Question

How do you factor $8{x^3} - 4{x^2} - 2x + 1 = 0$ ?

Answer 1

This is the equation with order 3. Thus it has three roots. That is there are three values of x that satisfy this equation. For that first we will take $4{x^2}$ common from the first two terms and then $- 1$ common from the last two terms. And after that we will proceed with the regular method of equating the brackets to zero. Then after that finding the values of x that satisfies the equation given above.

Complete step by step answer:
Given that,
$8{x^3} - 4{x^2} - 2x + 1 = 0$
Now we will take $4{x^2}$ common from the first two terms and then $- 1$ common from the last two terms.
$\Rightarrow 4{x^2}\left( {2x - 1} \right) - 1\left( {2x - 1} \right) = 0$
Now taking the brackets,
$\Rightarrow \left( {4{x^2} - 1} \right)\left( {2x - 1} \right) = 0$
Now equate the brackets separately to zero.
$\Rightarrow \left( {4{x^2} - 1} \right) = 0\& \left( {2x - 1} \right) = 0$
Now for first bracket: $\Rightarrow 4{x^2} - 1 = 0$
Taking 1 on other side we get,
$\Rightarrow 4{x^2} = 1$
Taking 4 on other side
$\Rightarrow {x^2} = \dfrac{1}{4}$
Taking root on both sides,
$\Rightarrow x = \pm \dfrac{1}{2}$
Now for second bracket: $\Rightarrow \left( {2x - 1} \right) = 0$
Taking 1 on other side,
$\Rightarrow 2x = 1$
Taking 2 on other side,
$\Rightarrow x = \dfrac{1}{2}$
This is the value of x.
Thus the factors are \Rightarrow x = \pm \dfrac{1}{2}\ & x = \dfrac{1}{2}.

Note: Note that the equation is with one variable but degree three. So there are three values of x. Also note that two values are the same $x = \dfrac{1}{2}$ but that is not to be worried. We also can solve this by a synthetic division method, where we will choose
the value of x randomly first that satisfies the given equation and then proceed with that value as one of the roots to find the remaining roots.