Question

http://awesomescreenshot.com/05f2ct6a63
"please solve step by step*

Answer 1

familiar wid factoring quadratics ?

Answer 2

To some degree, yes.

Answer 3

that wil do :)
u seeing FOUR quadratics in the given expression right ?

Answer 4

factor each of them first

Answer 5

\(\large \frac{d^2+d-30}{d^2+3d-40} + \frac{d^2+14d+48}{d^2-2d-48}\)

Answer 6

lets factor \(\large d^2+d-30\) :

Answer 7

can u ?

Answer 8

just find two numbers that satisfy below :
1) product = -30
2) sum = 1

Answer 9

Is product the answer when multiplied and the sum the answer to addition

Answer 10

yes !

Answer 11

Then would it just be -5 times 6 = -30
and -5 + 6 = 1

Answer 12

Correct !!

Answer 13

factor it now :
\(\large d^2+d-30 = (d+6)(d-5)\)

Answer 14

fine ?

Answer 15

Um... do I foil that?

Answer 16

nope. leave it like that

Answer 17

similarly if u factor other 3 quadratics also, the given expression becomes :

\(\large \frac{d^2+d-30}{d^2+3d-40} + \frac{d^2+14d+48}{d^2-2d-48}\)

after factoring all quadratic u wud get :

\(\large \frac{(d+6)(d-5)}{(d+8)(d-5)} + \frac{(d+8)(d+6)}{(d-8)(d+6)}\)

Answer 18

canceling watever u can :

\(\large \frac{d+6}{d+8} + \frac{d+8}{d-8}\)

Answer 19

see if you're okays still

Answer 20

\[\frac{ d+6 }{ d-8 }\] ?

Answer 21

noo, it wont cancel like that

Answer 22

next u need to find common

Answer 23

next u need to find common denominator,
and add numerators. let me show u

Answer 24

Ooooh... ok then

Answer 25

simplify the numerator

Answer 26

How did you get the second to the last part?

Answer 27

i just multiplied and divided first fraction wid (d-8)

Answer 28

and did the same to second fraction wid (d+8)

Answer 29

so that, we get the common denominator for both fractions

Answer 30

\(\large \frac{d^2+d-30}{d^2+3d-40} + \frac{d^2+14d+48}{d^2-2d-48}\)

after factoring all quadratic u wud get :

\(\large \frac{(d+6)(d-5)}{(d+8)(d-5)} + \frac{(d+8)(d+6)}{(d-8)(d+6)}\)

canceling watever u can :

\(\large \frac{d+6}{d+8} + \frac{d+8}{d-8}\)


\(\large \frac{(d+6)(d-8)}{(d+8)(d+8)} + \frac{(d+8)(d+8)}{(d-8)(d+8)}\)


\(\large \frac{(d+6)(d-8) + (d+8)(d+8)}{(d+8)(d-8)} \)


\(\large \frac{(d+6)(d-8) + (d+8)^2}{(d+8)(d-8)} \)

\(\large \frac{d^2 -2d -48 + d^2+16d + 64}{(d+8)(d-8)} \)

\(\large \frac{2d^2 +14d + 16}{(d+8)(d-8)} \)

Answer 31

see if this makes sense... ^^

Answer 32

Everything's making sense to me except I kind of got lost where it said 14d... I'm not entirealy sure as to where that came from.

Answer 33

good :)
u fine till below :
\(\large \frac{d^2 -2d -48 + d^2+16d + 64}{(d+8)(d-8)} \)

?

Answer 34

Yes sir

Answer 35

\(\large \frac{d^2 \color{green}{-2d} -48 + d^2\color{green}{+16d} + 64}{(d+8)(d-8)} \)

Answer 36

combine those two terms

Answer 37

Ohhhh I understand it now

Answer 38

good

Answer 39

Thank you kind stranger lol [:

Answer 40

lol you're welcome smart student :)