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)
+128406
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
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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
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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
