SOLUTION: Hi, I know I know this, but I am drawing a blank, and could use some help please. Factor completely. 9x^5-45x^4+27x^3. Thank you very much.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi, I know I know this, but I am drawing a blank, and could use some help please. Factor completely. 9x^5-45x^4+27x^3. Thank you very much.      Log On


   



Question 635633: Hi, I know I know this, but I am drawing a blank, and could use some help please.
Factor completely. 9x^5-45x^4+27x^3.
Thank you very much.

Found 2 solutions by jim_thompson5910, Kasandra24:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

9x%5E5-45x%5E4%2B27x%5E3 Start with the given expression.


9x%5E3%28x%5E2-5x%2B3%29 Factor out the GCF 9x%5E3.


Now let's try to factor the inner expression x%5E2-5x%2B3


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Looking at the expression x%5E2-5x%2B3, we can see that the first coefficient is 1, the second coefficient is -5, and the last term is 3.


Now multiply the first coefficient 1 by the last term 3 to get %281%29%283%29=3.


Now the question is: what two whole numbers multiply to 3 (the previous product) and add to the second coefficient -5?


To find these two numbers, we need to list all of the factors of 3 (the previous product).


Factors of 3:
1,3
-1,-3


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to 3.
1*3 = 3
(-1)*(-3) = 3

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -5:


First NumberSecond NumberSum
131+3=4
-1-3-1+(-3)=-4



From the table, we can see that there are no pairs of numbers which add to -5. So x%5E2-5x%2B3 cannot be factored.


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Answer:


So 9x%5E5-45x%5E4%2B27x%5E3 simply factors to 9x%5E3%28x%5E2-5x%2B3%29


In other words, 9x%5E5-45x%5E4%2B27x%5E3=9x%5E3%28x%5E2-5x%2B3%29.

Answer by
Kasandra24(4) About Me  (Show Source):
You can put this solution on YOUR website!
9x^5-45x^4+27x^3.
= 9x^3(x^2-5x+3)