Question

Questioun 13 pls I will mark brainlillest

Answer 1

$8x^3-6x^2-4x+4 \\ \\ 8x^3-6x^2-4x+3+1 \\ \\ 24x^2(\frac{x}{3}-\frac{1}{4})-12(\frac{x}{3}-\frac{1}{4})+1 \\ \\ (24x^2-12)(\frac{x}{3}-\frac{1}{4})+1 \\ \\ \\ \therefore\ \bold{1}\ is the remainder.$

$OR$

$\frac{x}{3}-\frac{1}{4}=0 \ \ \Rightarrow\ \ \frac{x}{3}=\frac{1}{4}\ \ \Rightarrow\ \ x=\frac{3}{4}$

$\therefore\ \ p(x)\ \ divided by \ \ g(x)\ \ leaves remainder \ \ p(\frac{3}{4}).$

$\therefore\ p(\frac{3}{4})=8(\frac{3}{4})^3-6(\frac{3}{4})^2-4(\frac{3}{4})+4 \\ \\ \Rightarrow p(\frac{3}{4})=8 \times \frac{27}{64}-6 \times \frac{9}{16}- 4\times \frac{3}{4} + 4 \\ \\ \Rightarrow p(\frac{3}{4})=\frac{27}{8}-\frac{27}{8}-3+4 \\ \\ \Rightarrow p(\frac{3}{4})=\bold{1}$

$\therefore\ \bold{1}\ is the remainder. \\ \\ \\ I think this is the right method for the question.$

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$\bold{OR}$

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$x^{1/3}+\frac{1}{x^{1/3}}=5 \\ \\ \\ (x^{1/3}+\frac{1}{x^{1/3}})^3=5^3 \\ \\ (x^{1/3})^3+(3 \cdot (x^{1/3})^2 \cdot \frac{1}{x^{1/3}})+(3 \cdot x^{1/3} \cdot (\frac{1}{x^{1/3}})^2)+(\frac{1}{x^{1/3}})^3=125 \\ \\ x+3 \cdot (x^{1/3}) + 3 \cdot (\frac{1}{x^{1/3}}) + \frac{1}{x}=125 \\ \\ x+\frac{1}{x}+3(x^{1/3}+\frac{1}{x^{1/3}})=125 \\ \\ x+\frac{1}{x}+3 \cdot 5=125 \\ \\ x+\frac{1}{x}+15=125 \\ \\ x+\frac{1}{x}=125-15 \\ \\ x+\frac{1}{x}=110$

$(x+\frac{1}{x})^2=110^2 \\ \\ x^2+(2 \cdot x \cdot \frac{1}{x})+\frac{1}{x^2}=12100 \\ \\ x^2+\frac{1}{x^2}+2=12100 \\ \\ x^2+\frac{1}{x^2}=12100-2 \\ \\ x^2+\frac{1}{x^2}=\bold{12098}$

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