# Thread:

1. ## Grade ten math

Factor completely:
40x^5y^2 - 32x^7y^6 + 28x^8y^4 - 36x^7y^5

I thought factor completely meant solve for the unknown variable. However, I have no idea what the answer is for this question. I know the greatest common denominator is 4x^5y^2. But do not know how to get on further. What do I do?

14. A rectangular field is to be enclosed on four sides and divided by another section of fence parallel tot he width using 4800m of fencing. What is the maximum area of the entire field? What are the dimensions of the field that maximums the area?

How would I even start this? I know that A = l x w. But other than that I have no idea. I thought it would be something like x (4800 -x) and then solve for x, but I don't think that will give me the correct answer. How do I do this question?

2. Originally Posted by Barthayn
Factor completely:
40x^5y^2 - 32x^7y^6 + 28x^8y^4 - 36x^7y^5

I thought factor completely meant solve for the unknown variable. However, I have no idea what the answer is for this question. I know the greatest common denominator is 4x^5y^2. But do not know how to get on further. What do I do?
Factor completely means take out common factors to get an expression into it's simplest form. You've already found the GCF (F=factor) so take that out front and divide each side by it

$4x^5y^2(10-...)$ - can you finish off?

3. Be careful with your terminology. Highest common factors and lowest common denominators are not the same thing.

$40x^5y^2 - 32x^7y^6 + 28x^8y^4 - 36x^7y^5$

Let's start with the constants. There is clearly a factor of 4 there.

$=4[10x^5y^2 - 8x^7y^6 + 7x^8y^4 - 9x^7y^5]$

As you've said, the highest common term of $x$ is $x^5$
So take this out next:

$=4x^5[10y^2 - 8x^2y^6 + 7x^3y^4 - 9x^2y^5]$

Now try to finish for the y terms. If you're still confused, reply and I'll go into detail.

4. Originally Posted by e^(i*pi)
Factor completely means take out common factors to get an expression into it's simplest form. You've already found the GCF (F=factor) so take that out front and divide each side by it

$4x^5y^2(10-...)$ - can you finish off?
I got $4x^5y^2(10-8x^5y^4+7x^3y^2-9x^2y^3)$

However, this cannot be the final answer. As well how do I do the second question that I asked?

Quacky I am still confused :S

5. Your answer is correct. That is as much as you can do with that function in terms of factorization. I'm trying to understand the second question at the moment - the wording is confusing me.

6. Q.14 is a Calculus question, but since this is Year 10 maths, I doubt the OP would know any. Have you been taught any Calculus yet?

7. No, calculus and vectors is taught in grade 12 math. How would I do the question though?

8. Originally Posted by Barthayn
14. A rectangular field is to be enclosed on four sides and divided by another section of fence parallel tot he width using 4800m of fencing. What is the maximum area of the entire field? What are the dimensions of the field that maximums the area?

How would I even start this? I know that A = l x w. But other than that I have no idea. I thought it would be something like x (4800 -x) and then solve for x, but I don't think that will give me the correct answer. How do I do this question?
I assume the "4800 m of fencing" is for the entire field- the way it is placed makes it look like it is the "fence parallel to the width" but that can't be right. Yes, area equals length times width: A= lw. The total length of fencing used is the two lengths, the two widths, and that additional piece of fencing "parallel to the width" and so the same length as the width: 2l+ 2w+ w= 2l+ 3w= 4800.

You can solve that equations for either l or w as a function of the other. For example, 2l= 4800- 3w so l= 2400- (3/2)w. Now put that into A= lw: $A= (2400- (3/2)w)w= 2400w- (3/2)w^2$. That's a quadratic function in w and you can find its maximum value (at the vertex of the parabola) by "completing the square".

9. Originally Posted by Barthayn
I got $4x^5y^2(10-8x^5y^4+7x^3y^2-9x^2y^3)$
Should be $8x^2y^4$, not $8x^5y^4$.
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