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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