Question

How do you factor: $ 2{x^3} - 3{x^2} - 2x + 3 $ ?

Answer 1

Hint: In the given problem, we have to factorise the given polynomial and represent it as a product of its factors. The given polynomial is of degree $ 3 $ and is thus called a cubic polynomial. For factoring the polynomial, first we have to simplify the given polynomial and then find its factors. The first zero of the cubic polynomial has to be found using the hit and trial method.

Complete step by step solution:
So, the given polynomial is $ 2{x^3} - 3{x^2} - 2x + 3 $ .
Now, using the hit and trial method, consider the given polynomial as $ p\left( x \right) $ .
 $ p\left( x \right) = 2{x^3} - 3{x^2} - 2x + 3 $
Putting $ x = 1 $ in the polynomial $ p\left( x \right) $ ,
 $ \Rightarrow $ \[p(x = 1) = 2{\left( 1 \right)^3} - 3{\left( 1 \right)^2} - 2\left( 1 \right) + 3\]
 $ \Rightarrow $ \[p(x = 1) = 2 - 3 - 2 + 3\]
On simplifying further, we get,
 $ \Rightarrow $ $ p\left( {x = 1} \right) = 0 $
So, $ p(1) = 0 $ .
So, using the factor theorem, $ \left( {x - 1} \right) $ is a factor of p(x).
Now, $ p(x) = (x - 1)(2{x^2} - x - 3) $
Now quadratic polynomials $ (2{x^2} - x - 3) $ can be factorised further as its Discriminant is positive. The quadratic equations having discriminant more than or equal to $ 0 $ have real roots.
Now we can factorise the obtained quadratic polynomial by splitting the middle term method, using quadratic formula and factorisation. Some of the polynomials can also be factorised using algebraic identities.
Now, let $ f(x) = (2{x^2} - x - 3) $
 $ \Rightarrow f(x) = (2{x^2} - 3x + 2x - 3) $
 $ \Rightarrow f(x) = x\left( {2x - 3} \right) + \left( {2x - 3} \right) $
 $ \Rightarrow f(x) = \left( {2x - 3} \right)\left( {x + 1} \right) $
Therefore, $ p(x) = (x - 1)\left( {2x - 3} \right)\left( {x + 1} \right) $ .
Hence, the factorised form of the polynomial $ p\left( x \right) = 2{x^3} - 3{x^2} - 2x + 3 $ is $ \left[ {(x - 1)\left( {2x - 3} \right)\left( {x + 1} \right)} \right] $ .
So, the correct answer is “$ \left[ {(x - 1)\left( {2x - 3} \right)\left( {x + 1} \right)} \right] $”.

Note: Such polynomials can be factorised by using the hit and trial method and then using factor theorem finding the factors of the polynomials. Bi-quadratic polynomials are polynomials with degree $ 4 $ . We can convert such bi-quadratic polynomials into quadratic polynomials by finding out $ 2 $ factors of the polynomial by hit and trial and then easily factorising the quadratic polynomial with splitting of middle term or using Quadratic formula.