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December 22, 2014

December 22, 2014

Posted by **Kent** on Thursday, June 18, 2009 at 10:47am.

a)5a^2b+ab^2

b)9k^2-1

c)2y^2-5y+2

If i can get an idea how to do these and i work the rest i have to do myself, thanks

- math -
**Reiny**, Thursday, June 18, 2009 at 11:34amThe 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.

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