Question

Divide. (10a4 – 5a3) ÷ 5a

A.
10a^4 – 5a^3 – 5a

B.
10a^4 – 5a^3 + 5a

C.
2a^3 – 5a^3

D.
2a^3 – a^2

Answer 1

hint: factor out 5a from the numerator

Answer 2

I got D..But I'm still confused, I don't think i got it right :/

Answer 3

Another way you can do it is to divide each term in the numerator by 5a in the denominator

First term:
\[\Large \frac{10a^4}{5a} = \frac{10}{5}\frac{a^4}{a^1} = 2a^{4-1} = 2a^3\]

Second term:
\[\Large \frac{-5a^3}{5a} = -\frac{5}{5}\frac{a^3}{a^1} = -1a^{3-1} = -a^2\]

Answer 4

So \[\Large 2a^3-a^2\] is correct

Answer 5

Really? :D Oh wow I thought I was wrong

Answer 6

did you follow those steps or did you take another route?

Answer 7

I did it in a way that I don't even understand I would explain it to you but I don't know how, I did half of it in my head and half of it on paper because I started over twice.

Answer 8

alright I'm just curious. Hopefully my method makes sense?

Answer 9

Yes it does :) I'm gonna try that next time

Answer 10

alright great

Answer 11

Thank you so much!

Answer 12

you're welcome

Answer 13

Can I ask you one more question?

Answer 14

sure

Answer 15

go for it

Answer 16

Determine whether the polynomial is a perfect square and if it is, factor it.

n^2 – 6n + 9

A.
is a perfect square: (n – 3)^2

B.
is not a perfect square

C.
is a perfect square: (n – 6)^2

D.
is a perfect square: (n + 3)^2

Would I do the same thing as the other problem?

Answer 17

how do you determine if a polynomial is a perfect square?

Answer 18

I think the first and the last term have to be squares, I'm not sure..

Answer 19

yes that's on the right track

Answer 20

is n^2 a perfect square?

Answer 21

um I think so..

Answer 22

it is because all perfect squares are of the form x^2 (which literally means "x squared")

Answer 23

how about 9, is that a perfect square?

Answer 24

no

Answer 25

why not?

Answer 26

Because it isn't a square root?

Answer 27

but since 3^2 = 9, this means 9 is actually a perfect square

Answer 28

put another way, \[\Large \sqrt{9} = 3\]

Answer 29

ohhh ok, I thought it had to look something like this x^9 to be a perfect square, but now I get what you mean.

Answer 30

no the exponents are 2 because we're dealing with squared terms

Answer 31

so we have n^2 is a perfect square, so is 9 (since 3^2 = 9)

Answer 32

what else do we need to check?

Answer 33

The 6n part?

Answer 34

yes, we need to see if it fits a very specific form

Answer 35

do you remember what the form is?

Answer 36

no

Answer 37

it turns out that when you expand (a-b)^2 you get a^2 - 2ab + b^2 using the FOIL rule or you can use various other methods

Answer 38

(a-b)^2 is a perfect square trinomial and so is a^2 - 2ab + b^2

Answer 39

notice how in a^2 - 2ab + b^2, we have a^2 and b^2, which are perfect squares themselves

the middle is -2ab

Answer 40

so the middle is formed by multiplying the terms 'a' and 'b' and doubling it (and it happens to be negative)

Answer 41

In this case, we have 'n' and 3 (notice how I'm using the bases), which multiply to n*3 = 3n

Now double this to get 2*3n = 6n

And finally make it negative to get -6n

Answer 42

the -6n fits the form so n^2 - 6n + 9 is indeed a perfect square trinomial

Answer 43

it is in the form (a-b)^2 = a^2 - 2ab + b^2 where a = n and b = 3

Answer 44

So would it be A..?

Answer 45

that is correct

Answer 46

here is another example to work with

z^2 - 10z + 25

we have z^2, a perfect square
25 is a perfect square since 5^2 = 25


so a = z, b = 5

Plug those into (a-b)^2 = a^2 - 2ab + b^2 to get...

(a-b)^2 = a^2 - 2ab + b^2

(z-5)^2 = z^2 - 2z*5 + 5^2

(z-5)^2 = z^2 - 2*5*z + 5^2

(z-5)^2 = z^2 - 10*z + 5^2

(z-5)^2 = z^2 - 10z + 25

Answer 47

Notice how if you change that -10z to anything else, then the trinomial won't be a perfect square

Answer 48

oh now I get it, But is their any possible number you could change it to? I'm just curious :)

Answer 49

which number are you referring to change?

Answer 50

The -10z I know you said that if you change it to anything else then the trinomial won't be a perfect square, but when you said that you can't change it to anything else, do you mean that if you changed it to any other number it wouldn't be a perfect square? Sorry if that's a confusing question

Answer 51

well if you fix the outer terms to be z^2 and 25, then the middle has to be either +10z or -10z for it to be a perfect square.

If it was +10z, then z^2 + 10z + 25 factors to (z+5)^2

If it was -10z, then z^2 - 10z + 25 factors to (z-5)^2

Answer 52

however, if you change either outer term, then you can change the middle term based on what the outer terms change to

Answer 53

for example, if you had z^2 for the first outer term and then had 36 instead of 25, then the middle term would be 2*z*6 = 12z (either plus or minus of that)

Answer 54

meaning that

z^2 - 12z + 36 = (z-6)^2

and

z^2 + 12z + 36 = (z+6)^2

Answer 55

ohh ok, sorry that I asked that was probably a dumb question XD But could y^2 be a perfect square?

Answer 56

y^2 is a perfect square

Answer 57

any variable squared is a perfect square

Answer 58

and don't be afraid to ask questions, there's no such thing as a dumb question

Answer 59

oh ok :) I'm doing a problem and it says y^2-16 and I know I'm supposed to see if the first term is a perfect square, which you said it is, and after that I/m supposed to see if the last ones a perfect square right?

Answer 60

so is 16 a perfect square?

Answer 61

yes

Answer 62

so what number replaces x in the equation x^2 = 16

Answer 63

in other words, what number do you square to get 16

Answer 64

ohh ok that makes more sense, um 2?

Answer 65

2 squared = 2^2 = 2*2 = 4

Answer 66

so that's not it

Answer 67

3 squared = 3^2 = 3*3 = 9

Answer 68

what about 8?

Answer 69

8^2 = 8*8 = 64

Answer 70

see how I'm squaring any given number?

Answer 71

yeah

Answer 72

visually, "squaring" means you find the area of a square with a given side length

example: find the area of a square with side length of 6