Answer :
To determine which expression is equivalent to [tex]\(\frac{2d - 6}{d^2 + 2d - 48} \div \frac{d - 3}{2d + 16}\)[/tex] if no denominator equals zero, we can simplify the given expression through a series of algebraic steps.
First, let's rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{2d - 6}{d^2 + 2d - 48} \div \frac{d - 3}{2d + 16} = \frac{2d - 6}{d^2 + 2d - 48} \cdot \frac{2d + 16}{d - 3} \][/tex]
Next, simplify the expression [tex]\(\frac{2d - 6}{d^2 + 2d - 48}\)[/tex].
Notice that [tex]\(2d - 6\)[/tex] can be factored:
[tex]\[ 2d - 6 = 2(d - 3) \][/tex]
Now factor [tex]\(d^2 + 2d - 48\)[/tex]. This expression can be factored using the method of factoring quadratic trinomials:
[tex]\[ d^2 + 2d - 48 = (d + 8)(d - 6) \][/tex]
Putting it all together, we get:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \][/tex]
Now consider the expression [tex]\(\frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2d + 16}{d - 3}\)[/tex].
Notice that [tex]\(2d + 16\)[/tex] can be factored:
[tex]\[ 2d + 16 = 2(d + 8) \][/tex]
So the product becomes:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2(d + 8)}{d - 3} \][/tex]
Next, cancel the common factors in the numerator and the denominator:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2(d + 8)}{d - 3} = \frac{2(d - 3) \cdot 2(d + 8)}{(d + 8)(d - 6) \cdot (d - 3)} \][/tex]
[tex]\[ = \frac{4(d + 8)}{(d - 6) \cdot (d + 8)} \][/tex]
Cancel the [tex]\((d + 8)\)[/tex] from the numerator and the denominator:
[tex]\[ = \frac{4}{d - 6} \][/tex]
Thus, the equivalent expression is:
[tex]\[ \boxed{\frac{4}{d - 6}} \][/tex]
Therefore, the correct answer is:
B. [tex]\(\frac{4}{d - 6}\)[/tex]
First, let's rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{2d - 6}{d^2 + 2d - 48} \div \frac{d - 3}{2d + 16} = \frac{2d - 6}{d^2 + 2d - 48} \cdot \frac{2d + 16}{d - 3} \][/tex]
Next, simplify the expression [tex]\(\frac{2d - 6}{d^2 + 2d - 48}\)[/tex].
Notice that [tex]\(2d - 6\)[/tex] can be factored:
[tex]\[ 2d - 6 = 2(d - 3) \][/tex]
Now factor [tex]\(d^2 + 2d - 48\)[/tex]. This expression can be factored using the method of factoring quadratic trinomials:
[tex]\[ d^2 + 2d - 48 = (d + 8)(d - 6) \][/tex]
Putting it all together, we get:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \][/tex]
Now consider the expression [tex]\(\frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2d + 16}{d - 3}\)[/tex].
Notice that [tex]\(2d + 16\)[/tex] can be factored:
[tex]\[ 2d + 16 = 2(d + 8) \][/tex]
So the product becomes:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2(d + 8)}{d - 3} \][/tex]
Next, cancel the common factors in the numerator and the denominator:
[tex]\[ \frac{2(d - 3)}{(d + 8)(d - 6)} \cdot \frac{2(d + 8)}{d - 3} = \frac{2(d - 3) \cdot 2(d + 8)}{(d + 8)(d - 6) \cdot (d - 3)} \][/tex]
[tex]\[ = \frac{4(d + 8)}{(d - 6) \cdot (d + 8)} \][/tex]
Cancel the [tex]\((d + 8)\)[/tex] from the numerator and the denominator:
[tex]\[ = \frac{4}{d - 6} \][/tex]
Thus, the equivalent expression is:
[tex]\[ \boxed{\frac{4}{d - 6}} \][/tex]
Therefore, the correct answer is:
B. [tex]\(\frac{4}{d - 6}\)[/tex]