Answer
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Hint: In the given problem, we have to factorize the given polynomial equation and represent it as a product of its factors. The given polynomial is of degree $3$ and is thus called a cubic polynomial. For factoring the polynomial, first, we have to simplify the given polynomial and then find its factors. For finding the factors of a cubic polynomial, we first find out the root of the polynomial by the hit and trial method. Hence, we get a one-factor using hit and trial and then find the remaining two roots by dividing the given polynomial by the factor already found.
Complete step-by-step solution:
We find the value of $x=a$ for which the function ${x^3} - 2{x^2} + 3$.
We take $x=a=-1$.
We can see \[f\left( -1 \right)={{\left( -1 \right)}^{3}}-2{{\left( -1 \right)}^{2}}+3=-1-2+3=0\].
So, the root of the ${x^3} - 2{x^2} + 3$ will be the function $\left( x+1 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x+1 \right)$ is a factor of the polynomial ${x^3} - 2{x^2} + 3$.
We can now divide the polynomial ${x^3} - 2{x^2} + 3$ by $\left( x+1 \right)$.
\[x+1\overset{{{x}^{2}}-3x+3}{\overline{\left){\begin{align}
& {{x}^{3}}-2{{x}^{2}}+3 \\
& \underline{{{x}^{3}}+{{x}^{2}}} \\
& -3{{x}^{2}}+3 \\
& \underline{-3{{x}^{2}}-3x} \\
& 3x+3 \\
& \underline{3x+3} \\
& 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with ${{x}^{2}}$. We get \[{{x}^{3}}+{{x}^{2}}\]. We subtract it to get \[-3{{x}^{2}}+3\]. We again equate with the highest power of the remaining terms. We multiply with $-3x$ and subtract to get \[-3{{x}^{2}}-3x\]. In the end, we had to multiply with 3 to complete the division. The quotient is \[{{x}^{2}}-3x+3\].
We cannot further factor \[{{x}^{2}}+5x+6\].
The final factorization of ${{x}^{3}}-2{{x}^{2}}+3=\left( x+1 \right)\left( {{x}^{2}}-3x+3 \right)$.
Note: Such polynomials can be factored by using the hit and trial method and then using factor theorem to find the factors of the polynomials. Cubic polynomials are polynomials with the degree of variable as $3$. The imaginary roots always occur in pairs. So, either there can be no imaginary root or two imaginary roots of a cubic polynomial. Conversely, there are either three or one root real root of a cubic polynomial.
Complete step-by-step solution:
We find the value of $x=a$ for which the function ${x^3} - 2{x^2} + 3$.
We take $x=a=-1$.
We can see \[f\left( -1 \right)={{\left( -1 \right)}^{3}}-2{{\left( -1 \right)}^{2}}+3=-1-2+3=0\].
So, the root of the ${x^3} - 2{x^2} + 3$ will be the function $\left( x+1 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x+1 \right)$ is a factor of the polynomial ${x^3} - 2{x^2} + 3$.
We can now divide the polynomial ${x^3} - 2{x^2} + 3$ by $\left( x+1 \right)$.
\[x+1\overset{{{x}^{2}}-3x+3}{\overline{\left){\begin{align}
& {{x}^{3}}-2{{x}^{2}}+3 \\
& \underline{{{x}^{3}}+{{x}^{2}}} \\
& -3{{x}^{2}}+3 \\
& \underline{-3{{x}^{2}}-3x} \\
& 3x+3 \\
& \underline{3x+3} \\
& 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with ${{x}^{2}}$. We get \[{{x}^{3}}+{{x}^{2}}\]. We subtract it to get \[-3{{x}^{2}}+3\]. We again equate with the highest power of the remaining terms. We multiply with $-3x$ and subtract to get \[-3{{x}^{2}}-3x\]. In the end, we had to multiply with 3 to complete the division. The quotient is \[{{x}^{2}}-3x+3\].
We cannot further factor \[{{x}^{2}}+5x+6\].
The final factorization of ${{x}^{3}}-2{{x}^{2}}+3=\left( x+1 \right)\left( {{x}^{2}}-3x+3 \right)$.
Note: Such polynomials can be factored by using the hit and trial method and then using factor theorem to find the factors of the polynomials. Cubic polynomials are polynomials with the degree of variable as $3$. The imaginary roots always occur in pairs. So, either there can be no imaginary root or two imaginary roots of a cubic polynomial. Conversely, there are either three or one root real root of a cubic polynomial.
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