MATH SOLVE

3 months ago

Q:
# What is the quotient in simplified form? State any restrictions on the variable? \frac{x^2-16}{x^2+5x+6} /\frac{x^2+5x+4}{x^2-2x-8}

Accepted Solution

A:

[tex]\frac{x^2-16}{x^2+5x+6} / \frac{x^2+5x+4}{x^2-2x-8}[/tex]

We can begin by rearranging this into multiplication:

[tex]\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}[/tex]

Now we can factor the numerators and denominators:

[tex]\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}[/tex]

The factors (x+4) and (x+2) cancel out, leaving us with:

[tex]\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}[/tex]

Our answer comes out to be:

[tex]\frac{(x-4)^{2} }{(x+3)(x+1)} [/tex] or [tex]\frac{ x^{2} -8x+16}{ x^{2}+4x+3 }[/tex]

Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4

We can begin by rearranging this into multiplication:

[tex]\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}[/tex]

Now we can factor the numerators and denominators:

[tex]\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}[/tex]

The factors (x+4) and (x+2) cancel out, leaving us with:

[tex]\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}[/tex]

Our answer comes out to be:

[tex]\frac{(x-4)^{2} }{(x+3)(x+1)} [/tex] or [tex]\frac{ x^{2} -8x+16}{ x^{2}+4x+3 }[/tex]

Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4