5 Marks Questions

Question 15 Marks

Simplify:
$\left(3 x^2+5 x-7\right)(x-1)-\left(x^2-3 x+3\right)(x+4)$

Answer

$\left(3 x^2+5 x-7\right)(x-1)$
By column method:
$3\text{x}^2+5\text{x}-7\\\underline{\ \ \ \ \ \ \times (\text{x}-1)\ \ }\\3\text{x}^3+5\text{x}^2-7\text{x}\\ \underline{\ \ \ \ \ \ -3\text{x}^2 -5\text{x}+7\ }\\3\text{x}^3+2\text{x}^2-12\text{x}+7$
$\left(x^2-3 x+3\right)(x+4)$
By column method:
$\text{x}^2-2\text{x}+3\\ \underline{\ \ \ \ \times(\text{x}+4)\ \ \ }\\\text{x}^3-2\text{x}^2+3\text{x}\\\underline{\ \ \ \ \ \ \ \ \ 4\text{x}^2-8\text{x}+12}\\\text{x}^3+2\text{x}^2-5\text{x}+12$
$\left(3 x^2+5 x-7\right)(x-1)-\left(x^2-2 x+3\right)(x+4)$
$=3 x^3+2 x^2-12 x+7-\left(x^3+2 x^2-5 x+12\right)$
$=3 x^3-x^3+2 x^2-2 x^2-12 x+5 x+7-12$
$=2 x^3-7 x-5$

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Question 25 Marks

Add the following expression: $\frac{3}{2}\text{x}^3-\frac{1}{4}\text{x}^2+\frac{5}{3},-\frac{5}{4}\text{x}^3+\frac{3}{5}\text{x}^2-\text{x}\\+\frac{1}{5},-\text{x}^2+\frac{3}{8}\text{x}-\frac{8}{15}$

Answer

$\Big(\frac{1}{2}\text{x}^3-\frac{1}{4}\text{x}^2+\frac{5}{3}\Big)+\Big(\frac{-5}{4}\text{x}^3+\frac{3}{5}\text{x}^2-\text{x}+\frac{1}{5}\Big)$
$+\Big(-\text{x}^2+\frac{3}{8}\text{x}-\frac{8}{15}\Big)$
$=\frac{3}{2}\text{x}^3-\frac{1}{4}\text{x}^2+\frac{5}{3}-\frac{5}{4}\text{x}^3+\frac{3}{5}\text{x}^2-\text{x}$
$+\frac{1}{5}-\text{x}^2+\frac{3}{8}\text{x}-\frac{8}{15}$
$=\frac{3}{2}\text{x}^3-\frac{5}{4}\text{x}^3+\Big(-\frac{1}{4}\Big)\text{x}^2+\frac{3}{5}\text{x}^2-\text{x}^2$
$-\text{x}+\frac{3}{8}\text{x}+\frac{5}{3}+\frac{1}{5}-\frac{8}{15}$
$=\Big(\frac{3}{2}-\frac{5}{4}\Big)\text{x}^3+\Big(-\frac{1}{4}+\frac{3}{5}-1\Big)\text{x}^2$
$+\Big(-1+\frac{3}{8}\Big)\text{x}+\Big(\frac{5}{3}+\frac{5}{3}+\frac{1}{5}-\frac{8}{15}\Big)$
$=\frac{6-5}{4}\text{x}^3+\frac{-5+12-20}{20}\text{x}^2+\frac{-8+3}{8}\text{x}+\frac{25+3-8}{15}$
$=\frac{1}{4}\text{x}^3+\Big(\frac{-13}{20}\Big)\text{x}^2+\Big(\frac{-5}{8}\Big)\text{x}+\frac{20}{15}$
$\Big(\because\frac{20}{15}=\frac{20\div5}{15\div5}=\frac{4}{3}\Big)$
$=\frac{1}{4}\text{x}^3-\frac{13}{20}\text{x}^2-\frac{5}{8}\text{x}+\frac{4}{3}$

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Question 35 Marks

Simplify: $(2x + 5y)(3x + 4y) - (7x + 3y)(2x + y)$

Answer

$(2 x+5 y)(3 x+4 y)$
$=2 x(3 x+4 y)+5 y(3 x+4 y)$
$=6 x^{(1+1)}+8 x y+15 y x+20 y^{(1+1)}$
$=6 x^2+23 x y+20 y^2$
$(7 x+3 y)(2 x+y)$
$=7 x(2 x+y)+3 y(2 x+y)$
$=14 x^{(1+1)}+7 x y+6 y x+3 y$
$=14 x^2+13 x y+3 y^2$
$\therefore(2 x+5 y)(3 x+4 y)-(7 x+3 y)(2 x-y)$
$=6 x^2+23 x y+20 y^2-\left(14 x^2+13 x y+3 y^2\right)$
$=6 x^2-14 x^2+23 x y-13 x y+20 y^2-3 y^2$
$=-8 x^2+10 x y+17 y^2$

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Question 45 Marks

Find the products given below and case verify the result for $a = 1, b = 2$ and $c = 3.$
$\Big(\frac{1}{4}\text{abc}\Big)\times(-6\text{b}^2\text{c})\times\Big(-\frac{1}{3}\text{c}^3\Big)$

Answer

$\Big(\frac{1}{4}\text{abc}\Big)\times(-6\text{b}^2\text{c})\times\Big(-\frac{1}{3}\text{c}^3\Big)$
$=\frac{1}{4}\times(-6)\times\Big(-\frac{1}{3}\Big)\text{a}\times \text{b}\times \text{b}^2\times \text{c}\times\text{c}\times \text{c}^3$
$=\frac{1}{2}\text{a}\times\text{b} ^{1+2}\times \text{c}^{1+1+3}$
$=\frac{1}{2}\text{ab}^3\text{c}^5$
Verification:
$\text{L.H.S}=\Big(\frac{1}{4}\text{abc}\Big)\times(-6\text{b}^2\text{c})\Big(-\frac{1}{3}\text{c}^3\Big)$
$=\Big[\frac{1}{4}(1)(2)(3)\Big]\times[-6(2)^2\times3]\Big[-\frac{1}{3}(3)^3\Big]$
$=\Big(\frac{3}{2}\Big)\times(-72)\times (-9)$
$=\frac{3\times648}{2}=3\times 324$
$=972$
$\text{R.H.S}=\frac{1}{2}\text{ab}^3\text{c}^5$
$=\frac{1}{2}\times1\times(2)^3\times(3)^5$
$=\frac{1}{2}\times8\times243$
$=4\times243$
$=972$
$\therefore \text{L.H.S}= \text{R.H.S}$

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Question 55 Marks

Find the products given below and case verify the result for $a = 1, b = 2$ and $c = 3$. $\Big(\frac{-4}{7}\text{a}^2\text{b}\Big)\times\Big(\frac{-2}{3}\text{b}^2\text{c}\Big)\times\Big(\frac{-7}{6}\text{c}^2\text{a}\Big)$

Answer

$\Big(\frac{-4}{7}\text{a}^2\text{b}\Big)\times\Big(\frac{-2}{3}\text{b}^2\text{c}\Big)\times\Big(\frac{-7}{6}\text{c}^2\text{a}\Big)$$=\Big(\frac{-4}{7}\Big)\times\Big(\frac{-2}{3}\Big)\times \Big(\frac{-7}{6}\Big)\text{a}^2\times \text{a}\times \text{b}\times \text{b}^2\times \text{c}\times \text{c}^2$
$=\frac{-4}{9}\text{a}^3\text{b}^3\text{c}^3$
Verification:
$\text{L.H.S}$
$\Big(\frac{-4}{7}\text{a}^2\text{b}\Big)\times\Big(\frac{-2}{3}\text{b}^2\text{c}\Big)\times \Big(-\frac{7}{6}\text{c}^2\text{a}\Big)$
$\Big[\frac{-4}{7}(1)^2(2)\Big]\times \Big[\frac{-2}{3}(2)^2\times3\Big]\times\Big[\frac{-7}{6}(3)^2\times1\Big]$
$\Big[\frac{-4}{7}\times 1\times2\Big]\times \Big[\frac{-2}{3}\times 4\times3\Big]\times\Big[\frac{-7}{6}\times 9\Big]$
$=\Big(\frac{-8}{7}\Big)\times(-8)\times\Big(-\frac{21}{2}\Big)$
$=-\Big[\frac{8}{7}\times 8\times\frac{21}{2}\Big]$
$=-96$
$\text{R.H.S}$
$=\frac{-4}{9}\text{a}^3\text{b}^3\text{c}^3$
$=\frac{-4}{9}(1)^3\times (2)^3\times(3)^3$
$=\frac{-4}{9}\times1\times 8\times27$
$=-96$
$\therefore \text{L.H.S}=\text{R.H.S}$

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Question 65 Marks

Find the products given below and case verify the result for $a = 1, b = 2$ and $c = 3$. $\Big(\frac{2}{5}\text{a}^2\text{b}\Big)\times (-15\text{b}^2\text{ac})\times\Big(-\frac{1}{2}\text{c}^2\Big)$

Answer

$\Big(\frac{2}{5}\text{a}^2\text{b}\Big)\times (-15\text{b}^2\text{ac})\times\Big(-\frac{1}{2}\text{c}^2\Big)$
$=\frac{2}{5}\times (-15)\times \Big(-\frac{1}{2}\Big)\text{a}^2\times\text{a}\times \text{b}\times\text{b}^2\times \text{c}\times\text{c}^2$
$=3\text{a}^{2+1}\times \text{b}^{1+2}\times \text{c}^{1+2}$
$=3\text{a}^3\text{b}^3\text{c}^3$
Verification: $=\Big(\frac{2}{5}\text{a}^2\text{b}\Big)\times (-15\text{b}^2\text{ac})\Big(-\frac{1}{2}\text{c}^2\Big)$
$=\frac{2}{5}(1)^2(2)\times (-15)\times(2)^2\times 1\times3\times\Big(-\frac{1}{2}(3)^2\Big)$
$=\Big(\frac{2}{5}\times1\times2\Big)\times(-15\times4\times3)\times\Big(-\frac{1}{2}\times9\Big)$
$=\frac{4}{5}\times(-180)\times\Big(-\frac{9}{2}\Big)$
$=\frac{360\times9}{5}=72\times9$
$=648$
$\text{R.H.S}=3\text{a}^3\text{b}^3\text{c}^3$
$=3(1)^3(2)^3(3)^3$
$=3\times1\times8\times27$
$=81\times8 $
$=648$
$\therefore \text{L.H.S}=\text{R.H.S}$

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Question 75 Marks

Find the products given below and case verify the result for $a = 1, b = 2$ and $c = 3$. $\Big(\frac{4}{9} \text{abc}^3\Big)\times \Big(\frac{-27}{5} \text{a}^3 \text{b}^2\Big)\times(-8 \text{b}^3 \text{c})$

Answer

$\Big(\frac{4}{9} \text{abc}^3\Big)\times \Big(\frac{-27}{5} \text{a}^3 \text{b}^2\Big)\times(-8 \text{b}^3 \text{c})$
$=\frac{4}{9}\times\Big(\frac{-27}{5}\Big)(-8)\text{a}\times\text{a}^3\times\text{b}\times\text{b}^2\times\text{b}^3\times \text{c}^3\times \text{c}$
$=\frac{96}{5}\text{a}^4\text{b}^6\text{c}^4$
Verification: $\text{L.H.S}=\Big(\frac{4}{9}\text{abc}^3\Big)\times \Big(\frac{-27}{5}\text{a}^3\text{b}^2\Big)\times (-8\text{b}^3\text{c})$
$=\Big[\frac{4}{9}\times1\times2\times(3)^3\Big]\times\Big[\frac{-27}{5}(1)^3(2)^2\Big]$
$\times[-8\times(2)^3\times3]$
$=\Big[\frac{4}{9}\times1\times2\times27\Big]\times\Big[\frac{-27}{5}1\times4\Big]\times[-8\times8\times3]$
$=(24)\times\Big(\frac{-108}{5}\Big)\times(-192)$
$=\frac{497664}{5}$
$\text{R.H.S}=\frac{96}{5}(\text{a}^4\text{b}^6\text{c}^4)$
$=\frac{96}{5}(1)^4(2)^6(3)^4$
$=\frac{96}{5}\times 1\times 64\times81$
$=\frac{497664}{5}$
$\therefore\text{L.H.S}=\text{R.H.S}$

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Question 85 Marks

Find the products: Multiply $-\frac{2}{3}\text{a}^2\text{b by }\frac{6}{5}\text{a}^3\text{b}^2$ and verify your result for $a = 2$ and $b = 3.$

Answer

$\frac{-2}{3}\text{a}^2\text{b}\times \frac{6}{5}\text{a}^3\text{b}^2$
$=\frac{-2}{3}\times \frac{6}{5}\text{a}^2\times \text{a}^3\times \text{b}\times \text{b}^2$
$=\frac{-4}{5}\text{a}^{2+3}\text{b}^{1+2}$
$=\frac{-4}{5}\text{a}^5\text{b}^3$
Verification:
Now if $a = 2, b = 3$, then $-\frac{2}{3}\text{a}^2\text{b}=\frac{-2}{3}\times (2)^2\times 3$
$=\frac{-2}{3}\times 4\times 3$
$=-8$
$\frac{6}{5}\text{a}^3\text{b}^2=\frac{6}{5}\times (2)^3\times (3)^2$
$=\frac{6}{5}\times 2\times 2\times 2\times 3\times 3$
$=\frac{6}{5}\times 8\times 9=\frac{432}{5}$
$\therefore \text{L.H.S} = -8 \times \frac{432}{5}$
$=\frac{-3456}{5}$
$\text{R.H.S}=\frac{-4}{5}\text{a}^5\text{b}^3=\frac{-4}{5}(2)^5(3)^3$
$=\frac{-4}{5}(2\times2\times2\times2\times2)\times3\times3\times3$
$=\frac{-4}{5}\times32\times27$
$=\frac{-4\times864}{5}=\frac{-3456}{5}$
$\therefore \frac{-3456}{5}=\frac{-3456}{5}$
$\text{L.H.S.}= \text{R.H.S}$

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Question 95 Marks

Find the products: Multiply $-\frac{8}{21}\text{x}^2\text{y}^3\text{ by }-\frac{7}{16}\text{xy}^2$ and varify you result for $x = 3$ and $y = 2.$

Answer

$-\frac{8}{21}\text{x}^2\text{y}^3\times\frac{-7}{16}\text{xy}^2$
$=\frac{-8}{21}\times \frac{-7}{16}\text{x}^2\times \text{x}\times \text{y}^3\times \text{y}^2$
$=\frac{-1}{3}\times \frac{-1}{2}\times \text{x}^{2+1}\times \text{y}^{3+2}$
$=\frac{1}{6}\text{x}^3\text{y}^5$ Verification: $x = 3, y = 2$
$\text{L.H.S} =-\frac{8}{21}\text{x}^2\text{y}^3\times \frac{-7}{16}\text{xy}^2$
$=\frac{-8}{21}(3)^2(2)^3\times \frac{-7}{16}(3)(2)^2$
$=\frac{-8}{21}\times9\times 8\times\frac{-7}{16}\times 3\times 4$
$=\frac{-8}{21}\times \frac{-7}{16}\times 9\times 8\times 3\times 4$
$=\frac{-1}{3}\times \frac{-1}{2}\times72\times12 $
$=12\times12=144$
$\text{R.H.S}=\frac{1}{6}\text{x}^3\text{y}^5$
$=\frac{1}{6}\times(3)^3(2)^5$
$=\frac{1}{6}\times 27\times 32$
$=9\times 16=144$
$\therefore\text{L.H.S}= \text{R.H.S}$

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