+118690 28n^4+16n^3-80n^2
\(28n^4+16n^3-80n^2\\ =4n^2(7n^2+4n-20)\)
I could use the quadrative formula to solve it but I will see if I can figure it out another way.
I need 2 numbers that multiply to 7*-20 and add to give 4
One will need to be positive and the other negative and since 4 is positive, the bigger number will be positive.
7*-20 = -7*2*2*5 = 14* -10 that works, the numbers are 14 and -10
change 4n into 14n-10n
\(=4n^2(7n^2+4n-20)\\ =4n^2(7n^2+14n-10n-20)\\ \text{Now factor in pairs}\\ =4n^2[7n(n+2)-10(n+2)]\\ =4n^2(7n-10)(n+2)\\ \)
