Question

Factorise $ 1 + {x^4} + {x^8} $

Answer 1

Hint: Use the formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ to factorise the given expression. The formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ states that the difference between the squares of the two numbers is equal to the product of the sum and difference of those two numbers.

Complete step-by-step answer:
The given expression is $ 1 + {x^4} + {x^8} $ .
We should add and subtract the term $ {x^4} $ from the equation $ 1 + {x^4} + {x^8} $ .
 $ 1 + {x^4} + {x^8} = 1 + {x^4} + {x^8} + {x^4} - {x^4} $
After rearranging the terms, we get,
 $
\Rightarrow 1 + {x^4} + {x^8} = 1 + {x^4} + {x^4} + {x^8} - {x^4}\\
 = \left( {1 + 2{x^4} + {x^8}} \right) - {x^4}
$
We know that, $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ for the polynomial.
 $ \left( {1 + 2{x^4} + {x^8}} \right) $ is in the form of $ {a^2} + 2ab + {b^2} $ .
So, we can compare the expression $ \left( {1 + 2{x^4} + {x^8}} \right) $ with $ {a^2} + 2ab + {b^2} $ .
After comparing, we get, $ a = 1,{\rm{ }}b = {x^4} $ .
Substitute the value of $ a = 1,{\rm{ }}b = {x^4} $ in the equation $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ .
After substituting, we get,
$\Rightarrow {\left( {1 + {x^4}} \right)^2} = 1 + 2{x^4} + {x^8}......\left( 1 \right) $
Now, we can substitute the equation (1) in the equation $ 1 + {x^4} + {x^8} = \left( {1 + 2{x^4} + {x^8}} \right) - {x^4} $ .
 \[1 + {x^4} + {x^8} = {\left( {1 + {x^4}} \right)^2} - {x^4}......\left( 2 \right)\]
For factorising the given expression further, we can use $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ .
 Now, compare the expression $ {\left( {1 + {x^4}} \right)^2} - {x^4} $ with $ {a^2} - {b^2} $ .
We get $ a = \left( {1 + {x^4}} \right) $ and $ b = {x^2} $ .
Substitute the value of $ a = \left( {1 + {x^4}} \right) $ and $ b = {x^2} $ in the formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ .
 $\Rightarrow {\left( {1 + {x^4}} \right)^2} - {x^4} = \left( {1 + {x^4} + {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right) $
We should add and subtract the term $ {x^2} $ in the term \[\left( {1 + {x^4} + {x^2}} \right)\].
\[\;{\left( {1 + {x^4}} \right)^2} - {\left( {{x^2}} \right)^2} = \left( {1 + {x^4} + {x^2} + {x^2} - {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)\]
\[
\;{\left( {1 + {x^4}} \right)^2} - {\left( {{x^2}} \right)^2} = \left( {1 + {x^4} + 2{x^2} - {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)\\
 = \left( {\left( {1 + {x^4} + 2{x^2}} \right) - {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)......\left( 3 \right)
\]
We know that, $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ .
Now, we can find the correct expression of factorisation for \[\left( {1 + {x^4} + 2{x^2}} \right)\].
On comparing the expressions \[\left( {1 + {x^4} + 2{x^2}} \right)\] and , we get, $ a = 1,{\rm{ }}b = {x^2} $ .
Now, we can substitute the value of $ a = 1,{\rm{ }}b = {x^2} $ in the equation $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ .
\[\left( {1 + {x^4} + 2{x^2}} \right) = {\left( {1 + {x^2}} \right)^2}......\left( 4 \right)\]
Consider the value of \[\left( {1 + {x^4} + 2{x^2}} \right) = {\left( {1 + {x^2}} \right)^2}\] and substitute it in the equation \[\;{\left( {1 + {x^4}} \right)^2} - {\left( {{x^2}} \right)^2} = \left( {\left( {1 + {x^4} + 2{x^2}} \right) - {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)\] .
On substitution, we get, \[\;{\left( {1 + {x^4}} \right)^2} - {\left( {{x^2}} \right)^2} = \left( {{{\left( {1 + {x^2}} \right)}^2} - {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)......\left( 5 \right)\]
In order to factorise \[\left( {{{\left( {1 + {x^2}} \right)}^2} - {x^2}} \right)\] further, we can use the formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ .
On comparing \[\left( {{{\left( {1 + {x^2}} \right)}^2} - {x^2}} \right)\] with , we get, $ a = 1 + {x^2},{\rm{ }}b = x $ .
Now, substitute the value of $ a = 1 + {x^2},{\rm{ }}b = x $ in the equation $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ .
\[\Rightarrow \left( {{{\left( {1 + {x^2}} \right)}^2} - {x^2}} \right) = \left( {1 + {x^2} + x} \right)\left( {1 + {x^2} - x} \right)......\left( 6 \right)\]
Now, on considering the equations (1), (2), (3), (4), (5) and (6), we can deduce the final answer.

Therefore, on factorising, we get \[1 + {x^4} + {x^8} = \left( {1 + {x^2} + x} \right)\left( {1 + {x^2} - x} \right)\left( {1 + {x^4} - {x^2}} \right)\].

Additional Information:
Factorisation of polynomials is done by separating it in the product of factors which canot be reduced further. In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials.
There are some of the important identities which are mostly used to either factorise of an expression.
 $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
The square of the sum of two numbers can be given as the sum of the squares of the individual numbers added to twice the product of those two numbers.
 $ {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab $
The square of the difference of two numbers can be given by subtracting twice the product of the numbers from the sum of the squares of the individual numbers.
 $ {a^2} - {b^2} = \left( {a + b} \right) \cdot \left( {a - b} \right) $
The difference between the squares of the two numbers is equal to the product of the sum and difference of those two numbers.

Note: Please take special care while considering the symbols of different operations i.e. addition, subtraction, multiplication. Students should follow BODMAS order of operations which starts with solving the bracket first, then division, multiplication, addition and at last subtraction.