 X^2 \times 2 X$
$= 24 X^2-24 \times 2 \times X^2 \times X$
$= 24 X^2-48 X^3$
When $x=2$,
$\text { L.H.S }=24 x^2(1-2 x)$
$=24 \times 2^2(1-2 \times 2)$
$=96(1-4)$
$=96 \times(-3)$
$=-288$
$\text { R.H.S. }=24 x^2-48 x^2$
$=24 \times 2^2-48 \times 2^3$
$=96-384$
$=-288$
$\text { L.H.S = R.H.S. }$
$\therefore 24 x^2(1-2 x)$
$=24 x^2-48 x^3$
View full question & answer→Question 143 Marks
Simplity:
$(\text{x}^2-\text{x})-\frac{1}{2}(\text{x}-3+3\text{x}^2)$
Answer$(\text{x}^2-\text{x})-\frac{1}{2}(\text{x}-3+3\text{x}^2)$
$=\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}-\frac{3}{2}\text{x}^2$
$=\text{x}^2-\frac{3}{2}\text{x}^2-\text{x}-\frac{1}{2}\text{x}+\frac{3}{2}$
$=\Big(1-\frac{3}{2}\Big)\text{x}^2-\Big(1+\frac{1}{2}\Big)\text{x}+\frac{3}{2}$
$=\frac{2-3}{2}\text{x}^2-\frac{2+1}{2}\text{x}+\frac{3}{2}$
$=-\frac{1}{2}\text{x}^2-\frac{3}{2}\text{x}+\frac{3}{2}$
View full question & answer→Question 153 Marks
Simplity:
$\Big(\frac{1}{3}\text{y}^2-\frac{4}{7}\text{y}+5\Big)-\Big(\frac{2}{7}\text{y}-\frac{2}{3}\text{y}^2+2\Big)\\-\Big(\frac{1}{7}\text{y}-3+2\text{y}^2\Big)$
Answer$\Big(\frac{1}{3}\text{y}^2-\frac{4}{7}\text{y}+5\Big)-\Big(\frac{2}{7}\text{y}-\frac{2}{3}\text{y}^2+2\Big)\\-\Big(\frac{1}{7}\text{y}-3+2\text{y}^2\Big)$
$=\frac{1}{3}\text{y}^2-\frac{4}{8}\text{y}+5-\frac{2}{7}\text{y}+\frac{2}{3}\text{y}^2\\-2-\frac{1}{7}\text{y}+3-2\text{y}^2$
$=\frac{1}{3}\text{y}^2+\frac{2}{3}\text{y}^2-2\text{y}^2-\frac{4}{7}\text{y}-\frac{2}{7}\text{y}-\frac{1}{7}\text{y}+5-2+3$
$=\Big(\frac{1}{3}+\frac{2}{3}-2\Big)\text{y}^2+\Big(\frac{-4}{8}\text{y}-\frac{2}{7}\text{y}-\frac{1}{7}\Big)+8-2$
$=\frac{1+2-6}{3}\text{y}^2+\frac{-4-2-1}{7}\text{y}+6$
$=\frac{-3}{3}\text{y}^2+\frac{-7}{8}\text{y}+6$
$=-\text{y}^2-\text{y}+6$
View full question & answer→Question 163 Marks
Simplify the following:
$2 x^2+3 x\left(1-2 x^3\right)+x(x+1)$
Answer$2 x^2+3 x\left(1-2 x^3\right)+x(x+1)$
$=2 x^2+3 x-3 x \times 2 x^3+x^2+x$
$=2 x^2+3 x-6 \times x^{(1+3)}+x^2+x$
$=2 x^2+3 x-6 x^4+x^2+x$
$=-6 x^4+3 x^2+4 x$
View full question & answer→Question 173 Marks
Find the products:
Find the value of $\left(-8 u^2 v^6\right) \times(-20 u v)$ for $u=2.5$ and $v=1$.
Answer$(-8\text{u}^2\text{v}^6)\times (-20\text{uv})$
$=(-8)\times (-20)\text{u}^{2+1}\times \text{v}^{6+1}$
$=160\text{u}^3\text{v}^7$
$=160(2.5)^3(1)^7$
$=160\times \Big(\frac{5}{2}\Big)^3\times (1)^7$
$=160\times \frac{5}{2}\times \frac{5}{2}\times \frac{5}{2}\times1\times1\times1\times1\times1\times1\times1$
$=2500$
View full question & answer→Question 183 Marks
Find the product $a b\left(a^2+b^2\right)$ and evaluate it fore $a =2$ and $b =\frac{1}{2}$.
Answer$a b\left(a^2+b^2\right)$
$=a b \times a^2+a b \times b^2$
$=a \times a^2 \times b+a \times b \times b^2$
$=a^{(1+2)} \times b+a \times b^{(1+2)}$
$=a^3 b+a b^3$
When a = 2 and $\text{b}=\frac{1}{2},$ we get,
$\text{L.H.S}=\text{ab}(\text{a}^2+\text{b}^2)$
$=2\times \frac{1}{2}(2^2+\frac{2}{2^2})$
$=4+\frac{1}{4}=\frac{17}{4}$
$\text{R.H.S}=\text{a}^3\text{b}+\text{ab}^3$
$=2^3\times \frac{1}{2}+2\Big(\frac{1}{2}\Big)^3$
$=4+\frac{1}{4}=\frac{17}{4}$
$\therefore \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 193 Marks
Simplity:
[7 - 2x + 5y - (x - y)] - (5x + 3y - 7)
Answer[7 - 2x + 5y - (x - y)] - (5x + 3y - 7)
= (7 - 2x + 5y - x + y) - (5x + 3y - 7)
= 7 - 2x + 5y - x + y - 5x - 3y + 7
= 7 + 7 - 2x - x - 5x + 5y + y - 3y
= 14 - 8x + 3y
View full question & answer→Question 203 Marks
Find the products:$\Big(\frac{-2}{7}\text{u}^4\text{v}\Big)\times\Big(\frac{-14}{5}\text{uv}^3\Big)\times \Big(\frac{-3}{4}\text{u}^2\text{v}^3\Big)$
Answer$\Big(\frac{-2}{7}\text{u}^4\text{v}\Big)\times\Big(\frac{-14}{5}\text{uv}^3\Big)\times \Big(\frac{-3}{4}\text{u}^2\text{v}^3\Big)$
$=\Big(\frac{-2}{7}\Big)\times \Big(\frac{-14}{5}\Big)\times \Big(\frac{-3}{4}\Big)\\\text{u}^4\times \text{u}\times \text{u}^2\times \text{v}\times \text{v}^3\times \text{v}^7$
$=\frac{-3}{5}\times \text{u}^{4+1+2}\times \text{x}^{1+3+3}$
$=\frac{-3}{5}\text{u}^7\text{v}^7$
View full question & answer→Question 213 Marks
Find the products:
$\left(a b^2\right) \times\left(-b^2 c\right) \times\left(-a^2 c^3\right) \times(-3 a b c)$
Answer$\left(a b^2\right) \times\left(-b^2 c\right) \times\left(-a^2 c^3\right) \times(-3 a b c)$
$=1\left(a b^2\right) \times(-1) b^2 c \times(-1) a^2 c^3 \times(-3) a b c$
$=1 \times(-1)(-1)(-3) a \times a^2 \times a \times b^2 \times b^2 \times b \times c \times c^3 \times c$
$=-3 \times a^{1+2+1} \times b^{2+2+1} \times c^{1+3+1}$
$=-3 a^4 b^5 c^5$
View full question & answer→Question 223 Marks
Simplify the following:
$a^2 b\left(a-b^2\right)+a b^2\left(4 a b-2 a^2\right)-a^3 b(1-2 b)$
Answer$a^2 b\left(a-b^2\right)+a b^2\left(4 a b-2 a^2\right)-a^3 b(1-2 b)$
$=a^2 b \times a-a^2 b \times b^2+a b 2 \times 4 a b-a b 2 \times 2 a b 2-a 3 b \times 2 b$
$=a^{(2+1)} \times b-a^2 \times b^{(1+2)}+4 \times a^{(1+1)} \times b^{(2+1)}-2 \times a^{(1+2)} \times b^2-a^3 b+2 \times a^3 \times b^{(1+1)}$
$=a^3 b-a^2 b^3+4 a^2 b^3-2 a^3 b^2-a^3 b+2 a^3 b^2$
$=3 a^2 b^3$
View full question & answer→Question 233 Marks
Subtract (2a - 3b + 4c) from the sum of (a + 3b - 4c), (4a - b + 9c) and (-2b + 3c - a).
AnswerSum of (a + 3b - 4c), (4a - b + 9c) and (-2b + 3c - a)
$=\text{a}+3\text{b}-4\text{c}\\\ \ 4\text{a}-\ \ \text{b}+9\text{c}\\-\text{a}-2\text{b}+3\text{c}\\\underline{\overline{\ \ 4\text{a}\ \ \ \ \ \ \ \ \ + 8\text{c}}}$
Now subtract (2a - 3b + 4c) from 4a + 8c
= 4a + 8c - (2a - 3b + 4c)
= 4a + 8c - 2a + 3b - 4c
= 4a - 2a + 3b + 8c - 4c
= 2a + 3b + 4c
View full question & answer→Question 243 Marks
Simplify the following:
$4 s t(s-t)-6 s^2\left(t-t^2\right)-3 t^2\left(2 s^2-s\right)+2 s t(s-t)$
Answer$4 s t(s-t)-6 s^2\left(t-t^2\right)-3 t^2\left(2 s^2-s\right)+2 s t(s-t)$
$=4 s t \times s-4 s t \times t-6 s^2 \times t-6 s^2 \times\left(-t^2\right)-3 t^2 \times 2 s^2-3 t^2 \times(-s)+2 s t \times s-2 s t \times t$
$=4 \times s^{(1+1)} \times t-4 \times s \times t^{(1+1)}-6 s^2 t+6 s^2 t^2-6 t^2 s^2+3 t^2 s+2 \times s^{(1+1)} \times t-2 \times s \times t^{(1+1)}$
$=4 s^2 t-4 s t^2-6 s^2 t+6 s^2 t^2-6 t^2 s^2+3 t^2 s+2 s^2 t-2 s t^2$
$=4 s^2 t-6 s^2 t+2 s^2 t-4 s t^2+3 s t^2-2 s t^2$
$=-3 s t^2$
View full question & answer→Question 253 Marks
Simplify:
$(3 x+4)(2 x-3)+(5 x-4)(x+2)$
Answer$(3 x+4)(2 x-3)$
$=3 x(2 x-3)+4(2 x-3)$
$=6 x^{(1+1)}-9 x+8 x-12$
$=6 x^2-x-12$
$(5 x-4)(x+2)$
$=5 x(x+2)-4(x+2)$
$=5 x^{(1+1)}+10 x-4 x-8$
$=5 x^2+6 x-8$
$\therefore(3 x+4)(2 x-3)+(5 x-4)(x+2)$
$=6 x^2-x-12+5 x^2+6 x-8$
$=11 x^2+5 x-20$
View full question & answer→Question 263 Marks
Simplify the following:
$x(x+4)+3 x\left(2 x^2-1\right)+4 x^2+4$
Answer$x(x+4)+3 x\left(2 x^2-1\right)+4 x^2+4$
$=x \times x+x \times 4+3 x \times 2 x^2-3 x+4 x^2+4$
$=x^{(1+1)}+4 x+6 \times x^{(1+2)}-3 x+4 x^2+4$
$=x^2+4 x+6 x^3-3 x+4 x^2+4$
$=6 x^3+5 x^2+x+4$
View full question & answer→