'); parent.frames[0].document.write('
'); parent.frames[0].document.write('

The problems you missed are listed below with answers and explanations.  Look over the explanation for each problem so you can understand why you missed it!

'); parent.frames[0].document.write('
'); resetdisplay(); for(var i = 0; i <5; i++) { ganswer[i] = prompt(question[i],""); if (ganswer[i] == "quit" || ganswer[i] == "QUIT" || ganswer[i] == "Quit") { break; } q++; if (ganswer[i] == answer[i]) { c++; } else { rightans[i] = 0; w++; } resetdisplay(); } for(var i = 0; i < 10; i++) { if (rightans[i] == 0) { expl = expl + explain[i]; } } parent.frames[0].document.write(expl); parent.frames[0].document.write("
"); } question[0] = 'What is the 4 letter word that you should remember when multiplying binomials? (use uppercase letters)'; question[1] = 'When combining like terms, can you combine a term with x and term with x^2?'; question[2] = 'The difference of squares is a formula that we use for simplifying. If you take a difference of two squares problem is there a middle term like you usually have from a normal FOIL?'; question[3] = 'The Sum of Cubes is used for what?'; question[4] = 'What is the other cubes formual (list only the important word)'; answer[0] = 'FOIL'; answer[1] = 'no'; answer[2] = 'no'; answer[3] = 'factoring'; answer[4] = 'differences'; explain[0] = '#1

FOIL stands for: First Outside Inside Last. These are the terms that must be multiplied when dealing with binomials. If you ignore one of the multiplications, you will lose part of the answer.

'; explain[1] = '#2

x and x^2 are not combinable. If you combine the two then you would have 2x^2 which if factored out is x^2 + x^2 which is not what you started with. So variables with different exponents are not considered "like terms"

'; explain[2] = '#3

The difference of squares creates two center terms that cancel each other out, so there is no center term at all like you normally have.

'; explain[3] = '#4

The Sum of Cubes formula helps to factor cubed terms

'; explain[4] = '#5

The other formula (not sum of cubes) that helps with factoring cubes, is the difference of cubes

'; rightans[0] = 1; rightans[1] = 1; rightans[2] = 1; rightans[3] = 1; rightans[4] = 1; //-->