Thread: Need help with a couple of questions

1. Need help with a couple of questions

Any help would be greatly appreciated.

1. When an expression f(x) is divided by 3x+1, the quotient is 2x-3 and the remainder is x+5. What is f(x)?

2. Factor the following:

(a) x^12n - y^20n
(b) x^2 + 4xy +4y^2 - m^2 - 6mn - 9n^2 [EDIT - SOLVED?]: (x+2y)^2 - (m+3n)^2
(c) 1/3x^-3 y^6 -1/4x^4y^3

2. Re: Need help with a couple of questions

1. Suppose when x is divided by 3, the quotient is 5 and the remainder is 1. Can you find x? Also, it is essential to know the definition of the remainder. Can you write it?

2.

(a) Use the formula $a^2 - b^2 = (a - b)(a + b)$.
(b) Use the formula $a^2 + 2ab+b^2=(a+b)^2$ and the one in (a).
(c) You can either factor out $x^{-3}y^6$ or use the formula in (a).

3. Re: Need help with a couple of questions

Originally Posted by emakarov
1. Suppose when x is divided by 3, the quotient is 5 and the remainder is 1. Can you find x? Also, it is essential to know the definition of the remainder. Can you write it?

2.

(a) Use the formula $a^2 - b^2 = (a - b)(a + b)$.
(b) Use the formula $a^2 + 2ab+b^2=(a+b)^2$ and the one in (a).
(c) You can either factor out $x^{-3}y^6$ or use the formula in (a).
Thanks, I understand #2 now, but I'm still having trouble with #1. (I haven't done dividing by hand since I was in Grade 6.) How do I find f(x)?

4. Re: Need help with a couple of questions

The quotient and remainder when a natural number x is divided by y are natural numbers q and r such that $x = qy + r$ and $0\le r. If you know y, q and r, you can find x from the equation above.

5. Re: Need help with a couple of questions

Originally Posted by emakarov
The quotient and remainder when a natural number x is divided by y are natural numbers q and r such that $x = qy + r$ and $0\le r. If you know y, q and r, you can find x from the equation above.
But how do I find y?

6. Re: Need help with a couple of questions

The number y in my post is what you divide by. In the original problem about polynomials, this is given.

7. Re: Need help with a couple of questions

Originally Posted by emakarov
The number y in my post is what you divide by. In the original problem about polynomials, this is given.
Thanks, one more question - how do you factor out x^-3 y^6?

8. Re: Need help with a couple of questions

one more question - how do you factor out x^-3 y^6?
As usual. When you factor out $a$ from $m\cdot a - n\cdot b$, you get $a(m - n\cdot b/a)$. Here $a = x^{-3} y^6$ and $b = x^4y^3$, so you need to find $(x^4y^3)/(x^{-3} y^6)$. Remember that $x^k/x^l = x^{k-l}$ and take care when $l$ is negative.

There are several ways to factor $m\cdot a - n\cdot b$. This also equals $b(m\cdot a/b-n)$ and $(\sqrt{m}\sqrt{a}+\sqrt{n}\sqrt{b})(\sqrt{m}\sqrt{ a}-\sqrt{n}\sqrt{b})$.