Question

Help!

Answer 1

Anyone?

Answer 2

Hey @lala2!
First, I'll write everything in the equation editor, to see what is going on:\[\frac{ d^2-9d+20 }{ d^2-3d-10 }+\frac{ d^2-2d-8 }{ d^2+4d-32 }=\]

Answer 3

Yes, I'm typing, but something went wrong and I lost everything I had already typed in :(

Answer 4

Now you have to factor everything in sight. Let me do the first one:

d²-9d+20 = (d-4)(d-5), because -4+-5=-9 and -4*-5=20.

Can you do this kind of factorisation?

Answer 5

d^2-3d-10 would = (d-5)(D+2)

Answer 6

@ZeHanz ? is that right

Answer 7

Yes, that is right!

Answer 8

d^2-2d-8 would equal (d-4)(d+2)

Answer 9

You're doing fine!

Answer 10

d^2+4d-32 = (d-4)(d+8)

Answer 11

NOw what?

Answer 12

@ZeHanz ?

Answer 13

Let's see how the fractions look now:\[\frac{ (d-4)(d-5) }{ (d-5)(d+2)}+\frac{ (d+2)(d-4) }{ (d-4)(d+8) }=\]

You see that there are factors that can be cancelled, in each fraction!

Answer 14

So we now have:\[\frac{ d-4 }{ d+2 }+\frac{ d+2 }{ d+8 }\]We have to write these as one fraction I think, so we have to find a new denominator.
What would that be?

Answer 15

(d+2)(d+8)

Answer 16

@ZeHanz

Answer 17

Yes, so we can write it as:\[\frac{ (d-4)(d+8)+(d+2)(d+2) }{ (d+2)(d+8) }=\]Now expand the numerator and collect terms.

Answer 18

Is it d^2+8d-28/d+2)(d+8)

Answer 19

Make that first term 2d², then I agree.

Answer 20

its 2d^2+8d-28

Answer 21

Only thing that could be done now is : 2(d²+4d-14)/((d+2)(d+8)), then we're done, because d²+4d-14 cannot be factorised with integer numbers.

Answer 22

its not the one i just told u

Answer 23

2d^2+8d-28

Answer 24

b/c 2(d²+4d-14)/((d+2)(d+8)) is not one of my choices

Answer 25

OK, didn't think about multiple-choice questions...

Answer 26

is it right

Answer 27

wasmy answer correct @ZeHanz

Answer 28

It was! Well done!

Answer 29

thanks! can i tag you in another question

Answer 30

You're welcome!
Tag me!