# What are the possible number of positive real, negative real, and complex zeros of f(x) = 4x^3 + x^2 + 10x – 14?

May 26, 2016

This cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

#### Explanation:

$f \left(x\right) = 4 {x}^{3} + {x}^{2} + 10 x - 14$

The signs of the coefficients follow the pattern: $+ + + -$

With one change of sign, we can tell that this cubic polynomial has one positive zero.

Consider $f \left(- x\right)$...

$f \left(- x\right) = - 4 {x}^{3} + {x}^{2} - 10 x - 14$

The signs of the coefficients follow the pattern: $- + - -$

With two changes of sign, we can tell that this cubic has $0$ or $2$ negative zeros.

Let us examine the discriminant:

$4 {x}^{3} + {x}^{2} + 10 x - 14$ is in the form $a {x}^{3} + b {x}^{2} + c x + d$ with $a = 4$, $b = 1$, $c = 10$ and $d = - 14$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

$= 100 - 16000 + 56 - 84672 - 10080$

$= - 110596$

Since $\Delta < 0$ we can deduce that $f \left(x\right)$ has one Real zero and a pair of Complex conjugate zeros.

graph{4x^3+x^2+10x-14 [-10, 10, -200, 200]}