Question

Factorize completely: $2{{x}^{3}}-{{x}^{2}}-2x+1$

Answer 1

Hint: Find the given equation is the perfect cube. If yes then factorize directly or try to get a factor to divide the expression into 2 parts try to find a common point in them both, then try to convert this cubic into the multiplication of a quadratic and a degree polynomial. So, next try to check if quadratic is a perfect square, if not use the general method and factorize the quadratic. By multiplying the 1-degree polynomial with the factored version on quadratic you get a complete factored cubic.

Complete step-by-step answer:

Given expression for which we need to factorize is:

$2{{x}^{3}}-{{x}^{2}}-2x+1$.

Checking for the perfect cube:

If \[2{{x}^{3}}\ =\ {{a}^{3}}\] then \[a\ =\ \sqrt[3]{2}x\] by we don’t have \[\sqrt[3]{2}\] anywhere else. So, we cannot convert this into a cube of another expression. So, by this we say that the given expression is not a perfect cube.

Given expression can be written as: 2 terms ass shown;

\[\left( 2{{x}^{3}}-{{x}^{2}} \right)+\left( -2x+1 \right)\].

Let the first two terms be A and next two term be B. By taking $-{{x}^{2}}$ common from the term A, we get:

\[-{{x}^{2}}\left( -2x+1 \right)+1\left( -2x+1 \right)\]

By taking $\left( 1-2x \right)$ term common from equation, we get:

\[\left( -2x+1 \right)\left( 1-{{x}^{2}} \right)\]

BY multiplying both terms with minus sign, we get:

\[\left( 2x-1 \right)\left( {{x}^{2}}-1 \right)\]

The second term can be easily factored into 2 parts:

By general knowledge of algebra, we can say that:

\[\left( {{x}^{2}}-1 \right)\ =\ \left( x-1 \right)\left( x+1 \right)\]

By substituting this into equation above, we get

\[\left( 2x-1 \right)\left( x-1 \right)\left( x+1 \right)\]

By equating it to given expression, we get:

$2{{x}^{3}}-{{x}^{2}}-2x+1\ =\ \left( 2x-1 \right)\left( x-1 \right)\left( x+1 \right)$

So, the polynomial is factored completely.

Therefore, the expression \[\left( 2x-1 \right)\left( x-1 \right)\left( x+1 \right)\] is the complete factorization of the cubic expression.

Note: Here, the quadratic was direct to solve. If not, we must find the roots a, b and write the quadratic as \[\left( x-a \right)\centerdot \left( x-b \right)\].