Question

Factorise \[{{x}^{3}}+13{{x}^{2}}+31x-45\] by using the factor theorem.

Answer 1

Hint: We solve this problem first by using the trial and error method, we find the value of \['x'\] for which the given polynomial becomes zero. If \['x=a'\] satisfies the above condition then, \[\left( x-a \right)\] is the factor of the given polynomial then by dividing the polynomial with \[\left( x-a \right)\] we get a quadratic polynomial which will be easy to factorise. This method is called factorisation using the factor theorem.

Complete step by step answer:
Let us assume the given polynomial as
\[p\left( x \right)={{x}^{3}}+13{{x}^{2}}+31x-45\]
Now, let us use the trial and error method to find the first zero of the polynomial.
Let us substitute \[x=0\] in above polynomial
\[\begin{align}
  & \Rightarrow p\left( 0 \right)={{0}^{3}}+13\left( {{0}^{2}} \right)+31\left( 0 \right)-45 \\
 & \Rightarrow \left( 0 \right)=-45 \\
\end{align}\]
Here, we can see that \[p\left( 0 \right)\ne 0\]. So, we can say that ‘0’ is not the zero of the given polynomial.
Now, by substituting \[x=1\] in the given polynomial we get
\[\begin{align}
  & \Rightarrow p\left( 1 \right)={{1}^{3}}+13{{\left( 1 \right)}^{2}}+31\left( 1 \right)-45 \\
 & \Rightarrow p\left( 1 \right)=1+13+31-45=0 \\
\end{align}\]
Here, we can see that \[p\left( 1 \right)=0\], so we can say that \[\left( x-1 \right)\] is the factor of a given polynomial \[p\left( x \right)\].
So, we can write the given polynomial as
\[p\left( x \right)=\left( x-1 \right).g\left( x \right)\]
Here, the polynomial \[g\left( x \right)\] is the polynomial which we get after dividing the given polynomial by its factor.
Now, let us calculate for the polynomial \[g\left( x \right)\].
\[\Rightarrow g\left( x \right)=\dfrac{p\left( x \right)}{\left( x-1 \right)}\]
By substituting the polynomial \[p\left( x \right)\] in above equation and by using the simple division we get
\[\begin{align}
  & \Rightarrow g\left( x \right)=\dfrac{{{x}^{3}}+13{{x}^{2}}+31x-45}{x-1} \\
 & \Rightarrow g\left( x \right)={{x}^{2}}+14x+45 \\
\end{align}\]
Now, let us factorise \[g\left( x \right)\] as we got \[g\left( x \right)\] as quadratic polynomial as follows
\[\begin{align}
  & \Rightarrow g\left( x \right)=\left( {{x}^{2}}+9x \right)+\left( 5x+45 \right) \\
 & \Rightarrow g\left( x \right)=x\left( x+9 \right)+5\left( x+9 \right) \\
 & \Rightarrow g\left( x \right)=\left( x+9 \right)\left( x+5 \right) \\
\end{align}\]
Now by substituting the value of \[g\left( x \right)\] in the given polynomial \[p\left( x \right)\] we get
\[\Rightarrow p\left( x \right)=\left( x-1 \right)\left( x+5 \right)\left( x+9 \right)\]

Therefore, the factorisation of given polynomial is
\[{{x}^{3}}+13{{x}^{2}}+31x-45=\left( x-1 \right)\left( x+5 \right)\left( x+9 \right)\]

Note: Students will make mistakes in finding the first zero of the given polynomial. While finding the first zero of any polynomial we substitute the value of \['x'\] as integers from \[-3\] to \[3\]. After substituting all these values, still any value does not result to zero then we can conclude that the given polynomial cannot be factored. But, students will think of some fractions and substitute then in the given polynomial. But, when there is a question to factorise there will be no possibility for fractions to substitute in the given polynomial. It will just increase the time for solving.