When factorising a quadratic that has a coefficient >1>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...