Find the LCM of $10 x^{3} + 6 x^{2} - 28 x, 9 x^{3} + 15 x^{2} - 6 x$

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Solution

We know that $L C M$ is the least common multiple.

Factorise $10 x^{3} + 6 x^{2} - 28 x$ as follows:

$10 x^{3} + 6 x^{2} - 28 x = 2 x (5 x^{2} + 3 x - 14) = 2 x (5 x^{2} + 10 x - 7 x - 14) = 2 x [5 x (x + 2) - 7 (x + 2)] = 2 x (5 x - 7) (x + 2)$

Now, factorise $9 x^{3} + 15 x^{2} - 6 x$ as follows:

$9 x^{3} + 15 x^{2} - 6 x = 3 x (3 x^{2} + 5 x - 2) = 3 x (3 x^{2} + 6 x - x - 2) = 3 x [3 x (x + 2) - 1 (x + 2)] = 3 x (3 x - 1) (x + 2)$

Therefore, the least common multiple between the polynomials $10 x^{3} + 6 x^{2} - 28 x$ and $9 x^{3} + 15 x^{2} - 6 x$ is:

$L C M = 2 \times 3 \times x \times (x + 2) \times (3 x - 1) \times (5 x - 7) = 6 x (x + 2) (3 x - 1) (5 x - 7)$

Hence, the $L C M$ is $6 x (x + 2) (3 x - 1) (5 x - 7)$ .

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