SOLUTION: Factor completely. 7x^7-56x^6+28x^5 Thanks!

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Factor completely. 7x^7-56x^6+28x^5 Thanks!      Log On


   



Question 222684: Factor completely. 7x^7-56x^6+28x^5
Thanks!

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


Jump to Answer


7x%5E7-56x%5E6%2B28x%5E5 Start with the given expression.


7x%5E5%28x%5E2-8x%2B4%29 Factor out the GCF 7x%5E5.


Now let's try to factor the inner expression x%5E2-8x%2B4


---------------------------------------------------------------


Looking at the expression x%5E2-8x%2B4, we can see that the first coefficient is 1, the second coefficient is -8, and the last term is 4.


Now multiply the first coefficient 1 by the last term 4 to get %281%29%284%29=4.


Now the question is: what two whole numbers multiply to 4 (the previous product) and add to the second coefficient -8?


To find these two numbers, we need to list all of the factors of 4 (the previous product).


Factors of 4:
1,2,4
-1,-2,-4


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


These factors pair up and multiply to 4.
1*4 = 4
2*2 = 4
(-1)*(-4) = 4
(-2)*(-2) = 4

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -8:


First NumberSecond NumberSum
141+4=5
222+2=4
-1-4-1+(-4)=-5
-2-2-2+(-2)=-4



From the table, we can see that there are no pairs of numbers which add to -8. So x%5E2-8x%2B4 cannot be factored.


===============================================================



Answer:


So 7x%5E7-56x%5E6%2B28x%5E5 simply factors to 7x%5E5%28x%5E2-8x%2B4%29


In other words, 7x%5E7-56x%5E6%2B28x%5E5=7x%5E5%28x%5E2-8x%2B4%29.


Jump to Top