Intermediate Algebra Factoring

In summary, according to EG, you can check prime numbers up to 30 to see if the other factor is less than 30. If it is, you can try one of the prime numbers up to 300. If that number is too small, try 313, 323, and so on.
  • #1
60
0
Does anybody know any tips on factoring large numbers.

For example- x[2] - 70x + 901=0

I keep running into large numbers like these and spend a lot of time trying factor them out. Is their any short cut other than just trial and error?

Thanks, EG
 
Physics news on Phys.org
  • #2
Poker-face said:
Does anybody know any tips on factoring large numbers.

For example- x[2] - 70x + 901=0

I keep running into large numbers like these and spend a lot of time trying factor them out. Is their any short cut other than just trial and error?

Thanks, EG
Start by factoring 901. You want two factors that multiply to make 901 and that add to make -70.
 
  • #3
I understand that, but how do you know which numbers equal 901 without randomly picking numbers, or is it truly trial and error?
 
  • #4
No, it's not a random process at all, and it isn't trial and error, either.
901 is odd, so you don't need to try 2, 4, 6, 8, and so on.
901 isn't a multiple of 3 (the digits don't add up to 3), so you don't need to try 3, 6, 9, 12, ...
901 doesn't end in a 0 or 5, so you don't need to try 5, 10, 15, 20, ...

All you need to do is check prime numbers up to 30, which is approximately the square root of 901. If you go past 30, the other factor will be less than 30, so you should already have caught it
 
  • #5
Their are also asking me to find the square root of numbers like 1/117,649. how do you break this numbers without a calculator?

Thanks, EG
 
  • #6
Mark44 said:
No, it's not a random process at all, and it isn't trial and error, either.
901 is odd, so you don't need to try 2, 4, 6, 8, and so on.
901 isn't a multiple of 3 (the digits don't add up to 3), so you don't need to try 3, 6, 9, 12, ...
901 doesn't end in a 0 or 5, so you don't need to try 5, 10, 15, 20, ...

All you need to do is check prime numbers up to 30, which is approximately the square root of 901. If you go past 30, the other factor will be less than 30, so you should already have caught it

I see, what do you mean bye (the digits don't add up to 3)?
 
  • #7
Poker-face said:
Their are also asking me to find the square root of numbers like 1/117,649. how do you break this numbers without a calculator?

Thanks, EG

Let's just look at the denominator - 117,649

100 * 100 = 10,000
300 * 300 = 90,000
400 * 400 = 160, 000
so for the square root of 117,649, you're looking for a number between 300 and 400, but closer to 300 than 400. The last digit of 117,649 is 9, so the last digit of its square root has to be 3. So try 303 (probably too small), 313, 323, and so on. Whatever answer you get, take the reciprocal for the square root of 1/117,649.
 
  • #8
Mark44 said:
Let's just look at the denominator - 117,649

100 * 100 = 10,000
300 * 300 = 90,000
400 * 400 = 160, 000
so for the square root of 117,649, you're looking for a number between 300 and 400, but closer to 300 than 400. The last digit of 117,649 is 9, so the last digit of its square root has to be 3. So try 303 (probably too small), 313, 323, and so on. Whatever answer you get, take the reciprocal for the square root of 1/117,649.

Thanks that clarifies that>
 
  • #9
Poker-face said:
I see, what do you mean bye (the digits don't add up to 3)?
For any number that's divisible by 3, the digits add up to 3 or a multiple of 3.

18 --> 1 + 8 = 9
48 --> 4 + 8 = 12

If the digits don't add up to 3 or a multiple of 3, the number is not divisible by 3. For example, 1,000,007 is not divisible by 3, since the digits add up to 1 + 7 = 8.

There's a similar rule for 9. For any number that is divisible by 9, the digits add up to a number that is divisible by 9.

243 --> 2 + 4 + 3 = 9
243 = 9*27

1,000,026 --> 1 + 2 + 6 = 9
1,000,026 = 9 * 111,114
 
  • #10
Awesome, thanks for the quick response.
 

1. What is factoring in intermediate algebra?

Factoring in intermediate algebra is the process of breaking down a mathematical expression into smaller, simpler expressions. It involves finding the factors, or numbers or expressions that can be multiplied together to get the original expression.

2. Why is factoring important in intermediate algebra?

Factoring is important in intermediate algebra because it allows us to simplify complex expressions and solve equations. It also helps us identify patterns and relationships between different expressions, which is useful in solving more advanced algebraic problems.

3. What are the different methods of factoring in intermediate algebra?

The three main methods of factoring in intermediate algebra are the greatest common factor (GCF) method, the difference of squares method, and the grouping method. Other methods include factoring by grouping, completing the square, and using the quadratic formula.

4. How do I know when to use each method of factoring?

The method of factoring to use depends on the type of expression you are given. For example, the GCF method is useful when the expression has common factors, the difference of squares method is used for binomials in the form of a^2 - b^2, and the grouping method is used for expressions with four or more terms.

5. What are some common mistakes to avoid when factoring in intermediate algebra?

Some common mistakes to avoid when factoring in intermediate algebra include not checking for common factors, using the wrong method, forgetting to factor out negative signs, and not checking the final answer for correctness. It is also important to remember to always double check your work and simplify your final answer.

Suggested for: Intermediate Algebra Factoring

Replies
5
Views
842
Replies
29
Views
2K
Replies
5
Views
662
Replies
21
Views
359
Replies
3
Views
1K
Replies
8
Views
1K
Replies
3
Views
1K
Replies
12
Views
2K
Back
Top