EquationsMCQMTP Dec 22 Series IIQuestion 1117 of 221
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The roots of the cubic eq. x3−7x+6=0\displaystyle x^3 - 7x + 6 = 0 are:

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — $1, 2and\displaystyle and-3$

All Options:

  • A$1, 2and\displaystyle and3$
  • B$1, -2and\displaystyle and3$
  • C$1, 2and\displaystyle and-3$
  • D$1, -2and\displaystyle and-3$

Detailed Solution & Explanation

Let us solve the cubic equation x3−7x+6=0\displaystyle x^3 - 7x + 6 = 0.

By trial and error, let us test x=1\displaystyle x = 1:
13−7(1)+6=1−7+6=01^3 - 7(1) + 6 = 1 - 7 + 6 = 0
Since x=1\displaystyle x = 1 is a root, (x−1)\displaystyle (x-1) is a factor.

Now, divide x3−7x+6\displaystyle x^3 - 7x + 6 by (x−1)\displaystyle (x-1):
x3−7x+6=(x−1)(x2+x−6)x^3 - 7x + 6 = (x-1)(x^2 + x - 6)
Factoring the quadratic term x2+x−6\displaystyle x^2 + x - 6:
x2+x−6=(x−2)(x+3)x^2 + x - 6 = (x-2)(x+3)

So the factored form is:
(x−1)(x−2)(x+3)=0(x-1)(x-2)(x+3) = 0
Which gives the roots:
x=1,x=2,x=−3\displaystyle x = 1, x = 2, x = -3

Thus, the roots are $1, 2and\displaystyle and-3$, which corresponds to Option (c).

Hence, the correct option is **Option (c)**.

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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