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Factorise (25-10y)^3 + y^2 (25-10y)^2
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High School Teacher
`(25 -10y)^3 + y^2 (25-10y)^2`
The common factor here (present on both the left and right hand sides of the negative (-) symbol is: `(25 - 10y)^2` .
`therefore =(25 - 10y)^2 [ (25-10y) +y^2 (1)]`
Now consolidate the bracket :
`therefore = (25-10y)^2( 25 - 10y +y^2)`
Now factorize the second bracket. Note that we need the factors of the first (`25 times 1 or 5 times 5)` and the factors of the third(`y times y)` which when combined must render the middle term. Thus
`5 times y and 5 times y ` give us `10y` which we need to be negative thus
`5 times - y and 5 times -y` when added together equal -10y
`therefore = (25 -10y)^2 ( 5-y)^2`
Posted by durbanville on April 23, 2013 at 2:48 PM (Answer #1)
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