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Question: Factorize \(2{x^3} - 128x.\)...

Factorize 2x3−128x.2{x^3} - 128x.

Explanation

Solution

We can use the formulaa2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right). This is the difference of square identity and by using this formula we can factorize the given equation. So in order to factorize we have to convert the question and then express it in the form of the difference of square identity.

Complete step by step solution:
Given
2x3−128x...........................(i)2{x^3} - 128x...........................\left( i \right)
Also a2−b2=(a+b)(a−b).....................(ii){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right).....................\left( {ii} \right)

So we need to express (ii) in terms of (i), for that we have to take the common factors from2x3−128x2{x^3} - 128x.

Such that:
⇒2x3−128x=2x(x2−64)......................(ii)\Rightarrow 2{x^3} - 128x = 2x\left( {{x^2} - 64} \right)......................\left( {ii} \right)

Now when we observe (ii) we observe that(x2−64)\left( {{x^2} - 64} \right)is in the form ofa2−b2{a^2} - {b^2}such that we can use the identitya2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right).
⇒(x2−64)=(x2−(8)2) ⇒(x2−(8)2)=((x+8)(x−8)).................(iii)  \Rightarrow \left( {{x^2} - 64} \right) = \left( {{x^2} - {{\left( 8 \right)}^2}} \right) \\\ \Rightarrow \left( {{x^2} - {{\left( 8 \right)}^2}} \right) = \left( {\left( {x + 8} \right)\left( {x - 8} \right)} \right).................\left( {iii} \right) \\\

Now substituting (x2−64)\left( {{x^2} - 64} \right) in (ii), we get:

\right).....................\left( {iv} \right)$$ **Therefore from (iv) we can write on factorization of $2{x^3} - 128x$ we get $$2x\left( {\left( {x + 8} \right)\left( {x - 8} \right)} \right)$$. ** **Additional Information:** Another technique for factoring is grouping, where we take common terms not from the full polynomial but only from the certain terms which have them. So it is used mainly when there is not a common factor for the whole polynomial but there is a common factor only for certain terms. However here we use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ because if we use here only grouping then we can’t fully factorize the term $2{x^3} - 128x$. **Note:** While approaching a question one should study it properly and accordingly should choose the method to factorize the polynomial. Similar questions which have real coefficients and some common terms should be approached using the same method as described above. Polynomial factorization is always done over some set of numbers which may be integers, real numbers or complex numbers.