When factorising a quadratic that has a coefficient #>1#, we always multiply the coefficient by the constant given at the end of the quadratic.
#therefore# #2y^2+29y+90 -> 2y^2+29y+180#
When factorising, we always want to find two numbers that add to make the second number #(29)# and multiply to make #(180)#. The best way to do this is to find the factors of the constant at the end of the quadratic #(180)# and see if they can add to make the middle number #(29)#
As this quadratic has a #+# and another #+# this means both numbers will be positive.
#180:#
#180xx1# cannot make to #29#...
#90xx2# cannot make #29#...
#60xx3# cannot make #29#...
#45xx4# cannot make #29#...
#36xx5# cannot make #29#...
#30xx6# cannot make #29#...
#20xx9# CAN make #29#... so therefore we use these two numbers.
Removing the #29# and plugging the #20# and #9# into the quadratic:
#2x^2+20x+9x+90#
Always remember to change the constant back to the original...
Split up into two parts:
#2x^2+20x# and #9x+90#
#2x^2+20x -> 2x(x+10)# taking out a factor of #2x#
#9x+90 -> 9(x+10)# taking out a factor of #9#
This leaves us with:
#2x(x+10)+9(x+10)# <- Notice that the brackets are the same, this is one of our solutions. The brackets should ALWAYS be the same...
#2xcancel((x+10))+9cancel((x+10))#
This leaves us with our other bracket of #(2x+9)#
#therefore# the final answer is #(x+10)(2x+9)#
We can always check by expanding out the brackets:
#2x^2+9x+20x+90 -> 2x^2+29x+90#
#therefore# #(x+10)(2x+9)# is correct...
