Question

How do you factor ${{y}^{3}}-3{{y}^{2}}+3y-1$ ?

Answer 1

Hint: In this question, we have to find the factors of a polynomial. Thus, we will use the hit-and trial method, long division and splitting the term method to get the solution. Therefore, to solve this problem, we will first find a root of the given polynomial by letting different values of y, whichever value gives the answer as 0, implies the value is the root of the polynomial. After that we will divide the polynomial by the root, to get a quotient. Then, we will solve the quotient using splitting the middle term method, to get the required solution to the problem.

Complete step by step solution:
According to the question, we have to find the factors of the cubic polynomial.
The polynomial given to us is ${{y}^{3}}-3{{y}^{2}}+3y-1$ ----------- (1)
Now, let $y=1$ and put this value in the equation (1), we get
$\begin{align}
  & \Rightarrow {{(1)}^{3}}-3.{{(1)}^{2}}+3.(1)-1 \\
 & \Rightarrow 1-3+3-1 \\
\end{align}$
As we know, the same terms with opposite signs cancel out each other, therefore we get
$\Rightarrow 0$
This implies that $y=1$ is the solution of the given polynomial.
Therefore, one of the roots of the given polynomial is $(y-1)$ ---- (2)
Now, we will divide equation (1) by equation (2) by long division method, we get
$\begin{align}
  & \text{ }{{\text{y}}^{2}}-2y+1 \\
 & \text{y-1}\left| \!{\overline {\,
 {{y}^{3}}-3{{y}^{2}}+3y-1 \,}} \right. \\
 & \text{ }\underline{{}_{-}{{y}^{3}}-{}_{+}{{y}^{2}}} \\
 & \text{ -2}{{\text{y}}^{2}}\text{+3y-1 } \\
 & \text{ }\underline{{}_{+}\text{-2}{{\text{y}}^{2}}\text{+}{}_{-}\text{2y }}\text{ } \\
 & \text{ y-1} \\
 & \text{ }\underline{{}_{-}\text{y-}{}_{+}1} \\
 & \text{ }\underline{\text{ 0}} \\
 & \text{ } \\
\end{align}$
Therefore, we get the remainder equals to 0 and the quotient is equal to ${{y}^{2}}-2y+1$ .
Now, we will use the splitting the middle term method in the quotient, which is
${{y}^{2}}-2y+1$
So, we will split the middle term as the sum of -y and y, because $a.c=1.1=1$ and $a+c=-1-1=-2$
Therefore, we get
$\Rightarrow {{y}^{2}}-y-y+1$
Now, we take y common from the first two terms and take common -1 from the last two terms, we get
$\Rightarrow y(y-1)-1(y-1)$
Now, take (y-1) common from the above equation, we get
$\Rightarrow(y-1)(y-1)$ ------- (3)
From equation (2) and (3), we get
$\Rightarrow {{(y-1)}^{3}}$

Therefore, for the equation ${{y}^{3}}-3{{y}^{2}}+3y-1$ , its factors are ${{(y-1)}^{3}}$.

Note: While solving this problem, do mention all the formulas and the methods you are using to avoid confusion and mathematical error. One of the alternative methods to solve this problem is to use the algebraic identity ${{(a-b)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$ in the given polynomial, and hence you will find the required solution to the problem.