Question

factorise the following by splitting the middle term 9(x-2y)^2-4(x-2y)-13​

Answer 1

9(x − 2y)^2 − 4(x − 2y) − 13

Put (x − 2y) = a

= 9a^2 − 4a − 13
= 9a^2 − 13a + 9a − 13
= a(9a − 13) + 1(9a − 13)
= (9a − 13)(a + 1)
{ putting the value of a}
= [9(x − 2y) − 13][(x − 2y) + 1]
= [9x − 18y − 13][x − 2y + 1]

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Answer 2

$\underline{\underline{QUESTION}}$➡

$Factorise \: \: the \: \: following \: \: by \: \: splitting \\ the \: \: middle \: \: term \: : \\ \\ \boxed{9(x - 2y) {}^{2} - 4(x - 2y) - 13}$

$\underline{\underline{SOLUTION}}$➡

$Let ,\\ \\ x - 2y = p$

$Then\: \: the \: \:polynomial\: \: will \: \: be \\ \\ \boxed{9p {}^{2} - 4 p - 13 \: \: }\\ \\ Now,\\ \\ \: \: We \: \: have \: \: to \: \: factorise \: \: it \: \: by \: \: \\ splitting \: \: middle \: \: term \: : \\ \\ 9p {}^{2} - 4p - 13 \\ \\ = 9p {}^{2} - (13 - 9)p - 13 \\ \\ = 9p {}^{2} - 13 p + 9p - 13 \\ \\ = p(9p - 13) + 1(9p - 13) \\ \\ = (p + 1)(9p -1 3) \: \: \: \: - - - - > (1) \\ \\ Here, \\ \\ \: \: We \: \:assume \: \: \:\: \boxed{p = x - 2y} \\ \\ So ,\\ \\ (1) = > ((x - 2y) + 1)(9(x - 2y) - 13) \\ \\ = > (x - 2y + 1)(9x - 18y - 13) \\ \\ \\ Hence, \\ \\ 9(x - 2y) {}^{2} - 4(x - 2y) - 13 \\ \\ = (x - 2y + 1)(9x - 18y - 13)$

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