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the area of a rectangle is ((x^3 - 5x^2 + 3x - 15)), and the width of t…

Question

the area of a rectangle is ((x^3 - 5x^2 + 3x - 15)), and the width of the rectangle is ((x^2 + 3)). if area = length × width, what is the length of the rectangle?
options: (x - 15), (x + 15), (x + 5), (x - 5)

Explanation:

Step1: Recall the formula for length

Since \( \text{Area} = \text{length} \times \text{width} \), we can solve for length by dividing the area by the width. So, \( \text{length} = \frac{\text{Area}}{\text{Width}} \). Substituting the given expressions, we have \( \text{length} = \frac{x^3 - 5x^2 + 3x - 15}{x^2 + 3} \).

Step2: Factor the numerator

We factor the numerator by grouping. Group the first two terms and the last two terms: \( (x^3 - 5x^2) + (3x - 15) \). Factor out the greatest common factor from each group: \( x^2(x - 5) + 3(x - 5) \). Now, we can factor out \( (x - 5) \) from both terms, resulting in \( (x - 5)(x^2 + 3) \).

Step3: Divide the factored numerator by the denominator

Substitute the factored numerator back into the fraction: \( \frac{(x - 5)(x^2 + 3)}{x^2 + 3} \). Since \( x^2 + 3
eq 0 \) for all real \( x \), we can cancel out the \( x^2 + 3 \) terms, leaving us with \( x - 5 \).

Answer:

\( x - 5 \) (corresponding to the option with \( x - 5 \))