Question

How do you simplify $\dfrac{{{x^3} - {x^2} - x + 1}}{{{x^2} - 2x + 1}}$?

Answer 1

Hint: Here we can use the synthetic division to reduce the cubic equation into three factors and use factorization to find the factors of the quadratic equation. And then cancel the common term. Finally we get the required answer.

Complete step-by-step solution:
Let us explain the method of synthetic division.
Consider $p(x) = {x^3} - {x^2} - x + 1$ be the dividend and assume the factor $q(x) = x - 1$ be the divisor.
It will give a factor for the given equation, if not try another factor. We shall find the quotient $s(x)$ and the remainder $r$ , by proceeding follows.
Arrange the dividend and the divisor according to the descending powers of $x$ and then write the coefficients of dividend in the first row.
Also, we have to find out the zero for the divisor. Put $0$ for the first entry in the second row. Complete the entries of the rows. Write down the quotient and the remainder accordingly. All the entries except the last one in the third row constitute the coefficients of the quotient. The above process will written follows,
$\begin{array}{*{20}{c}}
  {1\left| \!{\underline {\,
  \begin{gathered}
  1\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,1 \\
  0\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\, - 1 \\
\end{gathered} \,}} \right. } \\
  {\,\,\,\,\underline {1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,} \left| \!{\underline {\,
  0 \,}} \right. }
\end{array}$
Here $p(x)$ is divided by $x - 1$ , the quotient is ${x^2} - 1$ and the remainder is $0$.
 Therefore the factors are $x - 1$, ${x^2} - 1$
Now for ${x^2} - 2x + 1$ we can factorize the number,
\[\begin{array}{*{20}{c}}
  {\,1} \\
  {\, \swarrow - 2 \searrow } \\
  { - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1}
\end{array}\]
Since $ - 1 - 1 = - 2\,,\,\,\,\,\,( - 1)( - 1) = 1$
We get the factors, $\left( {x - 1} \right)\left( {x - 1} \right)$
Now replace these values in the given expression, $\dfrac{{{x^3} - {x^2} - x + 1}}{{{x^2} - 2x + 1}} = \dfrac{{\left( {{x^2} - 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$
We can write the numerator like this,
$ \Rightarrow \dfrac{{\left( {{x^2} - {1^2}} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$
Now using the formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ we can write the expression like this,
$ \Rightarrow \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$
Cancel the term we get,
$ \Rightarrow \left( {x + 1} \right)$

Hence the simplified form of given expression is $x + 1$.

Note: For any polynomial $p(x)$ , \[x = a\] is zero if and only if $p(a) = 0$ .
\[x - a\] is a factor for $p(x)$ if and only if $p(a) = 0$(factor theorem)
\[x - 1\] is a factor of $p(x)$ if and only if the sum of coefficients of $p(x)$ is $0$.
\[x + 1\] is a factor of $p(x)$ if and only if the sum of coefficients of even powers of $x$ including constant is equal to sum of the coefficients of odd powers of $x$.