Question

Please someone help me factoring...

Answer 1

\[64r ^{2}-144r-81\]

Answer 2

I would factor by decomposition.

Answer 3

http://www.youtube.com/watch?v=eF6zYNzlZKQ

Answer 4

Hello, Daniel! glad to see you online again.
At first I thought that your expression, \[64r ^{2}-144r-81\] would require more work than usual to factor, because of those "big" coefficients.

but then I noticed that both 64 and 81 are perfect squares, with roots 8 and 9 respectively.

Could you take this info and use it to factor the given expression? Remember, we're looking for two binomial factors. Are those factors going to be the same or different?

Answer 5

I really dont know @mathmale

Answer 6

I'm going to start with a much simpler example:

\[x ^{2} + 2x + 1\]Would you try your hand at factoring this?

Answer 7

\[(x+1)^{2}\]

Answer 8

@mathmale

Answer 9

@danielbarriosr1 I think you typed your problem wrong. Are you sure both the 2nd and 3rd terms are both negative?

Answer 10

I'm guessing the 3rd term is actually positive.

Answer 11

\[64r ^{2}-144r-81\] should be \[64r ^{2}-144r+81\]

Otherwise, this polynomial is not factorable with real numbers.

Answer 12

I agree with mathbrz. I completely overlooked the fact that the last term needs to be +81 (not -81) for this trinomial to be easily factored. Thank you, @mathbrz .

Answer 13

Daniel: Your (x+1)(x-1) is just fine. Where did you get that 1? Where did you get that x?
Go back and look at the problem you've posted. Where did I get that 8? that 9?

Answer 14

Yes it is positive... sorry @mathbrz @mathmale

Answer 15

Wasn't so smart of me to overlook that -81.

Now:

Daniel: Your (x+1)(x-1) is just fine. Where did you get that 1? Where did you get that x?
Go back and look at the problem you've posted. Where did I get that 8? that 9?

Answer 16

Sorry, my openstudy just stoped

Answer 17

8 and 9 are the roots of 64 and 81

Answer 18

@mathmale

Answer 19

Please
Someone

Answer 20

Here's what I'd we did before: x^2 + 2x + 1 = (x+1)(x+1).
new example: 9x^2 + 6x + 1 = (3x+1)(3x+1)

Daniel: Can you explain where the "3x" in each factor came from? The "1"?

Answer 21

You can pick up this discussion at any time. if you respond, I'll see your post and respond if at all possible. Best to you. MM

Answer 22

Yes... 3 is a factor of 9 as 1 is of 2

Answer 23

Yes: 9 separates into two equal factors: 3*3

Now try a similar attack for

\[64r ^{2}-144r+81\]

Answer 24

64r^2 separates into what two identical factors?

Answer 25

81 separates into which two identical factors?

Answer 26

9*9

Answer 27

@mathmale

Answer 28

Yes, Daniel, and 64r^2 separates into what two identical factors?

Answer 29

Got it (8r-9)(8r-9)

Answer 30

Would you please help me with another equation

Answer 31

So glad you 've gotten that problem right. Cool!! Bueno!
It's important that you review what we did here, so that you'll be well prepared to tackle test and quiz questions later on.
Perhaps the next equation will be a bit harder. But you can handle it.

Answer 32

Ok...
\[18h^3+45h^2-8h-20\]

Answer 33

We'll try to do this one using a method called "factoring by grouping." Have you heard of that?

Answer 34

Yes it is when you put two monomials in a parenthesis

Answer 35

like
\[(18h^3+45h^2)-(8h-20)\]

Answer 36

Then the G.C.F of them

Answer 37

But I got stock there

Answer 38

You're on the right track. Your use of parentheses is good! But consider this:
\[18h^3+45h^2-8h-20 = (18h^3+45h^2) - (8h + 20).\]

Answer 39

How does that differ from your response?

Answer 40

The 20 is positive

Answer 41

@mathmale

Answer 42

Right. Bueno. Now, please factor (8h+20). Hint: look for the single common factor.

Answer 43

4

Answer 44

4(2h+5)

Answer 45

Nice work. Now, would you please try the same factoring stunt on the expression inside the other set of parentheses? Hint: You'll need to factor out both a numeral and a literal (letter).

Answer 46

\[3h^2(6h+15)\]

Answer 47

Good start! But isn't 3 common to both of the numerals within the parentheses?

Answer 48

\[3h^2*3( ? )\]

Answer 49

What goes inside those parentheses here?

Answer 50

2h+5 ?

Answer 51

Right. isn't that the same thing you found to be a factor of the other half of this expression?

Answer 52

yes

Answer 53

Voila! You've got it. The factors, at this point, are (2h+5)( ?? )

Answer 54

(9h^2-4)(2h+5)

Answer 55

Yes! So let's now re-write that as (2h+5)(9h^2 - 4).
Does it appear that the 2nd factor could itself be factored? If so, how can you tell?

Answer 56

I do not know

Answer 57

Hint: this is a "difference of squares:"\[a ^{2}-b ^{2}=(a-b)(a+b)\]

Answer 58

I do not know yet

Answer 59

Therefore, \[9h^2 - 4 =(3h)^2-2^2=( ? ) * ( ? )\]

Answer 60

Daniel,
\[a ^{2}-b ^{2}=(a-b)(a+b)\] is an identity that's well worth learning and remembering. It's called the "difference of squares" formula, and is one of several such formulas that you'd find were you to do a search for "special products and factoring."

Answer 61

\[9h^2 - 4 =(3h)^2-2^2=( ? ) * ( ? )\]

Answer 62

Hint: Look at that (3h)^2. What's the square root of that?
Look at that 4. What's the square root of 4?

Answer 63

4=2
9=3

Answer 64

Would you please tell me whats wrong (or left) with this
\[8p^3-32p^2+28p-112\]
\[4(2p^3-4p^2)+(7p-28)\]
\[(4)2p^2(p-2)+7(p-4)\]

Answer 65

Sqrt of 4 is 2, yes.
Sqrt of 9 is 3, yes.
Sqrt of 9h^2 is what?

Sorry, I got disconnected. Please type in the result you got for the previous problem.
(2h+5)(( )*( )

Answer 66

In regard to your newest math problem:

You got\[4(2p^3-4p^2)+(7p-28)\]

Answer 67

The error has to do with your -4p^2. The numeral 4 is incorrect. Try again. Otherwise, your expression is fine.

Answer 68

Oh i got it

Answer 69

So: are you satisfied with your solution to this latest problem?
with your solution to the previous problem?

Answer 70

I've enjoyed working with you! Hope to see you again soon on OpenStudy.

Answer 71

Thank you so Much