Question 64454
Factor by grouping:
77)
{{{x^2+2xy+3x+6y}}} Group as shown by the parentheses.
{{{(x^2+2xy) + (3x+6y)}}} Factor an x from the 1st group and a 3 from the 2nd group.
{{{x(x+2y) + 3(x+2y)}}} Now factor (x+2y)
{{{(x+2y)(x+3)}}} ...Done!

79)
{{{x^2y-x^2-3y+3}}} Group as shown by the parentheses.
{{{(x^2y-x^2) - (3y-3)}}} Factor {{{x^2}}} from the 1st group and 3 from the 2nd group.
{{{x^2(y-1) - 3(y-1)}}} Now factor (y-1).
{{{(y-1)(x^2-3)}}} Normally this would be sufficient, but you could further factor the 2nd factor {{{(x^2-3) =(x+sqrt(3))(x-sqrt(3))}}} to get:
{{{(y-1)(x+sqrt(3))(x-sqrt(3))}}}

81)
{{{x^4-2x^3-x^2+x-1}}} Group as shown:
{{{(x^4-2x^3+x^2) + (x-1)}}} Factor {{{x^2}}} from the 1st group.
{{{x^2(x^2-2x+1) + (x-1)}}} Now factor the 1st group in parentheses.
{{{x^2(x-1)(x-1) + (x-1)}}} Now factor (x-1)
{{{x^2(x-1)((x-1) + 1)}}} Simplify.
{{{x^2(x-1)(x)}}}
{{{x^3(x-1)}}}