JAMB - Mathematics (1979 - No. 26)

If x\(^3\) - 12x - 16 = 0 has x = -2 as a solution then the equation has
x = -4 as a solution also
3 roots all different
3 roots with two equal and the third different
3 roots all equal
only one root

Explanation

Given the equation 

\[x^3 - 12x - 16 = 0\]

with \( x = -2 \) as a solution, we can factor it as follows:

 Divide by \( x + 2 \) gives:

\[x^3 - 12x - 16 = (x + 2)(x^2 - 2x - 8)\]

Finding Roots of \( x^2 - 2x - 8 = 0 \)

Using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{36}}{2}\]

This gives:\[x = 4 \quad \text{and} \quad x = -2 \text{ (with multiplicity 2)}\]

 

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