# How do you find all the zeros of f(x)= (x-1)(3x-4)(x^2+3)?

Jun 15, 2016

If the domain is real numbers, zeros are $x = 1$ and $x = \frac{4}{3}$. If domain is extended to complex numbers, we have two additional zeros $x = i \sqrt{3}$ and $x = - i \sqrt{3}$.

#### Explanation:

Zeros of a polynomial are those values of variable, for which value of the polynomial becomes zero.

As the given polynomial f(x)=(x-1)(3x-4)(x^2+3_ is already written as product of binomials, $f \left(x\right)$ will be zero if

$x - 1 = 0$ or $3 x - 4 = 0$ or ${x}^{2} + 3 = 0$.

If the domain for $x$ is real numbers, the two zeros are given by first two binomials i.e. $x = 1$ and $x = \frac{4}{3}$ (as in such a case ${x}^{2} + 3$ will always be non-zero.

However, if domain is extended to complex numbers, ${x}^{2} + 3 = 0$ gives us two more zeros given by $x = i \sqrt{3}$ and $x = - i \sqrt{3}$.

Jun 15, 2016

The zeros, which correspond to the factors are:

$x = 1$, $x = \frac{4}{3}$, $x = \pm \sqrt{3} i$

#### Explanation:

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

Note that if any one of the factors $\left(x - 1\right)$, $\left(3 x - 4\right)$ or $\left({x}^{2} + 3\right)$ is zero, then their product $f \left(x\right)$ is zero.

If all of the factors are non-zero, then their product $f \left(x\right)$ is also non-zero.

In symbols we could write:

$f \left(x\right) = 0 \iff \left(\left(x - 1\right) = 0 \vee \left(3 x - 4\right) = 0 \vee \left({x}^{2} + 3\right) = 0\right)$

Looking at each factor in turn:

$\left(x - 1\right) = 0$ if and only if $x = 1$

$\left(3 x - 4\right) = 0$ if and only if $x = \frac{4}{3}$

In general, note that each linear factor corresponds to a zero:

$\left(x - a\right) = 0$ if and only if $x = a$

$\left(q x - p\right) = 0$ if and only if $x = \frac{p}{q}$

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The last factor $\left({x}^{2} + 3\right)$ is slightly more complicated.

First note that it has no Real zeros, since for any Real value of $x$ we have ${x}^{2} \ge 0$, so $\left({x}^{2} + 3\right) \ge 3$.

If we look at Complex numbers, we find:

${\left(\sqrt{3} i\right)}^{2} = {\left(\sqrt{3}\right)}^{2} \cdot {i}^{2} = 3 \cdot \left(- 1\right) = - 3$

So:

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

We now have linear factors, corresponding to zeros:

$x = \sqrt{3} i \text{ }$ and $\text{ } x = - \sqrt{3} i$