Question

Factorise: ( p+q)2-20 (p+q)-125

Answer 1

Answer:

= > ( p + q )^2 - 20( p + q ) - 125

Splitting the middle term in such a manner that the product of those parts will be equal to the product of co.eff. of (p+q)^2 and the constant term.

= > ( p + q )^2 - ( 25 - 5 )( p + q ) - 125

= > ( p + q )^2 - 25( p + q ) + 5( p + q ) - 125

= > ( p + q )( p + q - 25 ) + 5( p + q - 25 )

= > ( p + q - 25 )( p + q + 5 )

It's factorized now.

Hence, (p+q)^2-20(p+q)-125=(p+q-25)(p+q+5)

Answer 2

$\huge\star \: {\sf{\underline{\orange{SolUtion:-}}}}$

$\star \: {\sf{\underline{\green{Given}}}}$

${\tt{\orange{Factorise : (p+q) 2- 20 (p+q) - 125}}}$

${\tt{\pink{By \:splitting \:middle\: term \: in \: such \: a \: manner \: to \: get \: the \: co - efficient\: of \: (p + q) ^2 }}}$

${\implies{\tt{\red{(p + q)^2 - (25 - 5) (p+q) - 125 }}}}$

${\implies{\tt{\red{(p + q)^2 - 25 (p+q) 5(p+q) - 125}}}}$

${\implies{\tt{\red{(p + q) - ( p + q -25) 5 (p+q - 125) }}}}$

${\implies{\tt{\red{(p + q - 25) (p+q+5) }}}}$

${\tt{\pink{ Factorized,}}}$

$\boxed {\green{(p+q)^2 - 20(p + q) - 125 = (p+q- 25) (p+q+5)}}$