SOLUTION: FACTOR BY GROUPING X^3 + 8X^2+6X +48 A)(X+8)(X^2-6) B)(X+8X)(X^2+6) C)(X+8)(X^2+6) D)(X-8)(X^3+6) SO I KNOW IT IS EASIER TO MAKE THEM TWO PROBLEMS BUT I DONT KNOW WHY I CA

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: FACTOR BY GROUPING X^3 + 8X^2+6X +48 A)(X+8)(X^2-6) B)(X+8X)(X^2+6) C)(X+8)(X^2+6) D)(X-8)(X^3+6) SO I KNOW IT IS EASIER TO MAKE THEM TWO PROBLEMS BUT I DONT KNOW WHY I CA      Log On


   

Question 335981: FACTOR BY GROUPING
X^3 + 8X^2+6X +48
A)(X+8)(X^2-6)
B)(X+8X)(X^2+6)
C)(X+8)(X^2+6)
D)(X-8)(X^3+6)
SO I KNOW IT IS EASIER TO MAKE THEM TWO PROBLEMS BUT I DONT KNOW WHY I CANT GET THE RIGHT ANSWER PS. WHEN I DO THE ^ SYMBOL IT TO THE POWER OF NOT TIMES...
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3%2B8x%5E2%2B6x%2B48 Start with the given expression


%28x%5E3%2B8x%5E2%29%2B%286x%2B48%29 Group like terms


x%5E2%28x%2B8%29%2B6%28x%2B8%29 Factor out the GCF x%5E2 out of the first group. Factor out the GCF 6 out of the second group


%28x%5E2%2B6%29%28x%2B8%29 Since we have the common term x%2B8, we can combine like terms


So x%5E3%2B8x%5E2%2B6x%2B48 factors to %28x%5E2%2B6%29%28x%2B8%29


In other words, x%5E3%2B8x%5E2%2B6x%2B48=%28x%5E2%2B6%29%28x%2B8%29


Since the order of the factors does not matter, this means that x%5E3%2B8x%5E2%2B6x%2B48=%28x%2B8%29%28x%5E2%2B6%29 as well


So the answer is C


If you need more help, email me at jim_thompson5910@hotmail.com

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Jim