Question

What is the third term in the binomial expansion (x – y^5)^3?
@phi

Answer 1

@tkhunny

Answer 2

@phi

Answer 3

@mathmale

Answer 4

do you have notes on the binomial expansion?

Answer 5

no but wouldnt the first 2 be x^3 - y^8?

Answer 6

or is that wrong

Answer 7

no. It's a bit complicated. you need the formula. Can you find it in your notes?

Answer 8

no sorry

Answer 9

could you help show me

Answer 10

@phi

Answer 11

wait i think i found it

Answer 12

https://assets.lincolninteractive.org/courses/NN.MAT.302511.1.PE/Unit4/Lesson10/Algebra_2_Part_1_U4L10_Image_1.jpg

Answer 13

the idea is if we start with (a+b)^n
each term will be
a b
with some number out front
and a and b will be to some power.

a will start with the power "n"
b will start with the power 0

the easiest way to start is write the powers
n 0 (n-1) 1, (n-2) 2
and stick in a and b:
\[ a^n b^0 + a^{n-1} b^1 + a^{n-2} b^2 +...\]
we can go on, but we only need the 3rd term

Answer 14

x^3 is the first term

Answer 15

in your problem, what is "n" ?

Answer 16

3?

Answer 17

yes.

also, we want the third term, and I listed the first 3 terms
\[ a^n b^0 + a^{n-1} b^1 + a^{n-2} b^2 +... \]
if we put in 3 for n, the first four terms are:
\[ a^3 b^0 + a^2b^2 + a^1 b^2 +a^0b^3\]
do you see which one is the 3rd term ?

Answer 18

the third term is a^1 b^2

Answer 19

btw, I have a typo in the 2nd term, it should read a^2 b^1

Answer 20

now we need what in your problem matches up with a and with b
any idea ?

Answer 21

no sorry

Answer 22

in other words
(x – y^5) matched with (a+b)
I would write it as
(x + (-y^5)) so it is easier to see

Answer 23

ok so what would the third term be

Answer 24

3xy^10?

Answer 25

part of the answer is
a^1 b^2
but that is the "rule" and we have to match your problem to the a and b
what should a and b be ?

in other words, match
(x + (-y^5)) with (a+b)
what matches with "a" ?

Answer 26

x

Answer 27

ok, and b is ?

Answer 28

for the third term it is 3ab^

Answer 29

y^5

Answer 30

I think the answer is 3xy10?

Answer 31

would that be right?

Answer 32

almost, b is -y^5

so ignoring the leading number the third term is
a^1 b^2 but with a replaced with x and b replaced with (-y^5)
we get
\[ x (-y^5)^2\]

Answer 33

-y^5 squared means -y^5 * -y^5 = +y^10

Answer 34

right so would the third term be 3xy^10

Answer 35

and putting in the 3 (which is a pain to find , but using pascal's triangle like in your notes works)

yes the 3rd term is
3xy^10

Answer 36

Ok i have one more problem could you help?

Answer 37

x^7- 2x^6 + 3x^2 - 2x +5

Answer 38

4 sign changes

Answer 39

so 4,2, or 0 positive roots

Answer 40

can you do the number of sign changes when x is -1 ?

Answer 41

you should get 1 sign change for x=-1 so one negative root
there are 5,3 or 1 real roots
there are a total of 7 roots. at least 2 are imaginary.

Answer 42

I like to build things vertically

(x – y^5)^3

Term 0) \(1(x)^{3}(-y^{5})^{0}\)

Term 1) \(3(x)^{2}(-y^{5})^{1}\)

Term 2) \(3(x)^{1}(-y^{5})^{2}\)

Term 3) \(1(x)^{0}(-y^{5})^{3}\)

Patterns prescribed in the formula are very nicely laid out, there.

Answer 43

so there would be 7 complex and 1,3 or 5 real roots @tkhunny

Answer 44

actally 5 complex

Answer 45

There's always 7 (counting multiplicities)

x^7- 2x^6 + 3x^2 - 2x +5
4, 2, or 0 Positive Real

-x^7- 2x^6 - 3x^2 - 2x +5
1 Negative Real

This leaves
2, 4, or 6 Complex

With Real Coefficients, Complex Roots MUST come in pairs. If there are 5 Positive Real and 2 Negative Real, this leaves room for only 1 Complex. That's no good.