
Kenneth G. answered 01/27/14
Tutor
New to Wyzant
Experienced Tutor of Mathematics and Statistics
Let's try the simplest approach here first. Let's suppose that P(x) has a first-degree factor with integer coefficients.
The possible degree-1 factors of P(x) are (x±1), (x±2), and (x±4) . Testing division by each of these factors, we find that (x-4) divides evenly into P(x), giving P(x) = (x - 4)(x2 + 2x - 1).
By the quadratic formula, the xeros of x2 + 2x - 1 are -1 ± √2
So the three zeros are x = 4, x = -1 + √2, and x = -1 - √2
And the ordering is -1 - √2 < -1 + √2 < 4
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Second solution:
This solution is basically the same, except instead of dividing polynomials we just substitute the possible values of x.
Let's start with the sentence "The possible degree-1 factors of P(x) are (x±1), (x±2), and (x±4)." If x-1 is a factor, then x = 1 is a zero. So we could just substitute ±1, ±2, and ±4 into P(x) and we would find that P(4) = 0 which means (x-4) is a factor. Following this path would mean we would not have to do the other 5 division tests and only divide by x-4.
So P(x) = (x - 4)(x2 + 2x - 1) as before.