+0

# Help me guys!!!

0
71
2
+4

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

BryanPual  Aug 9, 2017
Sort:

#1
0

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
+76222
+1

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

CPhill  Aug 9, 2017

### 28 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details