Question

solve and factorise 75(2x-10)/8(x^2-10x) ​

Answer 1

Answer:-

We have to factorise:-

$\sf \: \dfrac{75(2x - 10)}{8( {x}^{2} - 10x) } = 1 \\ \\ \\ \implies \sf \: 150x - 750 = 8 {x}^{2} - 80x \\ \\ \\ \implies \sf \:0 = 8 {x}^{2} - 80x - 150x + 750 \\ \\ \\ \implies \sf \:0 = 8 {x}^{2} - 230x + 750 \\ \\ \\ \implies \sf \:0 = 2(4 {x}^{2} - 115x + 375) \\ \\ \\ \implies \sf \:4 {x}^{2} - 115x + 375 = 0 \\ \\ \\ \implies \sf \:4 {x}^{2} - 100x - 15x + 375 = 0 \\ \\ \\ \implies \sf \:4x( {x} - 25) - 15(x - 25) = 0 \\ \\ \\ \implies \boxed{\sf \:(x - 25)(4x - 15) = 0}$

Answer 2

$\dag{\large{\red{\bold{\sf{Required \; Solution}}}}}$

${\small{\bold{\tt{\underline{In \: the \: question \: it \: is \: given \: that,}}}}}$

${\small{\bold{\tt{\underline{We \: have \: to \: factorise \: the \: following \: term -}}}}}$

${\bold{\pink{\dfrac{75(2x-10)}{8(x^{2}-10x)} \: \: \: \green \bigstar}}}$

$\begin{gathered} \rightarrow \bf \: \dfrac{75(2x - 10)}{8({x}^{2} - 10x)} = 1 \\ \\ \rightarrow \bf \: 150x - 750 = 8 {x}^{2} - 80x \\ \\ \rightarrow \bf \:0 = 8 {x}^{2} - 80x - 150x + 750 \\ \\ \rightarrow \bf \:0 = 8 {x}^{2} - 230x + 750 \\ \\ \rightarrow \bf \:0 = 2(4 {x}^{2} - 115x + 375) \\ \\ \rightarrow \bf \:4 {x}^{2} - 115x + 375 = 0 \\ \\ \rightarrow \bf \:4 {x}^{2} - 100x - 15x + 375 = 0 \\ \\ \rightarrow \bf \:4x({x} - 25) - 15(x - 25) = 0 \\ \\ \rightarrow {\bf \:(x - 25)(4x - 15) = 0}\end{gathered}$

$\; \; \; \; \; \; \; \; \; \; \; \; \;{\ddag{\rm{Hence, \: factorised}}}$

$\dag{\large{\red{\bold{\sf{Knowledge \; drink}}}}}$

These expressions can be easily factorised using identities !.

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$