Question

How do you find the zeros of \[f\left( x \right) = {x^3} + 10{x^2} - 13x - 22\] ?

Answer 1

Hint: To find the zeros of the polynomial function, Use the rational roots theorem to find possible roots, evaluate to find roots and next find out the factors of the equation by equating it to zero by this value of \[x\] is obtained and we find zeros at the value of \[x\].

Complete step-by-step solution:
The given polynomial equation is
 \[\Rightarrow f\left( x \right) = {x^3} + 10{x^2} - 13x - 22\],
Let us set the equation to zero i.e.,
\[\Rightarrow {x^3} + 10{x^2} - 13x - 22 = 0\] ……………………………. 1
We need to solve for x from equation 1.
\[\Rightarrow {x^3} + 10{x^2} - 13x - 22\]
If a polynomial function has integer coefficients, then any rational roots of \[f\left( x \right) = 0\]must be of the form \[\dfrac{p}{q}\] where p, q are integers, \[q \ne 0\], where p is a factor of the constant term 22 and q is a factor of the leading coefficient of the term of highest degree \[{x^3}\].
Factor \[{x^3} + 10{x^2} - 13x - 22\] using the Rationalize test.
So, the only possible rational roots are:
\[\Rightarrow p = \pm 1, \pm 2, \pm 11, \pm 22\]
\[\Rightarrow q = \pm 1\]
Substitute 2 and simplify the expression. In this case, the expression is equal to 0 so 2 is a root of the polynomial as
\[{x^3} + 10{x^2} - 13x - 22\]
\[\Rightarrow {2^3} + 10{\left( 2 \right)^2} - 13\left( 2 \right) - 22\]
Simplifying each term, we get
\[\Rightarrow {2^3} + 10\left( 4 \right) - 13\left( 2 \right) - 22\]
\[\Rightarrow 8 + 40 - 26 - 22\]
$\Rightarrow 0$
In this case, the expression is equal to 0 so 2 is a root of the polynomial, since 2 is a known root, divide the polynomial by \[x - 2\] to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
\[\Rightarrow \dfrac{{{x^3} + 10{x^2} - 13x - 22}}{{x - 2}}\]
Simplifying we get
\[\Rightarrow \left( {x - 2} \right)\left( {{x^2} + 12x + 11} \right) = 0\]
Now factorize \[\left( {{x^2} + 12x + 11} \right)\] using AC method we get
\[\Rightarrow \left( {x - 2} \right)\left( {\left( {x + 1} \right)\left( {x + 11} \right)} \right) = 0\]
\[\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right)\left( {x + 11} \right) = 0\] …………………………….. 2
The terms of equation 2 equating to zero as
\[\Rightarrow \left( {x - 2} \right) = 0\] …………………………………. 3
\[\Rightarrow \left( {x + 1} \right) = 0\] …………………………………. 4
\[\Rightarrow \left( {x + 11} \right) = 0\] ………………………………….. 5
Simplifying the equations 3, 4 and 5 to get the values of \[x\], as
\[\Rightarrow \left( {x - 2} \right) = 0\]
\[ \Rightarrow \]\[x = 2\]
\[\Rightarrow \left( {x + 1} \right) = 0\]
\[ \Rightarrow \]\[x = - 1\]
\[\Rightarrow \left( {x + 11} \right) = 0\]
\[ \Rightarrow \]\[x = - 11\]
Hence,
\[\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right)\left( {x + 11} \right) = 0\]
\[ \Rightarrow \]\[x = 2, - 1, - 11\]
Therefore, we find zeroes at \[x \in \left\{ {2, - 1, - 11} \right\}\].

Note: The key point to solve these equations is that all the linear polynomials have only one zero and the zeros of the polynomial depend on its degree. A polynomial of degree 2 is known as a quadratic polynomial and Standard form is \[a{x^2} + bx + c\], where a, b and c are real numbers and \[a \ne 0\]. Polynomial of degree 3 is known as a cubic polynomial and Standard form is \[a{x^3} + b{x^2} + cx + d\], where a, b, c and d are real numbers and \[a \ne 0\].