+160 a is a positive integer
equation: x^3 + (a+17)x^2 + (38-a)x - 56 = 0
The equation has 3 integer roots
what is the value of a?
what are the three roots?
hint: x=1 is one of the roots
+129849 If 1 is a root...let's perform some synthetic division
1 [ 1 a + 17 38 - a - 56 ]
1 a + 18 56
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1 a + 18 56 0
The remaining polynomial is
x^2 + (a + 18)x + 56
Factors of 56 are 1 2 4 7 8 14 28 56
Since a is positive...only the pairs 2, 28 and 1, 56 will give us what we want
So...the possibiliites for the factorization of this are
(x +28) ( x + 2) = x^2 + 30x + 56 so a = 12 or
( x + 56) ( x + 1) = x^2 + 57x + 56 so a = 39
Checking this
x^3 + 29x^2 + 26x - 56 = 0 produces the three roots of 1, -28 and -2
And
x^3 + 56x^2 -1x - 56 = 0 produces the three roots 1, -1, - 56
