4 Mark Question

Question 14 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:
$3 x^2+5 x-7$
$\quad \times(x-1)$
$\frac{x(x-1)}{3 x^3+5 x^2-7 x}$
$\frac{-3 x^2-5 x+7}{3 x^3+2 x^2-12 x+7}$
$\left(x^2-3 x+3\right)(x+4)$
By column method:
$x^2-2 x+3$
$\times(x+4)$
$x^3-2 x^2+3 x$
$\frac{4 x^2-8 x+12}{x^3+2 x^2-5 x+12}$
$\left(3 x^2+5 x-7\right)(x-1)-\left(x^2-2 x+3\right)(x+4)$
$=3 x^3+2 x^2-12 x+7-\left(x^3+2 x^2-5 x+12\right)$
$=3 x^3-x^3+2 x^2-2 x^2-12 x+5 x+7-12$
$=2 x^3-7 x-5$

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Question 24 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 34 Marks

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

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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