Factoring question?

2 (6x² + 17x + 5)
2 (3x + 1)(2x + 5)

There is one method, factoring by grouping...I'll show you how...it is easy after you do it a few times.
12x²+34x+10
Factor out a 2 first of all.
2(6x² + 17x + 5) aside from the 2 we have
6x² + 17x + 5
To factor by grouping we are going to rename the middle term, 17x. Do do this:
Multiply the coefficient of x², 6, and the constant, 5.
6 • 5 = 30. Find factors of 30 that multiply to give a product of 30 and a SUM of your coefficient in 17x, 17.
15 and 2 work (There will always be 2 numbers that work if it is factorable, by the way.)
15 • 2 = 30 and 15 +2 = 17
So, we restate the middle term as 15x + 2x

6x² + 2x +15x + 5 .....Group them as follows:
= (6x² + 2x) +(15x + 5)...factor each set of parentheses
= 2x(3x + 1) + 5(3x + 1)..Notice the matching parentheses
= (3x + 1)(2x + 5) (don't forget your 2 in your final statement.)

2(3x + 1)(2x + 5) = 12x² + 34x + 10

Try a few this way, you can't fail if it is factorable.

Example: 6x² + 23x + 21
6(21) = 126; 9(14) = 126 ; 9 + 14 = 23
6x² + 9x + 14x + 21
= (6x² + 9x) + (14x + 21)
= 3x(2x + 3) + 7(2x + 3)
= (2x + 3)(3x + 7)

Example: Even works for:

x² - x - 6; ...-6(1) = -6 ; (-3)(2) = -6 and -3 + 2 = -1
= x² + 2x + -3x - 6
= (x² + 2x) + (-3x - 6)
=x(x + 2) + -3(x + 2)
= (x + 2)(x - 3)
One does not usually factor this kind with grouping because it is so easy, but it does work. Kind of like trying to use the distributive property instead of the order of operations in arithmetic...it is slower. But for the other two above, it does take the guess work to a minimum.

1st take out the common factor
2(6x^2+17x+5)
as 6 X 5=30so find factorsof 30 which sum up to17 i.e. 2 and 15
2(6x^2+2x+15x+5)
=2[2x(3x+1)+5(3x+1)]
=2(3x+1)(2x+5)

All coefficients are even numbers and so you know you will at least be able to factor a two out of each term.

2(6x^2+17x +5) Now ignore the 2 for the time being:

(6x^2 +17x +5) Multiply the leading coefficient by the constant:

6 times 5 equals 30. Now list all the pairs of factors of 30:

1 x 30
2 x 15
3 x 10
5 x 6

Look for the pair of numbers above, that when added or subtracted equals the coefficient of the MIDDLE TERM'S COEFFICIENT: The 17 in this case.

You can see that one pair works. the 2 and 15
+2+15=+17 the middle term coefficient!

Now rewrite the whole middle term using these numbers:

+17x becomes +2x+15x Now rewrite with the rest of the terms:

6x^2 +17x + 5

becomes

6x^2 + 2x + 15x + 5 notice this is the same as above?

Now factor in groups of two like so:

2x(3x +1) +5 (3x+1) Now the BEAUTY OF THIS SYSTEM is that it is self confirming self checking if you will. If you see the same elements repeating within the brackets then you know you did it correctly! Notice the two (3x+1) ? Cool.

Now we can use the distributive law:
(2x+5)(3x+1) and we must remember the original 2

2(2x+5)(3x+1)

post note: Cory's Mom explains this method well!

"It minimizes guess work!" Indeed.
Use this method when there is a coefficient (other than 1) in front of the leading term . 3x^2 , (1/2)x^2 -5x^2 are examples of this.

If there are no coefficients other than 1 for the leading term, then you can factor by a simpler method which might involve some trial and error. There are several different methods to factor trinomial 2nd degree expressions like your example.

Cory's mom and I like the factor by grouping / rewriting the middle term and using the distributive law method best when the coefficient is not one.

ok i am not sure what method u are using to factor this equation, but this is how i was taught to do it

12x^2+34x+10

first...take the numerical coefficient of the quadratic term (12) and multiply it by the constant term (10)...you should end up with 120.

then...take the 120 and find the two factors that once added together you get the numerical coefficient of the linear term (34) and those two factors are 4 and 30.

now...replace the 34x with 4x and 30x
12x^2+4x+30x+10
and group together the terms by two
(12x^2+4x) + (30x+10)

then...you take the lowest common factor of the two groups and to know that you did it right, when you factor you should get the same thing for both...for example...
4x(3x+1) +10(3x+1)

finally...take just remove one of the 3x+1 and your and take the greatest common factor of 4x+10... and the complete factorization of 12x^2+34x+10 is....
2(2x+5)(3x+1)
thats your answer...try this one and mail me your answer...

3x^2+31x+36

12x^2 + 34x + 10
= 2(6x^2 + 17x + 5) (reduce the amount)
= 2(3x + 1)(2x + 5)

Use the AC method, the number in front of x^2 is A, the number in front of x is B, the constant term is C, you need to find two numbers that are factors of AC and that have a sum of B.

Go to the website included and look for AC method, good luck!