Factorize:
(a) [tex]x^3-3 x^2-x+3[/tex]
(b) [tex]2 x^3+3 x^2-1[/tex]

For (a), factor x 3 − 3 x 2 − x + 3 by grouping: ( x 3 − 3 x 2 ) + ( − x + 3 ) = x 2 ( x − 3 ) − ( x − 3 ) = ( x − 3 ) ( x 2 − 1 ) = ( x − 3 ) ( x − 1 ) ( x + 1 ) .
For (b), factor 2 x 3 + 3 x 2 − 1 using the Rational Root Theorem to find a root x = − 1 , so ( x + 1 ) is a factor. Divide 2 x 3 + 3 x 2 − 1 by ( x + 1 ) to get 2 x 2 + x − 1 . Factor the quadratic: 2 x 2 + x − 1 = ( x + 1 ) ( 2 x − 1 ) . Thus, 2 x 3 + 3 x 2 − 1 = ( x + 1 ) 2 ( 2 x − 1 ) .
The factorization of x 3 − 3 x 2 − x + 3 is ( x − 3 ) ( x − 1 ) ( x + 1 ) ​ .
The factorization of 2 x 3 + 3 x 2 − 1 is ( x + 1 ) 2 ( 2 x − 1 ) ​ .

Explanation

Problem Analysis We are asked to factorize two cubic polynomials. We will use factoring by grouping for the first polynomial and the rational root theorem combined with polynomial division for the second polynomial.

Factoring by Grouping (a) We have the polynomial x 3 − 3 x 2 − x + 3 . We can factor this by grouping. First, group the terms: ( x 3 − 3 x 2 ) + ( − x + 3 ) . Now, factor out the greatest common factor from each group: x 2 ( x − 3 ) − 1 ( x − 3 ) . Notice that ( x − 3 ) is a common factor, so we can factor it out: ( x − 3 ) ( x 2 − 1 ) . Finally, we can factor the difference of squares x 2 − 1 as ( x − 1 ) ( x + 1 ) . Therefore, the complete factorization is ( x − 3 ) ( x − 1 ) ( x + 1 ) .

Rational Root Theorem and Polynomial Division (b) We have the polynomial 2 x 3 + 3 x 2 − 1 . We can use the Rational Root Theorem to find possible rational roots. The possible rational roots are ± 1 , ± 2 1 ​ . Let's test these values. If x = − 1 , then 2 ( − 1 ) 3 + 3 ( − 1 ) 2 − 1 = − 2 + 3 − 1 = 0 . So, x = − 1 is a root, and ( x + 1 ) is a factor. Now, we can perform polynomial long division or synthetic division to divide 2 x 3 + 3 x 2 − 1 by ( x + 1 ) . The result is 2 x 2 + x − 1 . Now, we need to factor the quadratic 2 x 2 + x − 1 . We look for two numbers that multiply to − 2 and add to 1 . These numbers are 2 and − 1 . So, we can rewrite the middle term as 2 x 2 + 2 x − x − 1 . Now, we factor by grouping: 2 x ( x + 1 ) − 1 ( x + 1 ) . We can factor out the common binomial factor ( x + 1 ) : ( x + 1 ) ( 2 x − 1 ) . Thus, the complete factorization is ( x + 1 ) ( x + 1 ) ( 2 x − 1 ) = ( x + 1 ) 2 ( 2 x − 1 ) .

Final Answer Therefore, the factorizations are: (a) x 3 − 3 x 2 − x + 3 = ( x − 3 ) ( x − 1 ) ( x + 1 ) (b) 2 x 3 + 3 x 2 − 1 = ( x + 1 ) 2 ( 2 x − 1 )


Examples
Factoring polynomials is a fundamental skill in algebra and is used in many areas of mathematics and engineering. For example, in physics, you might need to factor a polynomial to find the possible energy levels of a quantum system. In engineering, you might use factoring to simplify an expression that describes the behavior of a circuit or a mechanical system. Factoring also helps in solving polynomial equations, which arise in various optimization problems and modeling scenarios.

Answered by GinnyAnswer | 2025-07-08