Question

Factorise 6a^5 - 21a^4 - 12a^2​

Answer 1

SOLUTION:-

Given: 6a^5 - 21a^4 - 12a².

$6 {a}^{5} - 21 {a}^{4} - 12 {a}^{2} \\ \\ = > 3 {a}^{2} [2 {a}^{3} - 7 {a}^{2} - 4] \\ \\ = > 3 {a}^{2} [2 {a}^{3} - 8 {a}^{2} + {a}^{2} - 4] \\ \\ = > 3 {a}^{2} [2 {a}^{3} ({a}^{2} - 4) + 1( {a}^{2} - 4)] \\ \\ = > 3 {a}^{2} ( {a}^{2} - 4)(2 {a}^{3} +1) \\ \\ = > 3 {a}^{2}(2 {a}^{3} + 1)(a + 2)(a - 2)$

Thank you.

Answer 2

Answer:

$3a^{2}(2a^{3}-7a^{2}-4)$

Step-by-step explanation:

$6a^{5}-21a^{4}-12a^{2}\\=3a^{2}(2a^{3}-7a^{2}-4)$

Therefore,

$6a^{5}-21a^{4}-12a^{2}\\=3a^{2}(2a^{3}-7a^{2}-4)$

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