What is the square root of $9 y^{4} - 6 y^{3} + 7 y^{2} - 2 y + 1$ using division method?

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Solution

$\div$ $3 y^{2} - y + 1$ <= Quotient
$3 y^{2}$
$9 y^{4} - 6 y^{3} + 7 y^{2} - 2 y + 1$
$9 y^{4}$
$6 y^{2} - y$
$- 6 y^{3} + 7 Y 62$
$- 6 y^{3} + y^{2}$
$6 y^{2} - 2 y + 1$
$6 y^{2} - 2 y + 1$
$6 y^{2} - 2 y + 1$
Remainder $\Rightarrow 0$
The square root of $9 y^{4} - 6 y^{3} + 7 y^{2} - 2 y + 1$ is $∣ ∣ 3 y^{2} - y + 1 ∣ ∣$

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