MyStudyPals.com MyStudyPals.com
  • Login
  • Feed

Resources

  • 📰
    Latest School News
  • 📚
    Scholarships
  • 🎓
    JAMB Syllabus
  • 📝
    Jamb Brochure

Information

  • 🔗
    Contact US
  • 🔗
    Privacy Policy
  • 🔗
    Terms and Conditions
  • 🔗
    About US

Find the roots of x\(^3\) – 2x\(^2\) – 5x + 6 = 0

  • 📑 Past Questions
  • ⌛ Take CBT exam
  • 📟 Referral
  • 🎫 Support
  • 📢 Earnings
Mathematics JAMB 2006

Find the roots of x\(^3\) – 2x\(^2\) – 5x + 6 = 0

  • 1, -2, 3 checkmark
  • 1, 2, -3,
  • -1, -2, 3
  • -1, 2, -3

The correct answer is: A

Explanation

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.

First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.

Then, a\(_0\) which is the coefficient void of x = 6.

The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.

The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).

= \(\pm (1, 2, 3, 6)\).

Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.

Therefore x = 1 is a root of the equation.

Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.

x\(^2\) - x - 6 = (x - 3)(x + 2).

x = -2, 3.

\(\therefore\) The roots of the equation = 1, -2 and 3.

Previous Question Next Question

Leave A Comment

© 2025 MyStudyPals.com