Question

Divide
${9a}^{4} - {4a}^{2} + 4 \: by \: {3 a}^{2} - 4a + 2$
​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\dfrac{9a^4-4a^2+4}{3a^2-4a+2}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\Large\boxed{\sf{3a^2+4a+2}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\dfrac{9a^4-4a^2+4}{3a^2-4a+2}}$

$\bigstar\:\:\bf{Divide \:\: \dfrac{9 a^{4}-4 a^{2}+4}{3 a^{2}-4 a+2}}$

$\begin{aligned}&\text { Divide the leading coefficients of the numerator } 9 a^{4}-4 a^{2}+4\\&\text { and the divisor } 3 a^{2}-4 a+2: \dfrac{9 a^{4}}{3 a^{2}}=3 a^{2}\end{aligned}$

$\boxed{\sf{\mathrm{Quotient}=3a^2}}$

$\mathrm{Multiply\:}3a^2-4a+2\mathrm{\:by\:}3a^2:\:9a^4-12a^3+6a^2$

$\mathrm{Subtract\:}9a^4-12a^3+6a^2\mathrm{\:from\:}9a^4-4a^2+4\mathrm{\:to\:get\:new\:remainder}$

$\boxed{\sf{\mathrm{Remainder}=12a^3-10a^2+4}}$

$\mathrm{Therefore}$

$\sf{\displaystyle\frac{9a^4-4a^2+4}{3a^2-4a+2}=3a^2+\frac{12a^3-10a^2+4}{3a^2-4a+2}}$

$\bigstar\:\:\bf{\displaystyle Divide\:\:\:\frac{12 a^{3}-10 a^{2}+4}{3 a^{2}-4 a+2}}$

$\begin{aligned}&\text { Divide the leading coefficients of the numerator } 12 a^{3}-10 a^{2}+4\\&\text { and the divisor } 3 a^{2}-4 a+2: \dfrac{12 a^{3}}{3 a^{2}}=4 a\end{aligned}$

$\boxed{\sf{\mathrm{Quotient}=4a}}$

$\mathrm{Multiply\:}3a^2-4a+2\mathrm{\:by\:}4a:\:12a^3-16a^2+8a$

$\mathrm{Subtract\:}12a^3-16a^2+8a\mathrm{\:from\:}12a^3-10a^2+4\mathrm{\:to\:get\:new\:remainder}$

$\boxed{\sf{\mathrm{Remainder}=6a^2-8a+4}}$

$\mathrm{Therefore}$

$\sf{\displaystyle\frac{12a^3-10a^2+4}{3a^2-4a+2}=4a+\frac{6a^2-8a+4}{3a^2-4a+2}}$

$\bigstar\:\:\bf{\displaystyle Divide\:\:\:\frac{6a^2-8a+4}{3a^2-4a+2}}$

$\begin{aligned}&\text { Divide the leading coefficients of the numerator } 6 a^{2}-8 a+4\\&\text { and the divisor } 3 a^{2}-4 a+2: \dfrac{6 a^{2}}{3 a^{2}}=2\end{aligned}$

$\boxed{\sf{\mathrm{Quotient}=2}}$

$\mathrm{Multiply\:}3a^2-4a+2\mathrm{\:by\:}2:\:6a^2-8a+4$

$\mathrm{Subtract\:}6a^2-8a+4\mathrm{\:from\:}6a^2-8a+4\mathrm{\:to\:get\:new\:remainder}$

$\boxed{\sf{\mathrm{Remainder}=0}}$

$\Large\boxed{\sf{=3a^2+4a+2}}$