Question

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

Answer 1

Hint: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.