How do you simplify \frac { x ^ { 2} - x - 72} { x ^ { 2} + 13x + 40} \div \frac { x ^ { 2} + 5x + 6} { x ^ { 2} + 7x + 10}x2x72x2+13x+40÷x2+5x+6x2+7x+10?

2 Answers
Sep 5, 2017

(x-9)/(x+3)x9x+3

Explanation:

color(blue)("Initial thoughts")Initial thoughts

We need to experiment with factorizing to see if there is anything we can cancel out.

x^2+5x+6" "->" "2xx3 = 6 and 2+3=5x2+5x+6 2×3=6and2+3=5

color(white)("ddd.dddddddd")->(x+2)(x+3)ddd.dddddddd(x+2)(x+3)
....................................................................................

x^2+7x+10" "->" "2xx5=10 and 2+5=7x2+7x+10 2×5=10and2+5=7

color(white)("dddddddddddd")->(x+2)(x+5)dddddddddddd(x+2)(x+5)
....................................................................................

x^2+13x+40" "->" "5xx8=40 and 5+8=13x2+13x+40 5×8=40and5+8=13

color(white)("ddddddddddddd")->(x+5)(x+8)ddddddddddddd(x+5)(x+8)
....................................................................................

x^2-x-72" "->" "9xx8=72 and 8-9=-1x2x72 9×8=72and89=1

color(white)("dddd.ddddddd")->(x-9)(x+8)dddd.ddddddd(x9)(x+8)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Putting it all together")Putting it all together

Invert and multiply the divisor:

((x-9)cancel((x+8)))/(cancel((x+5))cancel((x+8))) xx(cancel((x+2))cancel((x+5)))/(cancel((x+2))(x+3))

(x-9)/(x+3)

Sep 5, 2017

(x-9)/(x+3)

Explanation:

(x^2 - x - 72)/(x^2 + 13x + 40) div (x^2 + 5x + 6)/(x^2 + 7x + 10)

Lets start by rewriting the problem such that all of the terms are combined into a single numerator and denominator. We can eliminate the division symbol by inverting the second term and switching to multiplication.

(x^2 - x - 72)/(x^2 + 13x + 40) xx (x^2 + 7x + 10)/(x^2 + 5x + 6)

Now we can combine both terms into a single fraction.

((x^2 - x - 72)(x^2 + 7x + 10))/((x^2 + 13x + 40)(x^2 + 5x + 6))

We have four quadratic statements separated by parenthesis. By factoring each we may be able to find terms that cancel each other out.

((x+8)(x-9)(x+5)(x+2))/((x+8)(x+5)(x+2)(x+3))

Sure enough, we see that x+8, x+5, and x+2 are all present in both the numerator and denominator.

(color(red)cancel(color(black)((x+8)))(x-9)color(red)cancel(color(black)((x+5)))color(red)cancel(color(black)((x+2))))/(color(red)cancel(color(black)((x+8)))color(red)cancel(color(black)((x+5)))color(red)cancel(color(black)((x+2)))(x+3))

After eliminating the like terms, we are left with the reduced solution to the problem.

(x-9)/(x+3)