# How many zeroes does f(x)=(6x^7+4x^2+6)(x^4+5x-6) have?

Feb 19, 2016

$11$ counting multiplicity. $3$ are Real and $8$ non-Real Complex.

#### Explanation:

By the fundamental theorem of algebra, a non-constant polynomial in one variable has a zero in $\mathbb{C}$.

A simple corollary of this is that a polynomial of degree $n > 0$ has exactly $n$ zeros in $\mathbb{C}$ counting multiplicity. This follows since if a polynomial of degree $n$ has one zero $r \in \mathbb{C}$ then it will be divisible by $\left(x - r\right)$ resulting in a polynomial of degree $n - 1$. If $n - 1 > 0$ then we can use the fundamental theorem of algebra to deduce that this simpler polynomial has a root in $\mathbb{C}$ too, etc.

Some of the roots may coincide, but there will be $n$ roots counting multiplicity.

$f \left(x\right)$ is a non-constant polynomial of degree $7 + 4 = 11$, so it has $11$ zeros counting multiplicity.

Let us see what else we can find out about the zeros of this polynomial:

Since all of the coefficients are Real, any non-Real Complex zeros will occur in Complex conjugate pairs.

The septic factor is of odd degree, so will have at least $1$ Real zero, possibly $3$, $5$ or $7$. Since its coefficients have no changes of sign, none of the zeros are positive. If we reverse the sign of the terms of odd degree then there is only one change of sign, so a maximum of $1$ negative zero. So the septic has $1$ Real negative zero and $6$ non-Real Complex zeros (occuring as $3$ Complex conjugate pairs).

The quartic factor has an obvious zero in $x = 1$ (since the sum of its coefficients is zero), so a factor $\left(x - 1\right)$ resulting in a cubic:

${x}^{4} + 5 x - 6 = \left(x - 1\right) \left({x}^{3} + {x}^{2} + x + 6\right)$

By the rational root theorem, any rational zeros of this cubic will be factors of $6$, and since all of the coefficients are positive they will be negative factors of $6$. So the possible rational zeros are:

$- 1 , - 2 , - 3 , - 6$

We find $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} + {x}^{2} + x + 6 = \left(x + 2\right) \left({x}^{2} - x + 3\right)$

The remaining quadratic factor has negative discriminant, so a pair of Complex conjugate zeros given by the quadratic formula:

$x = \frac{1 \pm \sqrt{11} i}{2}$