It is given that \(\mathrm{f}(x)=4x^3+ax^2+bx+3\), where \(a\) and \(b\) are constants, has a factor of \(2x-1\) and leaves a remainder of 18 when divided by \(x+1\).
Find the values of \(a\) and \(b\).[6]
Using these values of \(a\) and \(b\), solve \(\mathrm{f}(x)=0\).[3]
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Answer:(a) \(3(3x-y)(3x^2+y^2)\). (b)(i) \(a=8\), \(b=-11\). (ii) \(x=\dfrac12,-3\).