# What are the possible integral zeros of P(y)=y^4-5y^3-7y^2+21y+4?

Jan 11, 2017

The "possible" integral zeros are $\pm 1$, $\pm 2$, $\pm 4$

None of these work, so $P \left(y\right)$ has no integral zeros.

#### Explanation:

$P \left(y\right) = {y}^{4} - 5 {y}^{3} - 7 {y}^{2} + 21 y + 4$

By the rational root theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm 1 , \pm 2 , \pm 4$

Trying each of these, we find:

$P \left(1\right) = 1 - 5 - 7 + 21 + 4 = 14$

$P \left(- 1\right) = 1 + 5 - 7 - 21 + 4 = - 18$

$P \left(2\right) = 16 - 40 - 28 + 42 + 4 = - 6$

$P \left(- 2\right) = 16 + 40 - 28 - 42 + 4 = - 10$

$P \left(4\right) = 256 - 320 - 112 + 84 + 4 = - 88$

$P \left(- 4\right) = 256 + 320 - 112 - 84 + 4 = 384$

So $P \left(y\right)$ has no rational, let alone integer, zeros.