What are the possible integral zeros of P(y)=y4−5y3−7y2+21y+4?

1 Answer
George C.
Jan 11, 2017

The "possible" integral zeros are ±1, ±2, ±4

None of these work, so P(y) has no integral zeros.

Explanation:

P(y)=y4−5y3−7y2+21y+4

By the rational root theorem, any rational zeros of P(x) are expressible in the form pq for integers p,q with p a divisor of the constant term 4 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are the possible integer zeros:

±1,±2,±4

Trying each of these, we find:

P(1)=1−5−7+21+4=14

P(−1)=1+5−7−21+4=−18

P(2)=16−40−28+42+4=−6

P(−2)=16+40−28−42+4=−10

P(4)=256−320−112+84+4=−88

P(−4)=256+320−112−84+4=384

So P(y) has no rational, let alone integer, zeros.

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