1.

Factorise `99x^(2)-202xy+99y^(2)`.

Answer» Here, a=99, b=-202, c=99
`therefore " " axx c=99xx99`
Now, we take two factors of `99xx99` whose sum is -202.
Let two such factors are m and n.
`therefore " " m+n=-202`
and `mn=99xx99`
V know that `(m-n)^(2)=(m+n)^(2)-4mn`
`=(-202)^(2)-4xx99xx99=40804-39204=1600`
`implies m-n=40`
Adding equations (1) and (2)
2m=-162
`implies m=-81`
Put this value in equation (1)
`-81+n=-202`
`implies n=-121`
The two factors of `99xx99` are -121 and -81.
Now, `99x^(2)-202xy+99y^(2)=99x^(2)+(-121-81)-xy+99y^(2)=99x^(2)-121 xy-81xy+99y^(2)`
`=11x(9x-11y)-9y(9x-11y)(11x-9y)`


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