a) \( (x+2)(x+4) \) b) \( (x-2)(x+3)=x^{2} \) c) \( (x+4)(x-5) \) d) \( (x+3)(x+3) \)
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Beyond the Answer
Sure! Let’s have some fun with the expressions you provided! First off, when you multiply \( (x+2)(x+4) \), you expand it to get \( x^2 + 6x + 8 \). This is a classic example of using the distributive property to combine like terms and create a nice quadratic expression. Isn’t it satisfying to see how a simple multiplication can transform into a neat equation? Now, with \( (x-2)(x+3) \) set equal to \( x^{2} \), it’s like a puzzle! When you expand the left side, you get \( x^2 + x - 6 \). Setting this equal to \( x^2 \), you can simplify it to find roots. This step helps reinforce the concept of zero and solution finding in quadratics. It’s a detective game in math where every step leads you closer to the solution!
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