The factors of x3+2x2−5x−6 is :
A(x + 1)(x - 2)(x + 3)
B(x - 1)(x - 2)(x + 3)
C(x - 1)(x - 2)(x - 3)
D(x + 1)(x - 2)(x - 3)
Answer:
A. (x + 1)(x - 2)(x + 3)
Read Explanation:
Polynomial Factorization Techniques
To find the factors of a cubic polynomial like $P(x) = x^3+2x^2-5x-6$, several methods can be employed, which are commonly tested in competitive exams.
1. Rational Root Theorem
- The Rational Root Theorem states that if a polynomial has integer coefficients, then any rational root, $ rac{p}{q}$, must have 'p' as a factor of the constant term and 'q' as a factor of the leading coefficient.
- In $x^3+2x^2-5x-6$, the constant term is -6 and the leading coefficient is 1.
- Factors of -6 are: $\pm 1, \pm 2, \pm 3, \pm 6$.
- Factors of 1 are: $\pm 1$.
- Possible rational roots are: $\pm 1, \pm 2, \pm 3, \pm 6$.
2. Testing Potential Roots
- We substitute these possible roots into the polynomial to find which ones make $P(x) = 0$.
- Test $x = -1$: $P(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0$. Thus, $(x+1)$ is a factor.
- Test $x = 2$: $P(2) = (2)^3 + 2(2)^2 - 5(2) - 6 = 8 + 8 - 10 - 6 = 0$. Thus, $(x-2)$ is a factor.
- Test $x = -3$: $P(-3) = (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 18 + 15 - 6 = 0$. Thus, $(x+3)$ is a factor.
3. Polynomial Division or Synthetic Division
- Once a root (say $x = -1$) is found, we know $(x+1)$ is a factor. We can then divide the original polynomial by $(x+1)$ to obtain a quadratic polynomial.
- Using synthetic division with root -1:
-1 | 1 2 -5 -6
| -1 -1 6
-----------------
1 1 -6 0
4. Alternative Factoring (Grouping)
- Sometimes, direct grouping can be applied if the terms align properly, though this is less common for general cubic polynomials.
Key Concepts for Exams
- Understanding the relationship between roots and factors of a polynomial.
- Proficiency in applying the Rational Root Theorem.
- Skill in performing polynomial division or synthetic division accurately.
- Ability to factor quadratic expressions.