Question 295782
It makes it a little more challenging but the method is the same. 
{{{3y^2+7y-6=(3y+a)(y+b)}}}
If you FOIL the right hand side you get,
{{{(3y+a)(y+b)=3y^2+3by+ay+ab}}}
{{{(3y+a)(y+b)=3y^2+(3b+a)y+ab}}}
So again look for factor of -6:
(1,-6)
(-1,6)
(2,-3)
(-2,3)
and try to solve for the y-coefficient until you hit the right one.
{{{a=1}}}, {{{b=-6}}}
{{{3b+a=7}}}
{{{3(-6)+1=7}}}
{{{-18+1=7}}}
{{{-17=7}}}
No
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{{{a=-6}}}, {{{b=1}}}
{{{3b+a=7}}}
{{{3-6=7}}}
{{{-3=7}}}
No
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{{{a=-1}}},{{{b=6}}} and {{{a=6}}},{{{b=-1}}} will give similar results.
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{{{ a=2}}}, {{{b=-3}}}
{{{3b+a=7}}}
{{{3(-3)+2=7}}}
{{{-9+2=7}}}
{{{-7=7}}}
No, but we only need to change the signs of a and b for this combination to work. 
{{{a=-2}}}, {{{b=3}}}.
{{{ 3y^2+7y-6=(3y-2)(y+3) }}}