Question

How do you determine the binomial factors of \[{{x}^{3}}-6{{x}^{2}}+11x-6=0\]?

Answer 1

Hint: In order to find the binomial factors of the given equation \[{{x}^{3}}-6{{x}^{2}}+11x-6=0\], test possible factors using synthetic division where Synthetic division is a shorthand, or shortcut method of polynomial division in the special case of dividing by a linear factor. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. After applying synthetic division, retain those which are true factors. Repeat with the factored polynomial until only binomial factors remain.

Complete step-by-step solution:
According to the question, given equation is as follows:
\[{{x}^{3}}-6{{x}^{2}}+11x-6=0\]
For \[\left( ax+b \right)\] to be a factor of a cubic polynomial \[A{{x}^{3}}+B{{x}^{2}}+Cx+D\], we need \[a\] to be a divisor of A, and \[b\] to be a divisor of D. So, as we're looking for integer-based factors, this reduces our options considerably.
For our cubic polynomial \[{{x}^{3}}-6{{x}^{2}}+11x-6=0\], we have A=1 and D=-6, so the possible factors we will check are \[x\pm 1,x\pm 2,x\pm 3,\text{ }\] and \[x\pm 6.\]
Let's start checking with the factors \[\left( x+1 \right)\] and \[\left( x-1 \right)\].
If \[\left( x+1 \right)\]is a factor, then dividing the polynomial by \[\left( x+1 \right)\] will leave 0 remainder. We divide using synthetic division:
\[x+1\overset{{{x}^{2}}-7x-18}{\overline{\left){\begin{align}
  & \text{ }\underline{\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+11x-6 \\
 & -{{x}^{3}}-{{x}^{2}} \\
\end{align}} \\
 & \text{ }\underline{\begin{align}
  & \text{ }0-7{{x}^{2}}+11x-6 \\
 & \text{ +}7{{x}^{2}}+7x \\
\end{align}} \\
 & \text{ }\underline{\begin{align}
  & \text{ 0 }+18x-6 \\
 & \text{ }-18x-18 \\
\end{align}} \\
 & \text{ }-24 \\
\end{align}}\right.}}\]
The -24 is our remainder after division. Since this is not zero, \[\left( x+1 \right)\] is not a factor of our polynomial.
Now check for \[\left( x-1 \right)\], using synthetic division we get:
\[x-1\overset{{{x}^{2}}-5x+6}{\overline{\left){\begin{align}
  & \text{ }\underline{\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+11x-6 \\
 & -{{x}^{3}}+{{x}^{2}} \\
\end{align}} \\
 & \text{ }\underline{\begin{align}
  & \text{ }0-5{{x}^{2}}+11x-6 \\
 & \text{ +5}{{x}^{2}}-5x \\
\end{align}} \\
 & \text{ }\underline{\begin{align}
  & \text{ 0 }+6x-6 \\
 & \text{ }-6x+6 \\
\end{align}} \\
 & \text{ }0 \\
\end{align}}\right.}}\]
Since the remainder is zero, we know \[\left( x-1 \right)\] is a factor of our polynomial. The other numbers on our bottom line \[\left( 1,-5,6 \right)\] are the coefficients of the factored polynomial. That is:
\[\Rightarrow {{x}^{3}}-6{{x}^{2}}+11x-6=\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)\]
We can determine the remaining factors by using synthetic division on other possible factors, or, since the remaining polynomial is a quadratic, we can use splitting the middle term method that is "find two numbers that add to -5 and multiply to 6" method.
Therefore, two numbers are -2 and -3:
\[\Rightarrow \left( {{x}^{2}}-5x+6 \right)=\left( x-2 \right)\left( x-3 \right)\]
So, we can substitute the \[\left( {{x}^{2}}-5x+6 \right)\] in our factorization of the polynomial with \[\left( x-2 \right)\left( x-3 \right)\] to get:
\[\Rightarrow {{x}^{3}}-6{{x}^{2}}+11x-6=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\]
Therefore, binomial factors of the equation \[{{x}^{3}}-6{{x}^{2}}+11x-6=0\] are \[\left( x-1 \right),\text{ }\left( x-2 \right)\] and \[\left( x-3 \right)\].

Note: Students generally make mistakes while applying the concept of splitting the middle term to solve the quadratic term, they make mistakes in identifying the common terms and take wrong terms in common which further leads to the wrong answer. Key point is to remember this concept of splitting the middle term that is in quadratic factorization, Splitting of Middle Term is variable term is the sum of two factors and product equal to last term and the concept of synthetic division where Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials.