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

Ask by Lane Schofield. in South Africa
Mar 11,2025

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Answer

Here are the factorizations for the given cubic polynomials: (a) \( x^{3}+4 x^{2}+x-6 = (x-1)(x+2)(x+3) \) (b) \( x^{3}-3 x^{2}-4 x+12 = (x-3)(x-2)(x+2) \) (c) \( x^{3}+8 x^{2}-11 x-18 = (x-2)(x+1)(x+9) \) (d) \( 2 x^{3}-3 x^{2}-2 x+3 = (x-1)(x+1)(2x-3) \) (e) \( 4 x^{3}-5 x^{2}-23 x+6 = (x-3)(x+2)(4x-1) \) (f) \( 2 x^{3}-16 x^{2}+10 x+28 = 2(x-7)(x-2)(x+1) \) (g) \( x^{3}-7 x-6 = (x-3)(x+1)(x+2) \) (h) \( x^{3}-13 x-12 = (x-4)(x+1)(x+3) \) (i) \( 2 x^{3}+7 x^{2}+4 x-4 = (x+2)^{2}(2x-1) \) (j) \( -4 x^{3}+3 x+1 = -(x-1)(2x+1)^{2} \)

Solution

Factor the expression by following steps: - step0: Factor: \(-4x^{3}+3x+1\) - step1: Calculate: \(-4x^{3}-4x^{2}-x+4x^{2}+4x+1\) - step2: Rewrite the expression: \(-x\times 4x^{2}-x\times 4x-x+4x^{2}+4x+1\) - step3: Factor the expression: \(-x\left(4x^{2}+4x+1\right)+4x^{2}+4x+1\) - step4: Factor the expression: \(\left(-x+1\right)\left(4x^{2}+4x+1\right)\) - step5: Factor the expression: \(\left(-x+1\right)\left(2x+1\right)^{2}\) - step6: Factor the expression: \(-\left(x-1\right)\left(2x+1\right)^{2}\) Factor the expression \( 4 x^{3}-5 x^{2}-23 x+6 \). Factor the expression by following steps: - step0: Factor: \(4x^{3}-5x^{2}-23x+6\) - step1: Calculate: \(4x^{3}+7x^{2}-2x-12x^{2}-21x+6\) - step2: Rewrite the expression: \(x\times 4x^{2}+x\times 7x-x\times 2-3\times 4x^{2}-3\times 7x+3\times 2\) - step3: Factor the expression: \(x\left(4x^{2}+7x-2\right)-3\left(4x^{2}+7x-2\right)\) - step4: Factor the expression: \(\left(x-3\right)\left(4x^{2}+7x-2\right)\) - step5: Factor the expression: \(\left(x-3\right)\left(x+2\right)\left(4x-1\right)\) Factor the expression \( x^{3}-3 x^{2}-4 x+12 \). Factor the expression by following steps: - step0: Factor: \(x^{3}-3x^{2}-4x+12\) - step1: Calculate: \(x^{3}-4x-3x^{2}+12\) - step2: Rewrite the expression: \(x\times x^{2}-x\times 4-3x^{2}+3\times 4\) - step3: Factor the expression: \(x\left(x^{2}-4\right)-3\left(x^{2}-4\right)\) - step4: Factor the expression: \(\left(x-3\right)\left(x^{2}-4\right)\) - step5: Factor the expression: \(\left(x-3\right)\left(x-2\right)\left(x+2\right)\) Factor the expression \( x^{3}-7 x-6 \). Factor the expression by following steps: - step0: Factor: \(x^{3}-7x-6\) - step1: Calculate: \(x^{3}+3x^{2}+2x-3x^{2}-9x-6\) - step2: Rewrite the expression: \(x\times x^{2}+x\times 3x+x\times 2-3x^{2}-3\times 3x-3\times 2\) - step3: Factor the expression: \(x\left(x^{2}+3x+2\right)-3\left(x^{2}+3x+2\right)\) - step4: Factor the expression: \(\left(x-3\right)\left(x^{2}+3x+2\right)\) - step5: Factor the expression: \(\left(x-3\right)\left(x+1\right)\left(x+2\right)\) Factor the expression \( 2 x^{3}-16 x^{2}+10 x+28 \). Factor the expression by following steps: - step0: Factor: \(2x^{3}-16x^{2}+10x+28\) - step1: Rewrite the expression: \(2x^{3}-2\times 8x^{2}+2\times 5x+2\times 14\) - step2: Factor the expression: \(2\left(x^{3}-8x^{2}+5x+14\right)\) - step3: Factor the expression: \(2\left(x-7\right)\left(x^{2}-x-2\right)\) - step4: Factor the expression: \(2\left(x-7\right)\left(x-2\right)\left(x+1\right)\) Factor the expression \( x^{3}+8 x^{2}-11 x-18 \). Factor the expression by following steps: - step0: Factor: \(x^{3}+8x^{2}-11x-18\) - step1: Calculate: \(x^{3}+10x^{2}+9x-2x^{2}-20x-18\) - step2: Rewrite the expression: \(x\times x^{2}+x\times 10x+x\times 9-2x^{2}-2\times 10x-2\times 9\) - step3: Factor the expression: \(x\left(x^{2}+10x+9\right)-2\left(x^{2}+10x+9\right)\) - step4: Factor the expression: \(\left(x-2\right)\left(x^{2}+10x+9\right)\) - step5: Factor the expression: \(\left(x-2\right)\left(x+1\right)\left(x+9\right)\) Factor the expression \( x^{3}-13 x-12 \). Factor the expression by following steps: - step0: Factor: \(x^{3}-13x-12\) - step1: Calculate: \(x^{3}+4x^{2}+3x-4x^{2}-16x-12\) - step2: Rewrite the expression: \(x\times x^{2}+x\times 4x+x\times 3-4x^{2}-4\times 4x-4\times 3\) - step3: Factor the expression: \(x\left(x^{2}+4x+3\right)-4\left(x^{2}+4x+3\right)\) - step4: Factor the expression: \(\left(x-4\right)\left(x^{2}+4x+3\right)\) - step5: Factor the expression: \(\left(x-4\right)\left(x+1\right)\left(x+3\right)\) Factor the expression \( x^{3}+4 x^{2}+x-6 \). Factor the expression by following steps: - step0: Factor: \(x^{3}+4x^{2}+x-6\) - step1: Calculate: \(x^{3}+5x^{2}+6x-x^{2}-5x-6\) - step2: Rewrite the expression: \(x\times x^{2}+x\times 5x+x\times 6-x^{2}-5x-6\) - step3: Factor the expression: \(x\left(x^{2}+5x+6\right)-\left(x^{2}+5x+6\right)\) - step4: Factor the expression: \(\left(x-1\right)\left(x^{2}+5x+6\right)\) - step5: Factor the expression: \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\) Factor the expression \( 2 x^{3}+7 x^{2}+4 x-4 \). Factor the expression by following steps: - step0: Factor: \(2x^{3}+7x^{2}+4x-4\) - step1: Calculate: \(2x^{3}-x^{2}+8x^{2}-4x+8x-4\) - step2: Rewrite the expression: \(x^{2}\times 2x-x^{2}+4x\times 2x-4x+4\times 2x-4\) - step3: Factor the expression: \(x^{2}\left(2x-1\right)+4x\left(2x-1\right)+4\left(2x-1\right)\) - step4: Factor the expression: \(\left(x^{2}+4x+4\right)\left(2x-1\right)\) - step5: Factor the expression: \(\left(x+2\right)^{2}\left(2x-1\right)\) Factor the expression \( 2 x^{3}-3 x^{2}-2 x+3 \). Factor the expression by following steps: - step0: Factor: \(2x^{3}-3x^{2}-2x+3\) - step1: Rewrite the expression: \(x^{2}\times 2x-x^{2}\times 3-2x+3\) - step2: Factor the expression: \(x^{2}\left(2x-3\right)-\left(2x-3\right)\) - step3: Factor the expression: \(\left(x^{2}-1\right)\left(2x-3\right)\) - step4: Factor the expression: \(\left(x-1\right)\left(x+1\right)\left(2x-3\right)\) Here are the factorizations for the given cubic polynomials: (a) \( x^{3}+4 x^{2}+x-6 \) **Factorization:** \( (x-1)(x+2)(x+3) \) (b) \( x^{3}-3 x^{2}-4 x+12 \) **Factorization:** \( (x-3)(x-2)(x+2) \) (c) \( x^{3}+8 x^{2}-11 x-18 \) **Factorization:** \( (x-2)(x+1)(x+9) \) (d) \( 2 x^{3}-3 x^{2}-2 x+3 \) **Factorization:** \( (x-1)(x+1)(2x-3) \) (e) \( 4 x^{3}-5 x^{2}-23 x+6 \) **Factorization:** \( (x-3)(x+2)(4x-1) \) (f) \( 2 x^{3}-16 x^{2}+10 x+28 \) **Factorization:** \( 2(x-7)(x-2)(x+1) \) (g) \( x^{3}-7 x-6 \) **Factorization:** \( (x-3)(x+1)(x+2) \) (h) \( x^{3}-13 x-12 \) **Factorization:** \( (x-4)(x+1)(x+3) \) (i) \( 2 x^{3}+7 x^{2}+4 x-4 \) **Factorization:** \( (x+2)^{2}(2x-1) \) (j) \( -4 x^{3}+3 x+1 \) **Factorization:** \( -(x-1)(2x+1)^{2} \) If you have any further questions or need additional assistance, feel free to ask!

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Bonus Knowledge

(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?

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