Posted by Kent on Thursday, June 18, 2009 at 10:47am.
The first is a simple case of "common factoring"
5a^2b+ab^2
= ab(5a+b)
these can be quickly verified by merely expanding your answers.
the second is a "difference of squares"
9k^2-1
= (3k+1)(3k-1)
the third is a standard "quadratic trinomial factoring"
the most popular method these days seems to be "decomposition"
2y^2-5y+2
multiply the coefficients of the first and last terms ... 2x2 = 4
now find two numbers so that when multiplied will give you 4, and when added will give you the coefficient of the middle term, namely -5
those 2 numbers clearly are -4 and -1
so rewrite the middle term of -5y as
-4y -y , thus the name "decomposition"
= 2y^2 - 4y - y + 2
now factor by finding the highest common factor of the first two terms
= 2y(y - 2) - y + 2
If you did everything correctly, that common factor of y-2 MUST also appear at the end, and surely ....
= 2y(y-2) - (y-2)
and a common factor once more ....
= (y-2)(2y-1)
again you can always check your answer by expanding it, you must get the original question.