Shon...write out the expressions right on top of each other. then factor each trinomial into binomials: x^2 + 3x + 2 should have factors that look something like this....( + )( + ) / ( + )( + ) then cancel like-terms. here's an example: try to solve your's...if you're still having difficulty...let me know...I don't want to answer the problems for you...but will show you how to solve it... here is a similar problem x^2 + 4x + 4 / x^2 + 7x + 14 ( x + 2 )( x + 2 )/( x + 2 )( x + 7 ) the numerator is ( x + 2 )( x + 2 ) and the denominator is ( x + 2 )( x + 7 ) since there is ( x + 2 )in the numerator and ( x + 2 )in the denominator they cancel out (because of the Property of One) so the factored answer would be ( x + 2 )/( x + 7 ) try it with your values...learn how to do this and your'll be the MASTER!

Remember that when multiplying (AX + B)(CX +D) you get (AC)X^2 + (AD + CB)X + BD

• So I look for the factors of the first and third terms then try to combine them to get the coefficient of the second term. (Hint: Prime Factors are very useful here)
• Factors of +1 are (1 * 1) or (-1 * -1)
• Factors of +2 are (2 * 1) or (-2 * -1)
• Factors of -4 are (-2 * 2) or (-4 * 1) or (4 * -1)
• So in the numerator (X^2+3x+2) we can see that 2+1 would equal 3
  
• and in the denominator (X^2-3x-4) we can see that -4 + 1 = -3
• Therefore
  

•  (x^2+3x+2)/(x^2-3x-4)
• (x+2)(x+1) / (x-4)(x+1)   Factoring
• (x+2) / (x-4)                   Canceling (x+1) from numerator and denomanator

Factor (x+1)(x+2)(x-4)(x+1)