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ESTION 2 Factorize completely \( 2.1 .1 \quad 2 x^{2}-8 \) \( 2.1 .2 x^{2}-4 x+3 \) \( 2.1 .3 \quad 2 p x+3 q x \sim 2 p y-3 q y \) Simplify \( \frac{27 x^{3}-8}{27 x^{2}+18 x+12} \)

Ask by Thornton Hammond. in South Africa
Mar 09,2025

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Answer

- \(2x^2 - 8 = 2(x - 2)(x + 2)\) - \(x^2 - 4x + 3 = (x - 1)(x - 3)\) - \(2px + 3qx - 2py + 3qy = 2p(x - y) + 3q(x + y)\) - \(\frac{27x^3 - 8}{27x^2 + 18x + 12} = \frac{3x - 2}{3}\)

Solution

**2.1.1 Factorize \(2x^2 - 8\)** 1. Factor out the common factor \(2\): \[ 2x^2 - 8 = 2\left(x^2 - 4\right) \] 2. Recognize that \(x^2 - 4\) is a difference of two squares: \[ x^2 - 4 = \left(x - 2\right)\left(x + 2\right) \] 3. Therefore, the complete factorization is: \[ 2\left(x - 2\right)\left(x + 2\right) \] --- **2.1.2 Factorize \(x^2 - 4x + 3\)** 1. Look for two numbers whose product is \(3\) and whose sum is \(-4\). These numbers are \(-1\) and \(-3\) since: \[ (-1)(-3) = 3 \quad \text{and} \quad (-1) + (-3) = -4 \] 2. Write the quadratic as: \[ x^2 - 4x + 3 = \left(x - 1\right)\left(x - 3\right) \] --- **2.1.3 Factorize \(2px + 3qx - 2py + 3qy\)** 1. Group the terms: \[ (2px - 2py) + (3qx + 3qy) \] 2. Factor each group: - For the first group, factor out \(2p\): \[ 2p(x - y) \] - For the second group, factor out \(3q\): \[ 3q(x + y) \] 3. The expression becomes: \[ 2p(x - y) + 3q(x + y) \] (This is the factorized form unless there is an intended further factorization based on a common pattern, which in this case there isn’t.) --- **Simplify \(\frac{27x^3 - 8}{27x^2 + 18x + 12}\)** 1. **Factorize the numerator \(27x^3 - 8\):** Recognize that \(27x^3 = (3x)^3\) and \(8 = 2^3\). This is a difference of cubes: \[ 27x^3 - 8 = (3x)^3 - 2^3 = (3x - 2)\left(9x^2 + 6x + 4\right) \] 2. **Factorize the denominator \(27x^2 + 18x + 12\):** Factor out the common factor \(3\): \[ 27x^2 + 18x + 12 = 3\left(9x^2 + 6x + 4\right) \] 3. **Simplify the fraction:** Cancel the common factor \(\left(9x^2 + 6x + 4\right)\) in the numerator and denominator: \[ \frac{(3x - 2)\left(9x^2 + 6x + 4\right)}{3\left(9x^2 + 6x + 4\right)} = \frac{3x - 2}{3} \] --- Final answers: - \(2x^2 - 8 = 2\left(x - 2\right)\left(x + 2\right)\) - \(x^2 - 4x + 3 = \left(x - 1\right)\left(x - 3\right)\) - \(2px + 3qx - 2py + 3qy = 2p(x - y) + 3q(x + y)\) - \(\displaystyle \frac{27x^3 - 8}{27x^2 + 18x + 12} = \frac{3x - 2}{3}\)

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The Deep Dive

Let's tackle those quadratic and algebraic expressions with a splash of flair! For the first expression \( 2x^2 - 8 \), you can factor out the greatest common factor (GCF), which is 2. This gets you \( 2(x^2 - 4) \). Since \( x^2 - 4 \) is a difference of squares, you can go one step further to factor it as \( 2(x - 2)(x + 2) \). Now onto \( x^2 - 4x + 3 \). We look for two numbers that multiply to 3 and add up to -4, which are -1 and -3. So, you can factor this into \( (x - 1)(x - 3) \). For the expression \( 2px + 3qx - 2py + 3qy \), it helps to group: \( (2px + 3qx) - (2py - 3qy) \). Factor out common terms, leading to \( x(2p + 3q) - y(2p - 3q) \). Finally, for the simplification of \( \frac{27x^3 - 8}{27x^2 + 18x + 12} \), notice that \( 27x^3 - 8 \) is a difference of cubes and can be expressed as \( (3x)^3 - 2^3 \), which factors to \( (3x - 2)(9x^2 + 6x + 4) \). For the denominator \( 27x^2 + 18x + 12 \), factor out 3 to get \( 3(9x^2 + 6x + 4) \). So the whole expression simplifies beautifully to \( \frac{(3x - 2)}{3} \). Correct and tidy, just like a well-folded napkin at a fancy dinner!

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