# Homework Help: Help with factor large numbers

1. Mar 30, 2010

### Poker-face

I am 31 and just started back in Math. My first class is Intermediate Algebra. I am sloving equations that force me to factor large numbers. Not sure if this is a skill that I was supposed to rember from high school, but nevertheless it is taking me a long time to do so. Can anyone tell me what the rules are when factor large roots. For example

1. w + 14 = 1458 to the 4th root.

EG

2. Mar 30, 2010

### Staff: Mentor

Re: Factoring?

I'm not sure what your equation is. Is it this?
$$w + 14 = \sqrt[4]{1458}$$

Click on the equation I wrote to see the LaTeX script I wrote for this equation.

If that's the equation you want to solve, there is no factoring needed. All you have to do to solve for w is to add -14 to both sides of the equation.

3. Mar 30, 2010

### Poker-face

Re: Factoring?

Yes. How do you simplfy the square root?

4. Mar 30, 2010

### Staff: Mentor

Re: Factoring?

That's a fourth root.
Simplify it by finding all factors and seeing if any are to the fourth or higher power. For this problem, 1458 = 2 * 729 = 2 * 9 * 81 = 2 * 93 = 2 * 36

The last expression can also be written as 34 * 2 * 9 = 34 * 18

Now use the property of square roots, cube roots, fourth roots, etc. that says
$$\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}$$

For even roots (square root, fourth root, etc.) in the equation above, both a and b have to be nonnegative. For odd roots (cube root, fifth root, etc.) a and be can be any real numbers.

5. Mar 30, 2010

### Poker-face

Re: Factoring?

I understand the rule but how you get to step two - 2 x 9 x 81

6. Mar 30, 2010

### Staff: Mentor

Re: Factoring?

729 = 9 * 81. I used the concept I mentioned to you in another thread - if the sum of the digits of a number is 9 or a multiple of 9, the number is divisible by 9.

So 1458 = 2 * 729 = 2 * 9 * 81

7. Mar 30, 2010

### Poker-face

Re: Factoring?

Thanks again both threads were a big help!!!

EG