How do you divide and simplify?

.x^2+7x+10.......x^2+4x+4
------------------- ÷-----------------
..x^2+8x+15........x+2

..x^2+7x+10........x+2
=----------------- x --------------
..x^2+8x+15.....x^2+4x+4

..(x+5)(x+2)............x+2
=----------------- x ------------------ cancel x+2 and x+5
..(x+3)(x+5)........(x+2)(x+2)

......1
=--------- answer//
....x+3

= ([x² + 7x + 10]/[x² + 8x + 15])/([x² + 4x + 4]/[x + 2])
= ([{x + 5}{x + 2}]/[{x + 5}{x + 3}])/([{x + 2}{x + 2}]/[x + 2]) cancel x + 2 & x + 5
= ([x + 2]/[x + 3])/(x + 2) cancel further x + 2
= 1/(x + 3)

Answer: 1/(x + 3)

(x^2 + 7x + 10)/(x^2 + 8x + 15) ÷ (x^2 + 4x + 4)/(x + 2)
= (x^2 + 5x + 2x + 10)/(x^2 + 5x + 3x + 15) ÷ (x^2 + 2x + 2x + 4)/(x + 2)
= [(x^2 + 5x) + (2x + 10)]/[(x^2 + 5x) + (3x + 15)] * (x + 2)/[(x^2 + 2x) + (2x + 4)]
= [x(x + 5) + 2(x + 5)]/[x(x + 5) + 3(x + 5)] * (x + 2)/[x(x + 2) + 2(x + 2)]
= [(x + 5)(x + 2)]/[(x + 5)(x + 3)] * (x + 2)/[(x + 2)(x + 2)]
= (x + 2)/(x + 3) * 1/(x + 2)
= 1/(x + 3)

We first factor each trinomial.

x^2 + 7x + 10 becomes (x + 2)(x + 5)

x^2 + 8x + 15 becomes (x + 3)(x + 5)

x^2 + 4x + 4 becomes (x + 2)(x + 2)

We now have to invert the right side fraction just like regular fractions.

[(x + 2)(x + 5)]/[(x + 3)(x + 5)] times (x + 2)/[ (x + 2)(x + 2)] =

1/(x = 3)

Solution:

Dividing a fraction by a fraction is the same as multiplying the fraction by its reciprocal. For instance, in this case, your problem becomes...

(x^2 + 7x + 10)/(x^2 + 8x + 15) * (x+2)/(x^2 + 4x + 4)

Let's factor some of the polynomials...

x^2 + 7x + 10 = (x + 5)(x + 2)
x^2 + 8x + 15 = (x + 5)(x + 3)
x^2 + 4x + 4 = (x + 2)(x + 2)

Substitute and multiply to get...

((x + 5)(x + 2)(x + 2))/((x + 5)(x + 3)(x + 2)(x + 2))

As long as x doesn't equal -5 or -2, we can cancel out to arive at our final simplification...

1/(x + 3)

(x^2+7x+10)/(x^2+8x+15)÷(x^2+ 4x+4)/(x+2)
(x+2)(x+5)/(x+5)(x+3) * (x+2)/(x+2)(x+2)

x+2/x+3 * 1/(x+2) =

1 / (x+3)