Question

How do you factor the expression \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\]?

Answer 1

Hint: In order to find the solution to the given question, that is to factor the expression \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\], use the rational zeros theorem and synthetic division.

Complete step-by-step solution:
According to the question, given expression in the question is as follows:
\[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\]
Since all coefficients are integers of the given expression, apply the rational zeros theorem.
The trailing coefficient (coefficient of the constant term) is 1.
Find its factors (with plus and minus): \[\pm 1\]. These are the possible values for p.
The leading coefficient (coefficient of the term with the highest degree) is 1.
Find its factors (with plus and minus): \[\pm 1\]. These are the possible values for q.
Find all possible values of pq: \[\pm 1\].
Simplify and remove duplicates (if any): \[\pm 1\].
If \[a\] is a root of the polynomial \[P\left( x \right)\], then the remainder from the division of \[P\left( x \right)\] by \[\left( x-a \right)\] should equal 0.
Now Check for \[1\]: Divide \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\] by \[x-1\] using synthetic division we get:
\[x-1\overset{{{x}^{3}}+5{{x}^{2}}+11x+15}{\overline{\left){\begin{align}
  & \text{ }{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1 \\
 & \underline{-{{x}^{4}}+{{x}^{3}}\text{ }} \\
 & \text{ }5{{x}^{3}}+6{{x}^{2}}+4x+1 \\
 & \text{ }\underline{\text{ }-5{{x}^{3}}+5{{x}^{2}}\text{ }} \\
 & \text{ }11{{x}^{2}}+4x+1 \\
 & \underline{\text{ }-11{{x}^{2}}+11x\text{ }} \\
 & \text{ }15x+\text{ }1 \\
 & \underline{\text{ }-15x+15} \\
 & \underline{\text{ }16} \\
\end{align}}\right.}}\]
Clearly, we can see that the quotient is \[{{x}^{3}}+5{{x}^{2}}+11x+15\], and the remainder is \[16\]. Therefore, \[x-1\] is not a factor.
After this Check for \[-1\]: divide \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\] by \[x+1\] using synthetic division method, we get:
\[x+1\overset{{{x}^{3}}+3{{x}^{2}}+3x+1}{\overline{\left){\begin{align}
  & \text{ }{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1 \\
 & \underline{-{{x}^{4}}-{{x}^{3}}\text{ }} \\
 & \text{ 3}{{x}^{3}}+6{{x}^{2}}+4x+1 \\
 & \text{ }\underline{\text{ }-3{{x}^{3}}-3{{x}^{2}}\text{ }} \\
 & \text{ 3}{{x}^{2}}+4x+1 \\
 & \underline{\text{ }-3{{x}^{2}}-3x\text{ }} \\
 & \text{ }x+1 \\
 & \underline{\text{ }-x+1} \\
 & \underline{\text{ 0}} \\
\end{align}}\right.}}\]
Here, we can see that the quotient is \[{{x}^{3}}+3{{x}^{2}}+3x+1\], and the remainder is \[0\].
Since the remainder is \[0\], then \[-1\] is the root, and \[x+1\] is the factor.
Hence, we can rewrite the given expression as:
\[\Rightarrow {{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1=\left( x+1 \right)\left( {{x}^{3}}+3{{x}^{2}}+3x+1 \right)\]
Now apply the cube of sum formula that is \[{{\alpha }^{3}}+3{{\alpha }^{2}}\beta +3\alpha {{\beta }^{2}}+{{\beta }^{3}}={{\left( \alpha +\beta \right)}^{3}}~\] where \[\alpha =1\] and \[\beta =x\] in the above equation, we get:
\[\Rightarrow {{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1=\left( x+1 \right){{\left( x+1 \right)}^{3}}\]
Thus, factor of the expression \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1={{\left( x+1 \right)}^{4}}\].

Note: Students generally make mistakes while identifying the concept of cube sum formula to solve the cubic expression, 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 cube sum formula that is \[{{\alpha }^{3}}+3{{\alpha }^{2}}\beta +3\alpha {{\beta }^{2}}+{{\beta }^{3}}={{\left( \alpha +\beta \right)}^{3}}~\]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.