Question

factorize (8x^3-4x+1)​

Answer 1

Let us introduce a new parameter such that $t=2x$.

Then $8x^3 - 4x + 1 = (2x)^3 - 2(2x) + 1 = t^3 - 2t + 1$.

 

However, $t^3-2t+1$ is 0 when $t=1$.

By remainder theorem, $t^3 - 2t + 1$ has a factor $t - 1$.

$\implies t^3 - 2t + 1 = ( t - 1 ) ( t^2 - t - 1 )$

 

After substituting $t=2x$ into the polynomial, the factorization is the following.

$\therefore ( 2x - 1 ) ( 4x^2 - 2x - 1 )$

 

More information:

A parameter is used to simplify the calculations. In this question, we chose $t=2x$ because each coefficient is a power of 2.