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# How do you find the zeros of ${x^3} - 3{x^2} + 6x - 18$ ?

Last updated date: 19th Jul 2024
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Hint: The given equation is a polynomial having degree three in one variable. In this type of problem, we substitute the different values of $x$ which satisfies the given polynomial. If the value gets satisfied then it will be the factor of the polynomial. Similarly we do until the factorization of polynomials is completely done. Write the different terms of the polynomial into brackets. One bracket should have only those terms of the polynomial which gives the common factor out of the brackets. Another method of solving this type of problem is to find the common factor of the polynomial. The common factor will also be the factor of the given polynomial and then we will apply the basic properties of simplification of equations.one can also simplify the given polynomial by substituting the different small values of $x$ in the polynomial. Here in the given polynomial first we will find a common factor in the equation.

Complete step by step solution:
Step: 1 the given equation of polynomial is,
${x^3} - 3{x^2} + 6x - 18$
So write the terms of the polynomial into brackets, containing each bracket two terms of the polynomial.
$\Rightarrow {x^3} - 3{x^2} + 6x - 18 = \left( {{x^3} - 3{x^2}} \right) + \left( {6x - 21} \right)$
Now take common from each of the brackets to simplify the given equation of polynomial.
In the first bracket ${x^2}$ will be the common, and in the second bracket $6$ will be the common.
$\Rightarrow \left( {{x^3} - 3{x^2}} \right) + \left( {6x - 21} \right) = {x^2}\left( {x - 3} \right) + 6\left( {x - 3} \right)$
So in all the brackets $\left( {x - 3} \right)$ is common in the equation, so we will take it as a common from the given equation.
$\Rightarrow {x^2}\left( {x - 3} \right) + 6\left( {x - 3} \right) = \left( {x - 3} \right)\left( {{x^2} + 6} \right)$

Therefore the factor of the given polynomial is $\left( {x + 3} \right)\left( {{x^2} + 6} \right)$.