Question

Factorize the following expression: \[8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}}\].
A. Cannot be determined
B. Does not exist
C. \[\left( {2p - \dfrac{1}{5}} \right)\left( {2p - \dfrac{1}{5}} \right)\left( {2p - \dfrac{1}{5}} \right)\]
D. None of these

Answer 1

Hint: Here, we are going to be use the most eccentric concept of solving the algebraic solution by means of various formulae such as expansion, indices property, etc. hence, relating the given expression with one of the expansion property that is ‘\[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]’ by adjusting the certain terms present in the given expression respectively. Then, by applying the rules of indices for the multiplication such as ‘\[{a^{m + n}} = {a^m} \times {a^n}\]’, the desired factor(s) are obtained.

Complete step by step solution:
Since, we have given the expression that
\[8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}}\]
As the highest power of the above given expression is ‘\[3\]’ ,
Hence, adjusting the expression so that to form the expansion of an algebraic expression or an identity that is
We know that,
\[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]
As a result, simplifying or adjusting the middle terms (that is to make square of the first term simultaneously for the second term, as justified below) of the given expression, we get
 \[ \Rightarrow 8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}} = 8{p^3} - 3{\left( {2p} \right)^2}\left( {\dfrac{1}{5}} \right) + 3\left( {2p} \right){\left( {\dfrac{1}{5}} \right)^2} - {\left( {\dfrac{1}{5}} \right)^3}\]
Hence, relating the above expanded equation with an algebraic identity (or a formula) explained above, we get
\[ \Rightarrow 8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}} = {\left( {2p - \dfrac{1}{5}} \right)^3}\]
\[ \Rightarrow 8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}} = {\left( {2p - \dfrac{1}{5}} \right)^{1 + 1 + 1}}\]
We also know that,
The respective rule for indices that is in term \[{a^m} \times {a^n} = {a^{m + n}}\]\[{a^{m + n}} = {a^m} \times {a^n}\],
Hence, applying the above rule (considering \[a = \left( {2p - \dfrac{1}{5}} \right)\] and here ‘\[m\]’ & ‘\[n\]’ are equal that is ‘\[1\]’ respectively), we get
\[ \Rightarrow 8{p^3} - \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p - \dfrac{1}{{125}} = \left( {2p - \dfrac{1}{5}} \right)\left( {2p - \dfrac{1}{5}} \right)\left( {2p - \dfrac{1}{5}} \right)\]
So, the correct answer is “Option C”.

Note: One must be able to distinguish the expansion formulae used in algebra to solve the respective solution such as ‘\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]’, ‘\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]’, ‘\[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\]’, ‘\[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]’, etc. (likewise here we have used the identity for ‘\[{\left( {a - b} \right)^3}\]’ respectively). Also, remember the rules for indices such as ‘\[\dfrac{1}{{{a^n}}} = {a^{ - n}}\]’, ‘\[{a^m} \times {a^n} = {a^{m + n}}\]’, ‘\[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]’, ‘\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]’, etc. where, base must be the same, so as to be sure of our final answer.