Fully factorise each of the following expressions. \( \begin{array}{ll}\text { 1. } x^{2}+7 x+12 & \text { 2. } 2 x^{2}-14 x+24 \\ \text { 3. } x^{2}+x-12 & \text { 4. } 2 x^{2}-2 x-24 \\ \text { 5. } 2 x^{2}-4 x-30 & \text { 6. } 2 x^{2}+4 x-16\end{array} \)
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The Deep Dive
Let's fully factorise each of the given expressions step by step. 1. **\( x^{2}+7x+12 \)** This expression can be factorised by finding two numbers that multiply to 12 and add up to 7. The numbers 3 and 4 do the trick! Thus, we have: \[ (x + 3)(x + 4) \] 2. **\( 2x^{2}-14x+24 \)** We can start by factoring out the common factor of 2: \[ 2(x^{2}-7x+12) \] Now, we can factor \( x^{2}-7x+12 \) the same way as before to get: \[ 2(x - 3)(x - 4) \] 3. **\( x^{2}+x-12 \)** Here, we need numbers that multiply to -12 and add up to 1. The numbers 4 and -3 are a perfect fit! Thus, we can factor it as: \[ (x + 4)(x - 3) \] 4. **\( 2x^{2}-2x-24 \)** We can factor out the common factor of 2 first: \[ 2(x^{2}-x-12) \] Now we need to factor \( x^{2}-x-12 \). The numbers that work are 4 and -3, which gives us: \[ 2(x - 4)(x + 3) \] 5. **\( 2x^{2}-4x-30 \)** Again, we start by factoring out the common factor of 2: \[ 2(x^{2}-2x-15) \] Now we need to factor \( x^{2}-2x-15 \). The numbers -5 and 3 will work, so: \[ 2(x - 5)(x + 3) \] 6. **\( 2x^{2}+4x-16 \)** We can factor out the common factor of 2 as well: \[ 2(x^{2}+2x-8) \] Now, to factor \( x^{2}+2x-8 \), we need numbers that multiply to -8 and add to 2. The pair 4 and -2 works! Thus, we have: \[ 2(x + 4)(x - 2) \] So the fully factorised forms are: 1. \( (x + 3)(x + 4) \) 2. \( 2(x - 3)(x - 4) \) 3. \( (x + 4)(x - 3) \) 4. \( 2(x - 4)(x + 3) \) 5. \( 2(x - 5)(x + 3) \) 6. \( 2(x + 4)(x - 2) \)
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