EXERCISE 5.4 Factorise the following cubic polynomials: \( \begin{array}{ll}\text { (a) } x^{3}+4 x^{2}+x-6 & \text { (b) } x^{3}-3 x^{2}-4 x+12 \\ \text { (c) } x^{3}+8 x^{2}-11 x-18 & \text { (d) } 2 x^{3}-3 x^{2}-2 x+3 \\ \text { (e) } 4 x^{3}-5 x^{2}-23 x+6 & \text { (f) } 2 x^{3}-16 x^{2}+10 x+28 \\ \text { (g) } x^{3}-7 x-6 & \text { (h) } x^{3}-13 x-12 \\ \text { (i) } 2 x^{3}+7 x^{2}+4 x-4 & \text { (i) }-4 x^{3}+3 x+1\end{array} \)

(a) To factor \(x^{3}+4x^{2}+x-6\), we can use the Rational Root Theorem to find potential roots, testing \(x=1\) which works. Thus, we can factor it as \((x-1)(x^{2}+5x+6)\), and further factor gives \((x-1)(x+2)(x+3)\). (b) For \(x^{3}-3x^{2}-4x+12\), testing \(x=2\) gives a root. This polynomial can be factored as \((x-2)(x^{2}-x-6)\), which can be further factored into \((x-2)(x-3)(x+2)\). (c) In \(x^{3}+8x^{2}-11x-18\), by the Rational Root Theorem, \(x=-2\) is a root. The factorization is \((x+2)(x^{2}+6x-9)\), which can be factored further as \((x+2)(x+3+\sqrt{18})(x+3-\sqrt{18})\). (d) For \(2x^{3}-3x^{2}-2x+3\), trying \(x=1\) works. We find \((2x-3)(x^{2}+1)\) since we have \((x-1)(\text{quadratic})\). (e) The polynomial \(4x^{3}-5x^{2}-23x+6\) is factored by testing the roots. We can find it to be \((x+3)(4x^{2}-17)\). (f) In \(2x^{3}-16x^{2}+10x+28\), testing leads us to \(x=4\) as the root, giving \(2(x-4)(x^{2}-3)\), which can be factored further as \(2(x-4)(x-\sqrt{3})(x+\sqrt{3})\). (g) For \(x^{3}-7x-6\), we find \(x+3\) is a root. Hence, it factors as \((x+3)(x^{2}-3x-2)\) and we can find further factored quadratic roots. (h) Finally, for \(x^{3}-13x-12\), testing reveals \(x=4\) as a root. Factor as \((x-4)(x^{2}+x+3)\), with no additional real roots. (i) For \(2x^{3}+7x^{2}+4x-4\), using \(x=1\) as a root gives us \(2(x-1)(x^{2}+5x+4)\), factoring down to \((x-1)(x+4)(x+1)\). (j) The polynomial \(-4x^{3}+3x+1\) gives us roots leading to \(-(4x+1)(x^2-x-1)\). Hope this serves you well! It's a fun whirligig of factors, isn't it?