Question

(2x+3)^5

Answer 1

@cupcake111

Answer 2

\[\Large\bf\sf (2x+3)^5\]\[\large\bf\sf =(2x)^5+5(2x)^4(3)+10(2x)^3(3)^2+10(2x)^2(3)^3+5(2x)(3)^4+3^5\]

Answer 3

We did this one the other day?
Did we make a boo boo somewhere?

Answer 4

\[\large\bf\sf =32x^5+240x^4+720x^3+10800x^2+810x+243\]

Answer 5

I think we get something like that.
Did we come up with something different last time? :U

Answer 6

woah... i got 96x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x^2 + 15

Answer 7

hmm

Answer 8

Look at the very last term a sec.
\[\Large\bf\sf 3^5\quad\ne\quad 3\cdot5\]

Answer 9

\[1(2x)^5(3)^0 + 5(2x)^4(3)^1 + 10(2x)^3(3)^2 + 10(2x)^2(3)^3 + 5x(2x)^1(3)^4 + 10(3)5\]

Answer 10

Maybe you made a few mistakes like that? D:

Answer 11

this is how i got it

Answer 12

Why is there a 10 on the last term? :U
Our Pythagorean Coefficients should be 1 5 10 10 5 1

Answer 13

So a 1 coefficient on the last term.

Answer 14

wait, its; 1^0(3)^5

Answer 15

Yes,

1*(2x)^0*(3)^5

Answer 16

Oh just a typo, i see :D

Answer 17

so.. i forgot the 2x?

Answer 18

ya i guess so.

Answer 19

how come there's no 2x in your equation

Answer 20

\[\Large\bf\sf 1(2x)^0(3)^5\quad=\quad 3^5\]Ya if you wanted to keep the pattern going, then (2x)^0 should be in your last term.
But see how it simplifies? That's why I didn't include it.

Answer 21

i see

Answer 22

now I need this one: (2x – 3y)^4 i only have a minute

Answer 23

Similarly let's look at the first term a sec:\[\Large\bf\sf 1(2x)^5(3)^0\quad=\quad 2^5x^5\]

Answer 24

Understand how to clean up that term?

Answer 25

yes

Answer 26

this is my work for the 2nd one.

Answer 27

\[\Large\bf\sf (2x)^4(-3y)^0\quad=\quad 2^4x^4\quad=\quad ?\]

Answer 28

did i get the first line right?

Answer 29

The second line? yes.

Answer 30

16x^4

Answer 31

Ok good.
For some reason you wrote 162x

Answer 32

yea

Answer 33

\[\Large\bf\sf 4(2x)^3(-3y)^1\quad=\quad 4(2x)^3(-3y)\quad=\quad -3\cdot4\cdot2^3x^3y\]

Answer 34

This is the second term.

Answer 35

-12 * 8 x^3y

Answer 36

= -96x^3y

Answer 37

wait that's my same answer...

Answer 38

Oh I see, I misread your answer key.
So for the first term you had \(\Large\bf\sf -48x^4y\).
So that was the one we wanted 16x^4 on.

I think you were just forgetting that:\[\Large\bf\sf (-3y)^0\quad=\quad 1\]

Answer 39

That's probably why you had the extra -3 and y in there.

Answer 40

Mmm 3rd term looks good.

Answer 41

wait... was 16x^4 for the first or last term??

Answer 42

The first. I was misreading your sheet.

Answer 43

ah

Answer 44

4th term:\[\Large\bf\sf 4(2x)^1(-3y)^3\quad=\quad 4(2x)(-3)^3y^3\quad=\quad ?\]

Answer 45

wait.. or is it 72xy^3?

Answer 46

4(2x)(−3)^3y^3 = 8x * 9y^3 = 72xy^3

Answer 47

\[\Large\bf\sf (-3)^3\quad=\quad -27\]

Answer 48

darn it

Answer 49

ok so, -216y^3

Answer 50

Woops there should be an x in there somewhere, right?

Answer 51

-216xy^3

Answer 52

ok good!

Answer 53

For the last term,
\[\Large\bf\sf (2x)^0(-3y)^4\quad=\quad (-3)^4y^4\]The zero gets rid of the 2 and x.

Answer 54

81y^4

Answer 55

yay good job.
So that will replace your 162x... I guess. :U

Answer 56

I have one more question. are u good with multiplicities?

Answer 57

yah i think so :U

Answer 58

ok this:

Answer 59

this is what i have:
-5, -1, 4, 7

Answer 60

for the zeroes^

Answer 61

my teacher said: What are the possible multiplicities (exponents) for those terms? Review "crossing" versus "touching" the x-axis to see!

Answer 62

also:

Answer 63

Remember, if the graph just touches the axis, the multiplicity will be even
If the graph crosses over the axis, then the multiplicity will be odd.
Recall that this is a seventh degree function, therefore there must be 7 zeros.

Answer 64

well, the graph touched the point -5, and that's not even... ?

Answer 65

(0,-5) i mean

Answer 66

Why would you say it's not even?
`Parabola` is an even function. That looks very much like a parabola right? :o

Answer 67

ohh

Answer 68

i thought it meant the NUMBER would be even

Answer 69

nvm

Answer 70

I don't really understand multiplicities

Answer 71

Multiplicity refers to the power on each factor that gives us our roots.
So for this problem, we might have:\[\Large\bf\sf (x+5)^2\]as one of our factors.
Did they want us to reconstruct the polynomials? lemme read the instructions again..

Answer 72

List the polynomial’s zeroes with possible multiplicities.

Answer 73

do we have to have 7 zeroes?

Answer 74

Yes we have to have 7 zeroes.

(at least) 2 of them occur at -5 because we have a parabola shape (even powers give us parabolas).

Answer 75

so do i write -5 twice?

Answer 76

So we could say for our first zeroes:\[\Large\bf\sf x=-5, \quad\text{multiplicity 2}\]

Answer 77

It could end up being 4 or 6, but as we go along, we'll see that it can only be 2.

Answer 78

i see 4 points that are crossed, and -5 is the only point that is touched

Answer 79

You see 4 zeroes, right?
That's why we need to look at multiplicity, to see where the other 3 zeroes are coming from. We're getting a double zero at -5. Maybe that's a poor way to say that but whatev.

Answer 80

wait, no, 5

Answer 81

ok

Answer 82

does the line going through -2,0 count?

Answer 83

We have another root at x=-1.
It crosses through, (line a straight line), what multiplicity should we get?

Answer 84

No that's not a zero of X.

Answer 85

I mean 0,-2

Answer 86

That's a zero of y, we'll ignore that.

Answer 87

It doesn't cross the x-axis at that point.

Answer 88

ok

Answer 89

um, 2?

Answer 90

oh, wait no, ! odd!

Answer 91

If you have a straight line, like y=mx+b.
What power do we have on the x?

Answer 92

im not sure

Answer 93

Straight lines have a power of 1.
\[\Large\bf\sf y=mx+b\qquad\to\qquad y=mx^1+b\]

Answer 94

i need to go in 1 minute

Answer 95

oh i didnt know that

Answer 96

This function behaves like a straight line through that zero.
So it has a multiplicity of 1.