How do you factor a number?

Answer 1

There are many ways, but here is one.

Example:

A. x2 + 5x + 6

coefficients: 1, 5, 6

You have to find two numbers that the two ends multiply to get the sum of the middle coefficient in the order of ax2 + bx + c (read on to understand)

1 x 6 = 6

Find two numbers that multiply to 6, but add to 5

1 6 : 1 x 6 = 6, 1 + 6 = 7 X

2 3 : 2 x 3 = 6, 2 + 3 = 5 <---

So, the special numbers are 2 and 3.

Now, add them into this: (x+_)(x+_)

Both are positive, so it works! (x+2)(x+3)

Let's check.

(x+2)(x+3)

Use FOIL Method (first, outer, inner, last)

= x2 + 2x + 3x + 6

= x2 + 5x + 6

It works! Now, if you have trouble understanding, reread it a few times, and you'll get it in a blink!

Answer 2

If you mean "factorize" than this is my favourite method. Take this standard equation:

X^2-12X+35

All you need to do is find two numbers which when added together equal -12 and when multiplied together equal 35.

-5 and -7 can be added to make -12

-5 and -7 can be multiplied to make 35

Now that you have your two numbers, this is how you write them in a factored equation:

Y=(X-5) (X-7) And that's your factored equation!

NOTE: Those 2 numbers are also your "zeroes" or "x-intercepts" on your parabola. But when you use them on your graph, YOU MUST CHANGE THEIR SIGNS! So as zeroes they would be positive 5 and 7 rather than negative. (5,0) (7,0) Good luck.

Answer 3

There are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 108

The prime factorization can be found by using a factor tree.

108

54,2

27,2,2

9,3,2,2

3,3,3,2,2

2^2 x 3^3 = 108

Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 10 and 11.

Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 2 and 3. Add one to each. 3 x 4 = 12

108 has 12 factors. Six of them are 10 or less, six of them are 11 or greater. All we have to do is divide the numbers one through ten into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier.

108 is divisible by...

1 because everything is.

2 because it's even.

3 because its digits add up to a multiple of 3.

4 because its last two digits are a multiple of 4.

6 because it's a multiple of 2 and 3.

9 because its digits add up to a multiple of 9.

That's six factors less than 10. Divide them into 108. That's the rest of them.

(108,1)(54,2)(36,3)(27,4)(18,6)(12,9)

1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108

Notice that all of those numbers, except for 1, can also be found in the prime factorization.

Answer 4

To solve a quadratic equation, you first need to make sure that you have zero on the right side. Then, in the simplest case, you would have an equation that starts with x squared - for example, x squared - 5x + 6 = 0. To factor this, in this case you write the left side as (x - ...) (x - ...); the minus sign in this example is due to the fact that the constant term is positive, but the linear term is negative. Then, to fill in the blanks, you need to find two numbers whose product is 6, and that add up to 5.In more complicated cases, it's better to use the quadratic formula.

For more details about factoring, check any standard high school algebra book.