Factorize the following expressions:
$6 x^{3} y - 12 x^{2} y + 15 x^{4}$

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Solution

Let us first find the HCF of all the terms of the given polynomial $6 x^{3} y - 12 x^{2} y + 15 x^{4}$ by factoring the terms as follows:

$6 x^{3} y = 2 \times 3 \times x \times x \times x \times y 12 x^{2} y = 2 \times 2 \times 3 \times x \times x \times y 15 x^{4} = 3 \times 5 \times x \times x \times x \times x$

Therefore, $H C F = 3 \times x \times x = 3 x^{2}$

Now, we factor out the HCF from each term of the polynomial $6 x^{3} y - 12 x^{2} y + 15 x^{4}$ as shown below:

$6 x^{3} y - 12 x^{2} y + 15 x^{4} = 3 x^{2} (2 x y - 4 y + 5 x^{2})$

Hence, $6 x^{3} y - 12 x^{2} y + 15 x^{4} = 3 x^{2} (2 x y - 4 y + 5 x^{2})$ .

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