Given that \(x-3\) and \(x+1\) are both factors of \(2x^3-3x^2-8x-3\), solve the equation \(2x^3-3x^2-8x-3=0\).[2]
The polynomial \(\mathrm{p}(x)=x^3+ax^2+bx+c\), where \(a\), \(b\) and \(c\) are constants, has remainder \(-5\) when divided by \(x-1\). The curve \(y=\mathrm{p}(x)\) has stationary points at \(x=\dfrac43\) and \(x=2\).
Find the values of \(a\), \(b\) and \(c\).[7]
Hence use the second derivative test to show that the stationary point at \(x=2\) is a minimum.[2]
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Answer:(a) \(x=3,-1,-\dfrac12\). (b)(i) \(a=-5,b=8,c=-9\). (ii) \(\mathrm{p}''(2)=2>0\), so a minimum.