MATH SOLVE

4 months ago

Q:
# Determine the product

Accepted Solution

A:

[tex] \frac{d^{2}}{d+6} * \frac{ d^{2}+8d+12 }{d^{2} +2d}[/tex]

We can begin by factoring the numerator of the second fraction and pulling a common factor out of the second denominator:

[tex]\frac{d*d}{d+6} * \frac{(d+6)(d+2)}{d(d+2)}[/tex]

To simplify the multiplication process, we can cancel out any common factors on the diagonals (numerator of one with denominator of the other). You can also cancel out any common factors in the numerator and denominator of the same fraction. We can cancel out the factors d, (d+6), and (d+2). When the remaining parts of the expression are multiplied out, we are left with d. Although this is not one of the options, the third option matches if you don't cancel out (d+2). It also simplifies because you can factor d out of the numerator, then cancel out (d+2) in the numerator and denominator. We know the restrictions are d ≠ -6, -2, and 0 because of the factors we canceled out, (d+6), (d+2), and d. When d is equal to any of these, it will leave a 0 in the denominator, which makes the answer undefined.

We can begin by factoring the numerator of the second fraction and pulling a common factor out of the second denominator:

[tex]\frac{d*d}{d+6} * \frac{(d+6)(d+2)}{d(d+2)}[/tex]

To simplify the multiplication process, we can cancel out any common factors on the diagonals (numerator of one with denominator of the other). You can also cancel out any common factors in the numerator and denominator of the same fraction. We can cancel out the factors d, (d+6), and (d+2). When the remaining parts of the expression are multiplied out, we are left with d. Although this is not one of the options, the third option matches if you don't cancel out (d+2). It also simplifies because you can factor d out of the numerator, then cancel out (d+2) in the numerator and denominator. We know the restrictions are d ≠ -6, -2, and 0 because of the factors we canceled out, (d+6), (d+2), and d. When d is equal to any of these, it will leave a 0 in the denominator, which makes the answer undefined.