<PRE><B>Question 89355</B>
In order to factor {{{5*y^2+7*y-6}}}, first multiply 5 and -6 to get -30 and we need to ask ourselves: What two numbers multiply to -30 and add to 7? Lets find out by listing all of the possible factors of -30


Factors:

1,2,3,5,6,10,15,30,

-1,-2,-3,-5,-6,-10,-15,-30, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -30.

(-1)*(30)=-30

(-2)*(15)=-30

(-3)*(10)=-30

(-5)*(6)=-30

 Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7

&lt;TABLE&gt;&lt;TR&gt;&lt;TD&gt;First Number&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;Second Number&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;Sum&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;-30&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1+(-30)=-29&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;-15&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;2+(-15)=-13&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;-10&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;3+(-10)=-7&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;5&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;5+(-6)=-1&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;-1&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;30&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;(-1)+30=29&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;-2&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;15&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;(-2)+15=13&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;-3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;10&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;(-3)+10=7&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;-5&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;6&lt;/TD&gt;|&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;(-5)+6=1&lt;/TD&gt;&lt;/TR&gt;&lt;TABLE&gt;We can see from the table that -3 and 10 add to 7. So the two numbers that multiply to -30 and add to 7 are: -3 and 10

 So the original quadratic


{{{5*y^2+7*y-6}}}


breaks down to this (just replace {{{7*y}}} with the two numbers that multiply to -30 and add to 7, which are: -3 and 10)


 {{{5*y^2-3y+10y-6}}}

Group the first two terms together and the last two terms together like this:

{{{(5*y^2-3y)+(10y-6)}}}

Factor a y out of the first group and factor a 2 out of the second group.


{{{y(5y-3)+2(5y-3)}}}


Now since we have a common term {{{5y-3}}} we can combine the two terms. 


{{{(y+2)(5y-3)}}} Combine like terms. 

Answer:

So the quadratic {{{5*y^2+7*y-6}}} factors to


{{{(y+2)(5y-3)}}}



Notice how {{{(y+2)(5y-3)}}} foils back to our original problem {{{5*y^2+7*y-6}}}. This verifies our answer.</PRE>