How do you factor #5y^2-7y-6#?

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Answer 1 · 2018-03-27T23:33:02

#(5y+3)(y-2)#

I made the assumption that all coefficients and factors were integers. Based on this, I started building a form of the factored equation:

#(ay+c)(by+d)=aby^2+(ad+bc)y+dc#

The expanded form allows us to make some guesses, again assuming integers.

We know that #ab=5# which means a or b has to equal 1 or 5. Let's plug that in:

#(5y+c)(y+d)=5y^2+(5d+c)y +dc#

Next, we need to figure out what d and c are. We know that #dc=-6# which gives us two integer options. Either d&c are 2&3, or they are 1&6.

The trick here is to figure out the values of d/c using the middle term:

#5d+c=-7#

If #d=-2# and #b=3#, it satisfies the middle equation:

#5(-2)+3 =-7 rArr -10+3=-7#

Now we know all of our variables, and can write out the final factored equation:

#color(red)((5y+3)(y-2))#