Question

How do you find the zeros of the polynomial function with equation $f(x) = {x^3} + 8{x^2} - x - 8$?

Answer 1

Hint: Zeros of a polynomial function are the values of $x$ for which $f(x) = 0$. For finding the zeros we substitute $f(x) = 0$ and then we solve the cubic equation $0 = {x^3} + 8{x^2} - x - 8$. The resulting values of $x$ will be the zeros of the given expression. For solving the cubic equation we first use a hit-and-trial method to find one value of $x$.

Complete step-by-step solution:
We have to find zeros of the polynomial function $f(x) = {x^3} + 8{x^2} - x - 8$.
For finding the zeros we substitute $f(x) = 0$. We get a cubic equation $0 = {x^3} + 8{x^2} - x - 8$.
Using hit-and-trial method we can find one value of $x$ satisfying the above equation.
For $x = 0$, we get RHS = $ - 8$
For $x = 1$, we get RHS = $0$ = LHS. Thus one value of $x$ is $1$.
By factor theorem, we can also say that $(x - 1)$ is a factor of the expression \[{x^3} + 8{x^2} - x - 8\].
We can divide \[({x^3} + 8{x^2} - x - 8)\] by $(x - 1)$. We get,
\[({x^3} + 8{x^2} - x - 8) = (x - 1)({x^2} + 9x - 8)\]
Thus we can write,
$
   \Rightarrow {x^3} + 8{x^2} - x - 8 = 0 \\
   \Rightarrow (x - 1)({x^2} + 9x - 8) = 0 \\
 $
We can further solve the quadratic expression in the above equation using factorization method.
$
 \Rightarrow {x^2} + 9x - 8 \\
 \Rightarrow {x^2} + x - 8x - 8 \\
 \Rightarrow x(x + 1) - 8(x + 1) \\
 \Rightarrow (x + 1)(x - 8) \\
 $
Thus, we can write,
$
 \Rightarrow (x - 1)({x^2} + 9x - 8) = 0 \\
  \Rightarrow (x - 1)(x + 1)(x - 8) = 0 \\
 $
From above equation we can write,
$x = 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} 8$
Thus, we get the values of $x$ as $1$, $ - 1$, or $8$.
Hence, the zeros of the given polynomial function are $1$, $ - 1$, or $8$. All the three values are the zeros of the given function as for any value of $x$ among these the given function will yield $f(x) = 0$.

Note: To find zeros of the function we substituted $f(x) = 0$. This we can also understand as the x-intercept of the graph of the given function, as $f(x) = 0$ means where the function is cutting the x-axis. Factor theorem states that $(x - a)$ is a factor of the polynomial function $f(x)$ if and only if $f(a) = 0$. Here $a$ takes all the values of zeros of the function.