Question

2 questions...

1. Use Pacal's Triangle to expand the binomial.
(8v + s)^5

2. Find a quadratic equation with roots -1 + 4i and -1 - 4i

Answer 1

the 2nd one is easier.
if 'a' is a root, then (x-a) is a factor.
so, for roots -1 + 4i and -1 - 4i, you have
f(x) = (x +1-4i)(x+1+4i)
can you simplify this ?

Answer 2

would you use distributive property to simplify it ?

Answer 3

you can use \((a+b)(a-b)=a^2-b^2\)
here a = x+1
b=4i

Answer 4

so if you were to substitute a and b in would it look like this ? ... \[(x + 1 - 4i)(x + 1 - 4i) = x + 1^{2} - 4i ^{2}\]

Answer 5

Oh the first one is suppose to be (x + 1 + 4i)

Answer 6

yes, but don't forget brackets \((x+1)^2-(4i)^2=....?\)

Answer 7

okay and Im not sure what to do now

Answer 8

\((a+b)^2=a^2+2ab+b^2\)
so, (x+1)^2=....?

Answer 9

Would it be \[x ^{2} + 2x + 1 \] ?

Answer 10

thats correct, now \((4i)^2=16i^2=....?\)

Answer 11

I have no clue :/

Answer 12

\(i=\sqrt{-1} \\i^2=...?\)

Answer 13

\[i ^{2} = -1\]

Answer 14

yes.
so 16i^2 = -16

Answer 15

so, x^2+2x+1-16=... ?

Answer 16

Oh okay....

\[x ^{2} + 2x - 15 \]

Answer 17

Would that be the answer for the second one ?

Answer 18

uhm, \(x^2+2x+1)-(-16) \dots\)

Answer 19

sorry, it was x^2+2x +1 -(-16) = x^2+2x+1+16 =... ?

Answer 20

Oh okay so it would be \[x ^{2} + 2x + 17 right ?

Answer 21

yes, thats correct now :)

Answer 22

Okay, so would that be the answer ?

Answer 23

x^2+2x+17.=0

Answer 24

Thanks, do you know about the first one ?

Answer 25

if you know the pascal triangle.....

Answer 26

I'm not really sure how you do it. It's confusing to me

Answer 27

See my explanation of this question:
http://openstudy.com/updates/50e5d8a5e4b058681f3f1833

Answer 28

nice explanation!
and if you have doubts ask here...

Answer 29

Could we work on it this problem together to see if I get the right answer ?

Answer 30

If you've read my explanation of the same kind of problem, you'll understand by now that if you expand (a+b)^5, you get 6 terms:\[a^5,a^4b,a^3b^2,a^2b^3,ab^4,b^5\]See? It's always the product of 5 numbers a and b, every possibility covered.

Answer 31

Yes, your explanation was very helpful. Would I substitute my answers into \[a ^{5} + 5a ^{4}b + 10^{3}b ^{2} + 10a ^{2}b ^{3} + b ^{5}\] ?

Answer 32

Right! Just put in the number 5ab^4 (fifth term).
The coefficients you get from Pascal's Triangle.

Now we have to set a=8v and b=s. This leaves some calculations to do:\[(8v)^5+5(8v)^4s+10(8v)^3s^2+10(8v)^2s^3+5 \cdot 8vs^4+s^5\]Now work out the brackets if you dare!

Answer 33

(doing it myself...large numbers...takes some time...)

Answer 34

Okay, so I got...\[32,768v ^{5} + 20,480sv ^{4} + 5,120sv ^{3} + 640s ^{3}v ^{2} + 40s ^{4}v + s ^{5}\] is this correct ?

Answer 35

Yes! I'm so proud!

Answer 36

Yay! lol thank you so much for both of your helps!

Answer 37

Welcome!