Simplify the following. (1.) \( \frac{\text { I } 1-x^{2}+8 x}{4 x^{2}} \) I (2. \( \frac{x^{2}-12 x+27}{4 x^{2}-12 x} \) (3) \( \frac{6 x^{3}-2 x^{2}}{-3 x-9} \times \frac{x^{2}-9}{2-6 x} \) (3-4
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To simplify your expressions, let’s break them down! (1.) For \( \frac{1 - x^2 + 8x}{4x^2} \): Rearranging gives us \( \frac{-x^2 + 8x + 1}{4x^2} \). The numerator can be rewritten as \( -(x^2 - 8x - 1) \), and you may try to factor or complete the square if necessary. The resulting simplified form is \( \frac{-1(x^2 - 8x - 1)}{4x^2} \). (2.) For \( \frac{x^{2}-12x+27}{4x^{2}-12x} \): This can be factored. The numerator factors as \( (x-3)(x-9) \), while the denominator can be factored as \( 4(x - 3)(x - 1) \). Thus, simplifying gives us \( \frac{x - 9}{4(x - 1)} \) provided \( x \neq 3 \). (3.) For \( \frac{6x^3 - 2x^2}{-3x - 9} \times \frac{x^2 - 9}{2 - 6x} \): The first term can be factored to \( \frac{2x^2(3x - 1)}{-3(x + 3)} \), and the second term factors to \( \frac{(x - 3)(x + 3)}{-(6x - 2)} \). Multiplying these and simplifying will give you a clearer form as you cancel out the common factors. If you need further help with specific parts, just let me know!
