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consider P(x) = ax^4 + bx^3 + cx^2 + dx + e, where a,b,c,d and e are integers.
Suppose w is an integer such that P(w) = 0.

(i) Prove that w divides e.

(ii) Prove that the polynomial Q(x) = 5x^4 + 2x^3 -3x^2 -x + 3 does not have an integer root.
 

Trebla

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consider P(x) = ax^4 + bx^3 + cx^2 + dx + e, where a,b,c,d and e are integers.
Suppose w is an integer such that P(w) = 0.

(i) Prove that w divides e.

(ii) Prove that the polynomial Q(x) = 5x^4 + 2x^3 -3x^2 -x + 3 does not have an integer root.
(i)
P(w) = 0
=> aw4 + bw3 + cw2 + dw + e = 0
=> e = - w(aw3 + bw2 + cw + d)
Since a, b, c , d and w are integers then e = w x integer => w divides e

(ii)
Suppose Q(x) does have an integer root w then from (i)
3 = - w(5w3 + 2w2 - 3w - 1)
This means that for integer roots to exist w must be a factor of 3. If this is true then w is either 1 or 3 (the only factors of 3). However Q(1) and Q(3) are non-zero therefore we have a contradication so the assumption is not true hence we conclude there is no integer root.
 
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(i)
P(w) = 0
=> aw4 + bw3 + cw2 + dw + e = 0
=> e = - w(aw3 + bw2 + cw + d)
Since a, b, c , d and w are integers then e = w x integer => w divides e

(ii)
Suppose Q(x) does have an integer root w then from (i)
3 = - w(5w3 + 2w2 - 3w - 1)
This means that for integer roots to exist w must be a factor of 3. If this is true then w is either 1 or 3 (the only factors of 3). However Q(1) and Q(3) are non-zero therefore we have a contradication so the assumption is not true hence we conclude there is no integer root.
For part (ii), could w also possibly be -1 or -3 (since they're also factors of 3)?
 

Trebla

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Yeah sorry, I forgot about them. Either way Q(-1) and Q(-3) should still be non-zero.
 

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