Factorize , \( {x}^{2}+2x-8-{y}^{2}-6y\)
Solution: we have \( 2x=2ab\) where \(a=x\) thus \(2x=2xb\) i.e \( b=1\)
so \({x}^{2}+2x+1-1={(x+1)}^{2}-1\) also we have \(-{y}^{2}-6y=-({y}^{2}+6y)\)
its clear that \(6y=2ab\) where \(a=y\) thus \(6y=2yb\) then \(b=3\)
so \(-({y}^{2}+6y) =-({y}^2+6y+9-9)={-(y+3)}^{2}+9\)
thus \( {x}^{2}+2x-8-{y}^{2}-6y = {(x+1)}^{2}-1-8-{(y+3)}^{2}+9 \)
\(={(x+1)}^{2}-{(y+3)}^{2} =(x+1-y-3)(x+1+y+3) = (x-y-2)(x+y+4) \)
Solution: we have \( 2x=2ab\) where \(a=x\) thus \(2x=2xb\) i.e \( b=1\)
so \({x}^{2}+2x+1-1={(x+1)}^{2}-1\) also we have \(-{y}^{2}-6y=-({y}^{2}+6y)\)
its clear that \(6y=2ab\) where \(a=y\) thus \(6y=2yb\) then \(b=3\)
so \(-({y}^{2}+6y) =-({y}^2+6y+9-9)={-(y+3)}^{2}+9\)
thus \( {x}^{2}+2x-8-{y}^{2}-6y = {(x+1)}^{2}-1-8-{(y+3)}^{2}+9 \)
\(={(x+1)}^{2}-{(y+3)}^{2} =(x+1-y-3)(x+1+y+3) = (x-y-2)(x+y+4) \)
nice
ReplyDeletethank you sir im so happy since i catch the key of typing
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