# Help! :)

#### sportywarbz

##### New member
Hi,

1. Here is the problem:
?(27x³y?)+?(48x³y?)
This is what I have for a solution... I am wondering if I am correct?
9xy?+16xy?
25x²y?

2. Here is the problem (and description):
Determine the quotient and remainder when the first polynomial is divided by the second.
(3x²-5x+4)/(x+3)
This is what I have for the solution... am I correct?
3x²-4x+7

#### mmm4444bot

##### Super Moderator
Staff member

sportywarbz said:
Here is the problem:

?(27x³y?) + ?(48x³y?)

This given expression is only one part of the problem; the other part is the unposted instruction that came with it.

You're asked to simplify and combine these two radical terms, yes? Or, in other words, "Complete the given operation".

So, first we simplify each radical term; only then can we combine the results.

Let's look at the first radical term above.

Your work seems to indicate that you simplified sqrt(27x^3y^7) to 9xy^4.

That's not correct.

It helps to be familiar with a property of radicals that allows us to write:

sqrt(27x^3y^7) = sqrt(27) * sqrt(x^3) * sqrt(y^7)

This way, we can simplify each individual factor.

Now, the square root of 27 does not simplify to 9. (9 is the square root of 81.)

sqrt(27) simplifies to 3 sqrt(3)

sqrt(x^3) simplifies to x sqrt(x)

sqrt(y^7) simplifies to y^3 sqrt(y)

In other words, the radical term sqrt(27x^3y^7) simplifies to 3xy^3 sqrt(3xy)

Try to simplify the second radical term sqrt(48x^3y^7) the same way. Then see if you understand how to add the two simplified terms.

Please show us what you get. Otherwise, ask specific questions about any parts that you do not understand.

#### mmm4444bot

##### Super Moderator
Staff member

sportywarbz said:
Determine the quotient and remainder

This is what I have for the solution... am I correct?

3x² - 4x + 7

I'm sorry, but this is not even close.

Do you realize that this exercise asks for two things ?

Or, are you trying to say that the quotient is 3x^2-4x+7 and the remainder is also 3x^2-4x+7 ?

Let's start over.

What have you learned so far about dividing one polynomial by another ?

I mean, are you learning polynomial division, synthetic division, or some other method ?

There are some lessons on polynomial division at this site.

If you would like to watch a 10-minute video, instead, check out this one at the Khan Academy.

After you review some lessons, we welcome your specific questions.

#### soroban

##### Elite Member
Hello, sportywarbz!

$$\displaystyle \text{Simplify: }\:\sqrt{27x^3y^7} + \sqrt{48x^3y^7}$$

$$\displaystyle \text{I got: }\:9xy^4+16xy^4 \:=\:25x^2y^8$$ . . . . Wow!

I really stretched my imagination
. . . and still couldn't believe what I saw.

First of all, those square roots are probably cube roots: .$$\displaystyle \sqrt[3]{27x^3y^7} + \sqrt[3]{48x^3y^7}$$

$$\displaystyle \text{Then }\sqrt[3]{27x^3y^7}\text{ was simplified like this . . .}$$

$$\displaystyle \sqrt[3]{27x^3y^7} \;=\; \sqrt[3]{27}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{y^7}$$

. . $$\displaystyle \sqrt[3]{27} \:=\:27 \div 3 \:=\:9$$ ?

. . $$\displaystyle \sqrt[3]{x^3} \:=\:x^{3-3} \:=\:x$$ ?

. . $$\displaystyle \sqrt[3]{y^7} \:=\:y^{7-3} \:=\:y^4$$ ?

$$\displaystyle \text{Hence: }\:\sqrt[3]{27x^3y^7} \;=\;9xy^4$$ ?

$$\displaystyle \text{And }\sqrt[3]{48x^3y^7}\text{ was simplified like this . . .}$$

$$\displaystyle \sqrt[3]{48x^3y^7} \;=\;\sqrt[3]{48}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{y^7}$$

. . $$\displaystyle \sqrt[3]{48} \:=\:48 \div 3 \:=\: 16$$ ?

. . $$\displaystyle \sqrt[3]{x^3} \:=\:x^{3-3} \:=\:x$$ ?

. . $$\displaystyle \sqrt[3]{y^7} \:=\:y^{7-3} \:=\:y^4$$ ?

$$\displaystyle \text{Hence: }\:\sqrt[3]{48x^3y^7} \:=\:16xy^4$$ ?

$$\displaystyle \text{Then: }\:\sqrt[3]{27x^3y^7} + \sqrt[3]{48x^3y^7} \;=\;9xy^4 + 16xy^4$$ ?

$$\displaystyle \text{And evidently: }\:9xy^4 + 16xy^4 \;=\;(9+16)(x+x)(y^4+y^4) \;=\;25x^2y^8$$ ??

This solution should be in the Guinness Book for "The Most Errors in One Solution".

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

And did you see what Sporty did with the second problem?

$$\displaystyle \frac{3x^2 - 5x + 4}{x+3} \;=\;\frac{3x^2}{0x^2} + \frac{-5x}{1x} + \frac{4}{3} \;=\;3x^2 - 4x + 7$$

#### sportywarbz

##### New member
Okay so for part one.... here's what I have now...

sqrt(27x^3y^7) + sqrt(48x^3y^7)
(sqrt(27)sqrt(x^3)sqrt(y^7)) + (sqrt(48)(x^3)(y^7))
(3sqrt(3))(xsqrt(x))(y^3 sqrt(x)) + (2sqrt(12))(xsqrt(x))(y^3sqrt(y))
(3xy^3sqrt(3xy)) + (2xy^3sqrt(12xy))
(5xy^3)sqrt(15xy)

Am I closer?

#### Denis

##### Senior Member
sportywarbz said:
1. Here is the problem:
?(27x³y?)+?(48x³y?)
This is what I have for a solution... I am wondering if I am correct?
9xy?+16xy?
25x²y?
WHY were you given that problem?

J

#### JeffM

##### Guest
sportywarbz said:
Okay so for part one.... here's what I have now...

sqrt(27x^3y^7) + sqrt(48x^3y^7) This would be clearer if you wrote sqrt(27(x^3)(y^7)) + sqrt(48(x^3)(y^7)).
(sqrt(27)sqrt(x^3)sqrt(y^7)) + (sqrt(48)(x^3)(y^7)) I presume you mean (sqrt(27)sqrt(x^3)sqrt(y^7)) + (sqrt(48)sqrt(x^3)sqrt(y^7))
(3sqrt(3))(xsqrt(x))(y^3 sqrt(x)) + (2sqrt(12))(xsqrt(x))(y^3sqrt(y)) Yes although 2sqrt(12) = 4sqrt(3).
(3xy^3sqrt(3xy)) + (2xy^3sqrt(12xy)) Yes, but it would be clearer if you wrote (3x(y^3)sqrt(3xy)) + (2x(y^3)sqrt(12xy))
(5xy^3)sqrt(15xy) Here is where you go off the rails. Your next step should be
x(y^3)(3sqrt(3xy) + 2sqrt(12xy)) because x and y^3 are common factors. It is generally false that (a * sqrt(b)) + (c * sqrt(d)) = (a + c)(sqrt(b + d)). Try it with a couple of numbers (2 * sqrt(16)) + (7 * sqrt(9)) = (2 * 4) + (7 * 3) = 8 + 21 = 29, but (2 + 7) * sqrt(16 + 9) = 9 * sqrt(25) = 9 * 5 = 45.

Am I closer?
Yes, you are quite a bit closer. You can get there if you substitute 4sqrt(3) for 2sqrt(12). Do you see why this is both necessary and helpful?
Try it again. I suspect you will get it right this time.

#### mmm4444bot

##### Super Moderator
Staff member

sportywarbz said:
3xy^3 sqrt(3xy) + 2xy^3 sqrt(12xy)

I removed the outer sets of parentheses because they're not needed, and I typed a space in front of each sqrt(). This makes the expression easier to read.

Am I closer?

If the original exercise involves square roots instead of cube roots, then yes you are much closer now. See my comment at the end about square-root notation versus cube-root notation.

You simplified sqrt(48) to 2 sqrt(12). It can be simplified further than that because the number 12 contains a perfect-square factor.

If you first consider the prime factorization of 48, it will help you to simplify sqrt(48) all the way.

48 = (2^4)(3)

Therefore, sqrt(48) = sqrt(2^4)*sqrt(3) = 4 sqrt(3)

Also, it looks like you considered sqrt(3xy) and sqrt(12xy) to be like-terms, which you combined to make sqrt(15xy). They are not like-terms, so you cannot add them that way.

Your like-terms at the end of this exercise should be 3xy sqrt(3xy) and 4xy sqrt(3xy).

$$\displaystyle \sqrt{48x^3y^7}$$ ? This radical sign means "square root"
$$\displaystyle \sqrt[3]{48x^3y^7}$$ ? This radical sign means "cube root"; note the index 3