Question

HELLLP!!!! VERY EASY QUESTION!!

Divide (x^4 + x^3y - xy^3 - y^4)/(x^2-y^2)

How to do??

Answer 1

i'm stuck :/

Answer 2

You need to factor the numerator and the denominator.

Answer 3

how would i factor it?

Answer 4

If it's a very easy question, why do you need help? :-)

You can do this just like ordinary long division.

\(x^2 - y^2\) | \( x^4 + x^3y - xy^3 - y^4\)

divide the first term of the dividend \((x^4 + x^3y-xy^3-y^4)\) by the first term of the divisor \((x^2-y^2)\) What is the result?

Answer 5

i really don't get it:/ can u show me the first step?

Answer 6

how would i do it by long division?

Answer 7

I did. What is \(x^4/x^2\)?

Answer 8

x^2, but what about y?

Answer 9

You'll see :-)

Now multiply the first term of the quotient that we just got (\(x^2)\) by the entire divisor \((x^2-y^2)\), and subtract that from the dividend.

Answer 10

where would i put that first term (x^2) though?

Answer 11

just keep track of it. it's the first term of the quotient. you can write it down over the top of the problem if you like, just as you would in doing long division on numbers.

Answer 12

Factor the numerator by grouping.
Factor the denominator by difference of squares.

\(\dfrac{x^4 + x^3y - xy^3 - y^4}{x^2-y^2} \)

\(\dfrac{(x^4 + x^3y) - (xy^3 + y^4)}{x^2-y^2} \)

\(= \dfrac{x^3(x + y) - y^3(x + y)}{x^2-y^2} \)

\(= \dfrac{(x^3-y^3)(x + y)}{(x+y)(x - y)} \)

\(= \dfrac{(x-y)(x^2 + xy + y^2)(x + y)}{(x+y)(x - y)} \)

\(=x^2 + xy + y^2\)

Answer 13

So \(x^2(x^2-y^2) = x^4-x^2y^2\)

If we subtract that from our original dividend, we get

\[x^4 + x^3y - xy^3 - y^4 - (x^4-x^2y^2) = \cancel{x^4} + x^3y-xy^3-y^4-\cancel{x^4}+x^2y^2\]\[=x^3y-xy^3-y^4+x^2y^2\]Now we repeat the process.
What is \(x^3y/x^2\)?

Answer 14

xy?

Answer 15

Answer: \(x^3y/x^2 = xy\) so \(xy\) is the next term of our quotient.

\[xy(x^2-y^2) = x^3y-xy^3\]We subtract that from our remaining polynomial:
\[x^3y-xy^3-y^4+x^2y^2-(x^3y-xy^3) =\cancel{ x^3y}-\cancel{xy^3}-y^4+x^2y^2-\cancel{x^3y}+\cancel{xy^3}\]\[=x^2y^2-y^4\]

Answer 16

If we divide \(x^2y^2/x^2\) we get our final term: \(y^2\)
\[y^2(x^2-y^2) = x^2y^2-y^4\]
subtracting that from the remaining polynomial we are left with
\[x^2y^2-y^4-(x^2y^2-y^4) = x^2y^2-y^4-x^2y^2+y^4 = 0\]

So our answer is \(x^2+xy+y^2\)

Answer 17

If you're good at factoring, maybe factoring works faster. If the problem doesn't divide out evenly, and you need to find the remainder, you need to do it this way. Mistakes can be made with either approach if you aren't careful :-)