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Question 119
If $$(x^2 - x - 6)$$ is a factor of $$x^4 + ax^3 - x^2 - bx - 12 = 0$$, what are the values of constants a and b, respectively, in the above equation ?
Solution
$$(x^2 - x - 6)$$ is a factor of the fourth degree equation. Hence, the roots of $$(x^2 - x - 6)$$ (3 and -2) will satisfy the fourth degree equation.
Putting x=3 in $$x^4 + ax^3 - x^2 - bx - 12 = 0$$, we get 81+27a-9-3b-12=0, or 27a-3b= -60..............(1)
Putting x= -2 in the equation, we get 16-8a-4+2b-12=0, or 8a-2b=0............ (2)
Solving the two, we get a as -4 and b as -16
