Question

How do you find all zeros of $f(x) = {x^3} + 5{x^2} - 4x - 20$?

Answer 1

Hint: To find the eventual zeros of a function it is necessary to solve the equation system between the equation of the function and the equation of the $X - $ axis ($y = 0$).
The given equation is $f(x) = {x^3} + 5{x^2} - 4x - 20$
The zeros of $f(x)$ are the values of $x$ that make $f(x) = 0$
Factoring the common terms and group the terms in pairs. After that we simplify the function.
Finally we get the $x$ values.

Complete step-by-step solution:
The given equation is $f(x) = {x^3} + 5{x^2} - 4x - 20$
The zeros of $f(x)$ are the values of $x$ that make $f(x) = 0$
Now we solve: ${x^3} + 5{x^2} - 4x - 20 = 0$
Begin by factoring the left side. Group the terms in pairs
$[{x^3} + 5{x^2}] + [ - 4x - 20]$ And factoring each pair gives
We take the common term in the first term, hence we get
$\Rightarrow$${x^2}(x + 5) + [ - 4x - 20] = 0$
We take the common term in the second term, hence we get
$\Rightarrow$${x^2}(x + 5) - 4(x + 5) = 0$
Now we take common term $(x + 5)$ in the two terms, hence we get
$\Rightarrow$$(x + 5)({x^2} - 4) = 0$
Factorize the two terms, hence we get
$\Rightarrow$$x + 5 = 0$ or ${x^2} - 4 = 0$
We rewrite ${x^2} - 4 = 0$ is $(x - 2)(x + 2)$
To Solve: $x + 5 = 0$ or ${x^2} - 4 = 0$
$\Rightarrow$$(x + 5)(x - 2)(x + 2) = 0$
$\Rightarrow$$x = - 5$,$x = \pm 2$

The zeros are $x = - 5$,$x = \pm 2$

Note: A polynomial of degree $n$ will have $n$ roots, some of which may be multiple roots.
 The fundamental theorem of algebra tells you that the polynomial has at least one root.
 The Factor Theorem tells you that if $r$ is a root then$(x - r)$ is a factor. But if you divide a polynomial of degree $n$ by a factor $(x - r)$, whose degree is $1$, you get a polynomial of degree $n - 1$.
Repeatedly applying the fundamental theorem and factor theorem gives you $n$ factors. When a given factor $(x - r)$ occurs $m$ times in a polynomial, $r$ is called a multiple roots or a root of multiplicity $m$