Question

Simply the sum

Answer 1

\[\frac{ d^2+d-30 }{ d^2+3d-40 }+\frac{ d^2+14d+48 }{ d^2-2d-48 }\]

Answer 2

d^2+d-30 = (d+6)(d-5). Right?

Answer 3

d^2+3d-40 = (d+8)(d-5)
d^2+14d+48 = (d+8)(d+6)
d^2-2d-48 = (d-8)(d+6)
Right?

Answer 4

Put the factors of numerator and denominator and simplify.

Answer 5

Does it help ?

Answer 6

No i dont understand

Answer 7

Do you know how to factorize ?

Answer 8

can you show me an example of how to do that?

Answer 9

For instance, d^2+d-30 can be written as d^2 +6d - 5d - 30. I have chosen 6 and -5 for their sum is 6+(-5) = 1 and their product is 6(-5) = -30.
So, d^2 +6d - 5d - 30 = d(d +6)- 5(d+6)= (d+6)(d-5)

Is it okay ?

Answer 10

Do you still need some help with factorization?

Answer 11

Eh, I'll walk you through it whether you need it or not. We're going to factor\[x^4-1\]

Answer 12

Here's a hint: Express x^4-1 as a difference of squares.

Answer 13

You're going to get: \[x^4-1=(x^2)^2-1^2\]

Answer 14

\[(x^2)^2-1^2\]

Answer 15

Factor the difference of the two squares.

Answer 16

The two squaresyou're factoring asre \[(x^2)^2-1^2-1^2=(x^2-1)(x^2+1)\]

Answer 17

\[(x^2-1)(x^2+1)\]

Answer 18

Write 1 as a square in order to express x^2-1 as a difference of squares.

Answer 19

i've been on this question for a day...im going to just skip it thanks for the effort

Answer 20

\[(x^2-1^2) (x^2+1)\] Focuse on the 1^2

Answer 21

Eh, I'll go ahead and finish what I started :P I hope you do well ^-^

Answer 22

Factor the difference of two squares. \[x^2-1^2=(x-1)(x+1)\]

Answer 23

Answer: \[(x-1)(x+1) (x^2+1)\]

Answer 24

For the original question, however, your answer will be \[2 d^2+\frac{ 18 }{ d^2 } +16 d-88 \]

Answer 25

Good luck with your studies ^-^

Answer 26

thanks :)