Question

simplify. x^2+4x-21 over x+7 times x-3 over x^2-5x+6

Answer 1

\[(x^2+4x-21)/(x+7)*(x-3)/(x^2-5x+6)\]
In general,
\[(x+a)(x+b)=x^2+ax+bx+ab = x^2 + (a+b)x + ab\]
We need to find a pair of numbers which when added gives us 4 and when multiplied gives us -21, and another pair where the sum is -5 and the product is 6. A little thought tells us that -3,7 is the first set, and -2,-3 is the second set. So we can rewrite our problem as
\[((x-3)(x+7))/(x+7)*(x-3)/((x-2)(x-3))\] and after we cancel out terms that appear on both the top and the bottom, we are left with
\[(x-3)/(x-2)\]