polynomials (1 Viewer)

[ashokkumar]

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]

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.

 

[ashokkumar]

(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]

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