Math, asked by pallavideshmukh0610, 10 days ago

(y+2) (y-3) (y-7) (y-2) +64
Factorise​

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Answered by rathiputta
1

Answer:

={(y-2)(y-3)}{(y+2)(y-7)}+64

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=a

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20=a^2-10a+2a -20

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20=a^2-10a+2a -20=a(a-10)+2(a-10)

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20=a^2-10a+2a -20=a(a-10)+2(a-10)=(a-10)(a+2)

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20=a^2-10a+2a -20=a(a-10)+2(a-10)=(a-10)(a+2)putting the value of a

={(y-2)(y-3)}{(y+2)(y-7)}+64=(y^2-5y+6)(y^2-5y-14)+64let y^2-5y=athen,(a+6)(a-14)+64=a^2+6a-14a-84+64=a^2-8a-20=a^2-10a+2a -20=a(a-10)+2(a-10)=(a-10)(a+2)putting the value of a(y^2-5y-10)(y^2-5y+2)

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