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x^4-x^3-19x^2+49x-30

BryanPual  Aug 9, 2017
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 #1
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Simplify the following:
x^4 - x^3 - 19 x^2 + 49 x - 30

The possible rational roots of x^4 - x^3 - 19 x^2 + 49 x - 30 are x = ± 1, x = ± 2, x = ± 3, x = ± 5, x = ± 6, x = ± 10, x = ± 15, x = ± 30. Of these, x = 1, x = 2, x = 3 and x = -5 are roots. This gives x - 1, x - 2, x - 3 and x + 5 as all factors:
Answer: | (x - 1) (x - 2) (x - 3) (x + 5)

Guest Aug 9, 2017
 #2
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I assume that you want to find the roots of this  ??

 

If so, we have that 

 

x^4-x^3-19x^2+49x-30 = 0

 

Note that if we can add the coefficients  and get 0, then 1 is a root

 

So  .....1 -1 -19 + 49 - 30  = 0

 

So  1  is a root

 

And we can perform some synthetic division to find a reduced polynomial factor of the original polynomial

 

 

1  [   1    -1      -19      49    -30  ]

               1         0     -19     30

       _____________________

        1    0       -19     30       0

 

So...a remaining polynomial  is

 

x^3  - 19x  + 30

 

And  testing some possible roots using the Rational Zeroes Theorem, we can note that 3  is a root

 

So.....performing synthetic division one more time we have

 

 

3   [  1     0     -19     30  ]

               3       9     -30

      ________________

       1      3      -10     0

 

So.....a remaining polynomial is

 

x^2 + 3x  - 10        which we can factor as   ( x + 5) ( x - 2)

 

And setting these factors to 0  and solving for x  gives us the other two roots of

 

x = -5   and x = 2

 

So......the roots of the original polynomial are    x =1, x = 2, x = 3  and  x = -5

 

 

cool cool cool

CPhill  Aug 9, 2017

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