Question

What is the quotient in simplified form? State any restrictions on the variable.
x^2-16/x^3+5x+6 divided by x^3+5x+4/x^2-2x-8

Answer 1

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

Answer 2

@johnweldon1993 @jim_thompson5910

Answer 3

invert and multiply, but really factor and cancel

Answer 4

can you walk me through it??

Answer 5

\[\frac{ x^{2}-16 }{ x^{2}+5x+6 } \div \frac{ a^{2}+5x+4 }{ x^{2}-2x-8 }\]
\[=\frac{ x^{2}-16 }{ x^{2}+5x+6 } \times \frac{x^{2}-2x-8 }{ x^{2}+5x+4 }\]

Answer 6

i assume the \(a\) was a typo, and it should be an \(x\) right?

Answer 7

now this problem was cooked so you don't actually multiply anything
factor each binomial , then cancel common factors

Answer 8

\[=\frac{ x^{2}-16 }{ x^{2}+5x+6 } \times \frac{x^{2}-2x-8 }{ x^{2}+5x+4 }\]
\[=\frac{(x+4)(x-4)(x-6)(x+2)}{(x+3)(x+2)(x+4)(x+1)}\] all factored out

Answer 9

ach i made a mistake!!

Answer 10

\[\large =\frac{(x+4)(x-4)(x-4)(x+2)}{(x+3)(x+2)(x+4)(x+1)}\] that's better

Answer 11

okay hold on. let me look at this and see if i can catch on.&&yes that z was supposed to be an x

Answer 12

ok take your time
step one was to flip the second one
step two was to factor each quadratic
the last step, which we didn't get to yet, is to cancel common factors

Answer 13

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

Answer 14

is this what i should get after canceling out common factors?

Answer 15

yes

Answer 16

and finally you need all the "restrictions"
find the zeros of the original denominator
\[(x+3)(x+2)(x+4)(x+1)\] and those are the restricted values
i.e.
\[-3,-2,-1,-4\]

Answer 17

would 4 be one?

Answer 18

no

Answer 19

there is no \(x-4\) in the denominator