Question

Help me FOIL 5x^2+6x+1

Answer 1

\[5x^{2}+6x+1\] multiply the leading coefiicient with the constant at the end.\[\color{red}{5}x^{2}+6x+\color{red}{1}\]\[x^2+6x+5\] find two numbers that multiply to give 5, and add to give 6. \[(x+5)(x+1)\] Now you have to divide again by the leading coefficient because you multiplied by it the first time.\[\frac{(x+5)}{\color{red}{5}}*\frac{(x+1)}{\color{red}{5}}\] you will see in one of these that the 5's can cancel out with each other by dividing. \[(x+\frac{5}{5})*(x+\frac{1}{5})\] because (1/5) cannot cancel out, you will multiply the "5" to the "x" to recreate a simplified form. \[(x+1)(5x+1)\] that is your factored form.

Answer 2

Does it make sense?

Answer 3

Yes, I needed to explain this to my nephew, and this is great ^^

Answer 4

Wonderful!!