To simplify \( \frac{5x^2 + 10x^2 + 7x + 12}{x^2 - 4x^2 + x - 6} \), we must first simplify both the numerator and the denominator.
Step 1: Simplify the Numerator
The numerator is: \[ 5x^2 + 10x^2 + 7x + 12 = (5x^2 + 10x^2) + 7x + 12 = 15x^2 + 7x + 12 \]
Step 2: Simplify the Denominator
The denominator is: \[ x^2 - 4x^2 + x - 6 = (x^2 - 4x^2) + x - 6 = -3x^2 + x - 6 \]
So the expression simplifies to: \[ \frac{15x^2 + 7x + 12}{-3x^2 + x - 6} \]
Step 3: Factor the Numerator and Denominator (if possible)
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Numerator: \( 15x^2 + 7x + 12 \)
To factor \( 15x^2 + 7x + 12 \), we look for two numbers that multiply to \( 15 \times 12 = 180 \) (the product of \( a \) and \( c \)) and add up to \( 7 \) (the coefficient of \( b \)). This can be challenging, and it is not easily factorable with rational coefficients.
Thus, we will leave it as \( 15x^2 + 7x + 12 \).
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Denominator: \( -3x^2 + x - 6 \)
We will factor out -1 to make it easier: \[ -1(3x^2 - x + 6) \]
Now, we can look for factors of \( 3x^2 - x + 6 \). However, this also does not seem to factor easily. The discriminant here ( \( b^2 - 4ac = (-1)^2 - 4 \cdot 3 \cdot 6 \) ) is negative, meaning this quadratic does not have real roots.
Step 4: Rewrite Simplified Form
So we end up with: \[ \frac{15x^2 + 7x + 12}{-3x^2 + x - 6} = -\frac{15x^2 + 7x + 12}{3x^2 - x + 6} \]
Result
Since we cannot factor further, the simplest form remains: \[ -\frac{15x^2 + 7x + 12}{3x^2 - x + 6} \]
As per your options, if we were to look for a clear linear form, you'd have to use numerical or graphical methods to find specific x-values for given scenarios to rewrite or simplify it further.
Given your multiple-choice answers and without specific factoring yielding a clear linear completion, it seems they don't provide an exact matching response based on straightforward algebraic simplification. However, you might have meant to present additional context or transformation rules for ultimate choices in responses.
There is no exact match from the options provided based on the simplifications carried here.