Question

How do you factor $27{{x}^{3}}+54{{x}^{2}}+36x+8$ by grouping?

Answer 1

Hint: Since the ratio of first and second coefficients is not equal to the ratio of third and fourth coefficients, we cannot factorize the given expression by grouping.
$\dfrac{27}{54}\ne \dfrac{36}{8}$
Instead, we are going to use the following identity to solve the given question,
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ .

Complete step by step answer:
Here, we know that the first term “27” is the cube of 3 and the last term “8” is the cube of 2.
So, we can write the given equation as
\[\Rightarrow {{\left( 3x \right)}^{3}}+54{{x}^{2}}+36x+{{\left( 2 \right)}^{3}}.....(i)\]
Now, let’s prime factorize the second term “54” and third term “36” to simplify the above expression even more,
$\begin{align}
  & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$ $\begin{align}
  & 2\left| \!{\underline {\,
  36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Prime factors of $54={{3}^{3}}\times 2$
$\Rightarrow 54=3\times {{3}^{2}}\times 2.....(ii)$
Prime factors of $36={{3}^{2}}\times {{2}^{2}}$
$\Rightarrow 36=3\times 3\times {{2}^{2}}.......(iii)$
Now, let’s put the values of 54 and 36 from (ii) and (iii) in the expression written in (i),
\[\Rightarrow {{\left( 3x \right)}^{3}}+3\times {{3}^{2}}\times 2\times {{x}^{2}}+3\times 3\times {{2}^{2}}\times x+{{\left( 2 \right)}^{3}}\]
Or we can write it as,
\[\Rightarrow {{\left( 3x \right)}^{3}}+3\times {{\left( 3x \right)}^{2}}\times 2+3\times \left( 3x \right)\times {{2}^{2}}+{{\left( 2 \right)}^{3}}\] .
Now, we can see that the above expression resembles the R.H.S. of the identity
 ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ , if we take $a=3x\text{ and }b=2$ .
Thus, we can write the given expression as
$\therefore {{\left( 3x+2 \right)}^{3}}={{\left( 3x \right)}^{3}}+{{2}^{3}}+3{{\left( 3x \right)}^{2}}2+3\left( 3x \right){{2}^{2}}$ .
Hence, the factors of the given expression in the question are $\left( 3x+2 \right),\text{ }\left( 3x+2 \right)\text{ and }\left( 3x+2 \right).$

Note:
To check whether a cubic expression can be factored by grouping, find whether the ratio of the first and second terms is equal to the third and last terms.
We can also solve this cubic polynomial by using a long division method. Just find the first factor by hit and trial method, here, put $x=\dfrac{-2}{3}$ in the polynomial to get $P\left( x \right)=0$ , thus getting the first factor $\left( 3x+2 \right)$ . Then use the long division method with
$\text{Divisor}=\left( 3x+2 \right)\text{, Dividend=}27{{x}^{3}}+54{{x}^{2}}+36x+8$ ,
to get the quotient as a quadratic polynomial. Factorize this quadratic polynomial to get the other two factors.