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F3/02/25 Grade 10 Factorise \( \begin{array}{ll}\text { a) } x^{2}-7 x+12 & \text { (h) } 40 x^{2}+6 x y-18 y^{2} \\ \begin{array}{ll}\text { (b) } 2 x^{2}+12 x+18 & \text { (b) } 24 x^{2}-2 x y-y^{2} \\ \text { c) } x^{2}-9 x+20 & \text { (j) } 3 x^{2}+19 x+20 \\ \text { (d) } 3 x^{2}-9 x+24 & \\ \text { (e) } 2 x^{2}+5 x+3 & \\ \text { (f) } 3 x^{2}+4 x-4 & \\ \text { (g) } 5 x^{3} y+25 x^{2} y-30 x y & \end{array}\end{array}> \)

Ask by Reeves Mullins. in South Africa
Feb 03,2025

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Answer

Here are the factorizations: a) \( x^{2}-7x+12 = (x-4)(x-3) \) b) \( 2x^{2}+12x+18 = 2(x+3)^{2} \) c) \( x^{2}-9x+20 = (x-5)(x-4) \) d) \( 3x^{2}-9x+24 = 3(x^{2}-3x+8) \) e) \( 2x^{2}+5x+3 = (x+1)(2x+3) \) f) \( 3x^{2}+4x-4 = (x+2)(3x-2) \) g) \( 5x^{3}y+25x^{2}y-30xy = 5xy(x-1)(x+6) \) h) \( 40x^{2}+6xy-18y^{2} = 2(5x-3y)(4x+3y) \) i) \( 24x^{2}-2xy-y^{2} = (6x+y)(4x-y) \) j) \( 3x^{2}+19x+20 = (x+5)(3x+4) \)

Solution

Factor the expression by following steps: - step0: Factor: \(2x^{2}+12x+18\) - step1: Factor the expression: \(2\left(x+3\right)^{2}\) Factor the expression \( 3 x^{2}-9 x+24 \). Factor the expression by following steps: - step0: Factor: \(3x^{2}-9x+24\) - step1: Factor the expression: \(3\left(x^{2}-3x+8\right)\) Factor the expression \( 3 x^{2}+19 x+20 \). Factor the expression by following steps: - step0: Factor: \(3x^{2}+19x+20\) - step1: Rewrite the expression: \(3x^{2}+\left(4+15\right)x+20\) - step2: Rewrite the expression: \(3x^{2}+4x+15x+20\) - step3: Factor the expression: \(x\left(3x+4\right)+5\left(3x+4\right)\) - step4: Factor the expression: \(\left(x+5\right)\left(3x+4\right)\) Factor the expression \( 2 x^{2}+5 x+3 \). Factor the expression by following steps: - step0: Factor: \(2x^{2}+5x+3\) - step1: Rewrite the expression: \(2x^{2}+\left(3+2\right)x+3\) - step2: Rewrite the expression: \(2x^{2}+3x+2x+3\) - step3: Factor the expression: \(x\left(2x+3\right)+2x+3\) - step4: Factor the expression: \(\left(x+1\right)\left(2x+3\right)\) Factor the expression \( x^{2}-9 x+20 \). Factor the expression by following steps: - step0: Factor: \(x^{2}-9x+20\) - step1: Rewrite the expression: \(x^{2}+\left(-4-5\right)x+20\) - step2: Rewrite the expression: \(x^{2}-4x-5x+20\) - step3: Factor the expression: \(x\left(x-4\right)-5\left(x-4\right)\) - step4: Factor the expression: \(\left(x-5\right)\left(x-4\right)\) Factor the expression \( x^{2}-7 x+12 \). Factor the expression by following steps: - step0: Factor: \(x^{2}-7x+12\) - step1: Rewrite the expression: \(x^{2}+\left(-3-4\right)x+12\) - step2: Rewrite the expression: \(x^{2}-3x-4x+12\) - step3: Factor the expression: \(x\left(x-3\right)-4\left(x-3\right)\) - step4: Factor the expression: \(\left(x-4\right)\left(x-3\right)\) Factor the expression \( 3 x^{2}+4 x-4 \). Factor the expression by following steps: - step0: Factor: \(3x^{2}+4x-4\) - step1: Rewrite the expression: \(3x^{2}+\left(-2+6\right)x-4\) - step2: Rewrite the expression: \(3x^{2}-2x+6x-4\) - step3: Factor the expression: \(x\left(3x-2\right)+2\left(3x-2\right)\) - step4: Factor the expression: \(\left(x+2\right)\left(3x-2\right)\) Factor the expression \( 40 x^{2}+6 x y-18 y^{2 \). Factor the expression by following steps: - step0: Factor: \(40x^{2}+6xy-18y^{2}\) - step1: Factor the expression: \(2\left(20x^{2}+15xy-12yx-9y^{2}\right)\) - step2: Factor the expression: \(2\left(5x\left(4x+3y\right)-3y\left(4x+3y\right)\right)\) - step3: Factor the expression: \(2\left(5x-3y\right)\left(4x+3y\right)\) Factor the expression \( 24 x^{2}-2 x y-y^{2 \). Factor the expression by following steps: - step0: Factor: \(24x^{2}-2xy-y^{2}\) - step1: Rewrite the expression: \(24x^{2}+\left(-6+4\right)xy-y^{2}\) - step2: Rewrite the expression: \(24x^{2}-6xy+4yx-y^{2}\) - step3: Factor the expression: \(6x\left(4x-y\right)+y\left(4x-y\right)\) - step4: Factor the expression: \(\left(6x+y\right)\left(4x-y\right)\) Factor the expression \( 5 x^{3} y+25 x^{2} y-30 x y \). Factor the expression by following steps: - step0: Factor: \(5x^{3}y+25x^{2}y-30xy\) - step1: Factor the expression: \(5\left(x^{3}y+5x^{2}y-6xy\right)\) - step2: Factor the expression: \(5xy\left(x^{2}+5x-6\right)\) - step3: Factor the expression: \(5xy\left(x-1\right)\left(x+6\right)\) Here are the factorizations for each expression: a) \( x^{2}-7x+12 \) **Factorization:** \( (x-4)(x-3) \) b) \( 2x^{2}+12x+18 \) **Factorization:** \( 2(x+3)^{2} \) c) \( x^{2}-9x+20 \) **Factorization:** \( (x-5)(x-4) \) d) \( 3x^{2}-9x+24 \) **Factorization:** \( 3(x^{2}-3x+8) \) (This does not factor further over the reals) e) \( 2x^{2}+5x+3 \) **Factorization:** \( (x+1)(2x+3) \) f) \( 3x^{2}+4x-4 \) **Factorization:** \( (x+2)(3x-2) \) g) \( 5x^{3}y+25x^{2}y-30xy \) **Factorization:** \( 5xy(x-1)(x+6) \) h) \( 40x^{2}+6xy-18y^{2} \) **Factorization:** \( 2(5x-3y)(4x+3y) \) i) \( 24x^{2}-2xy-y^{2} \) **Factorization:** \( (6x+y)(4x-y) \) j) \( 3x^{2}+19x+20 \) **Factorization:** \( (x+5)(3x+4) \) If you have any further questions or need additional assistance, feel free to ask!

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Extra Insights

For \( x^{2}-7x+12 \), we seek two numbers that multiply to 12 (the constant) and add up to -7 (the coefficient of x). Those numbers are -3 and -4. Thus, the factorization is \( (x-3)(x-4) \). For \( 40x^{2}+6xy-18y^{2} \), we can first factor out the greatest common factor, which is 2. This gives us \( 2(20x^{2}+3xy-9y^{2}) \). Next, we look for factors of \( 20 \cdot -9 = -180 \) that add up to 3. The pair (15, -12) fits, enabling us to rewrite and factor it further, giving \( 2(20x^{2}+15xy-12xy-9y^{2}) \) which can be factored into \( 2(5x-3y)(4x+3y) \).

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