Question

SOLVE QUESTION NO. 15 IN THE GIVEN IMAGE.

Answer 1

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✴ Required Answer:

✏ To Factorise:

• $\large{ \rm{5 - (3 {a}^{2} - 2a)(6 - 3 {a}^{2} + 2a)}}$

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✴ Solution:

The polynomial can be written as,

$\large{ \rm{ = 5 - (3 {a}^{2} - 2a)(6 - (3 {a}^{2} - 2a)}}$

Let's consider 3a² - 2a be t

$\large{ \rm{ = 5 - t(6 - t)}} \\ \\ \large{ \rm{ = 5 - 6t + {t}^{2} }} \\ \\ \large{ \rm{ = {t}^{2} - 6t + 5}}$

By middle term factorisation,

$\large{ \rm{ = {t}^{2} - 5t - t + 5}} \\ \\ \large{ \rm{ = t(t - 5) - 1(t - 5)}} \\ \\ \large{ \rm{ = (t - 1)(t - 5)}} \\ \\ \large{ \rm{ = (3 {a}^{2} - 2a - 1)(3 {a}^{2} - 2a - 5)}}$

This can be further factorised:

$\large{ \rm{ = 3 {a}^{2} - 2a - 1}} \\ \\ \large{ \rm{ = 3 {a}^{2} - 3a + a - 1}} \\ \\ \large{ \rm{ = 3a(a - 1) + 1(a - 1)}} \\ \\ \large{ \rm{ = (3a + 1)(a - 1) \: \: and}} \\ \\ \\ \large{ \rm{ = 3 {a}^{2} - 2a - 5}} \\ \\ \large{ \rm{ = 3 {a}^{2} - 5a + 3a - 5}} \\ \\ \large{ \rm{ = a(3a - 5) + 1(3a - 5)}} \\ \\ \large{ \rm{ = (a + 1)(3a - 5)}}$

Hence, Our final value:

$\large{ \rm{ = (3a + 1)(a - 1)(a + 1)(3a - 5)}} \\ \\ \large{ \rm{ = \boxed{ \red{ \rm{ (3a + 1)(3a - 5)( {a}^{2} - 1)}}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \green{ \rm{Hence \: solved \: \dag}}}}}}$

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