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

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

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Answer 1 · 2016-06-15T06:25:34

If the domain is real numbers, zeros are $x = 1$ $x=1$ and $x = \frac{4}{3}$ $x=4/3$ . If domain is extended to complex numbers, we have two additional zeros $x = i \sqrt{3}$ $x=isqrt3$ and $x = - i \sqrt{3}$ $x=-isqrt3$ .

Explanation:

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

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

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

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

However, if domain is extended to complex numbers, $x^{2} + 3 = 0$ $x^2+3=0$ gives us two more zeros given by $x = i \sqrt{3}$ $x=isqrt3$ and $x = - i \sqrt{3}$ $x=-isqrt3$ .

Answer 2 · 2016-06-15T06:27:36

The zeros, which correspond to the factors are:

$x = 1$ $x = 1$ , $x = \frac{4}{3}$ $x = 4/3$ , $x = \pm \sqrt{3} i$ $x = +-sqrt(3)i$

Explanation:

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

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

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

In symbols we could write:

$f (x) = 0 \Leftrightarrow ((x - 1) = 0 \lor (3 x - 4) = 0 \lor (x^{2} + 3) = 0)$ $f(x) = 0 <=> ((x-1) = 0 vv (3x-4) = 0 vv (x^2+3) = 0)$

Looking at each factor in turn:

$(x - 1) = 0$ $(x-1) = 0$ if and only if $x = 1$ $x = 1$

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

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

$(x - a) = 0$ $(x-a) = 0$ if and only if $x = a$ $x = a$

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

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

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

If we look at Complex numbers, we find:

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

So:

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

We now have linear factors, corresponding to zeros:

$x = \sqrt{3} i$ $x = sqrt(3)i" "$ and $x = - \sqrt{3} i$ $" "x = -sqrt(3)i$

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

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