+0  
 
-1
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2
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The polynomial$$f(x) = x^3+10x^2+27x+10$$has one integer root. What is it?

 

my go at it, I use the integer root theorem thing and get that p/q as q=+-1, and p=+-1, +-2, +-5, +-10, now I find all possible values so

p/q=+-1, +-1/2, +-1/5, +-1/10, i plug in one of them and should get 0. I am right?

 Dec 10, 2020
 #1
avatar+117546 
+1

1x^3  + 10x^2  + 27x  + 10

q                                     p

 

It's clear that the  root  must  be  negative

 

You have it a little confused

 

Remember  that p/q  =  all  the factors of 10/ all the  factors of  1

 

Possibilities

-1, -2. -5. -10

 

It can't be  -1

 

-2  [  1    10       27       10 ]

               -2      -16      -22

       __________________

       1        8      11       -12

 

 

-5  [ 1   10     27      10  ]

              -5    -25    -10

    __________________

       1       5    2          0

 

 

-5 is the integer  root

 

cool cool cool

 Dec 10, 2020
edited by CPhill  Dec 10, 2020
 #2
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+1

Thank you bro!

 Dec 10, 2020

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