how do you completely factor this?

Hi,

6x² - x - 5 = (x - 1)(6x + 5) <==ANSWER

This is how to factor trinomials the way that I teach it, which may be different from what you've done. Try to follow the steps. This method will work on any trinomial.

6x² - x - 5 Look for a GCF. There is none this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(6x.......)(6x..........) First sign goes in first parentheses.
(6x -....)(6x.........) Product of signs goes in 2nd parentheses.
(6x -....)(6x +.....) <== pos is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 6 x 5 = 30 So, out to the side list pairs of factors of 30.

30
------
1, 30
2, 15
3, 10
5, 6

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(6x -....)(6x +.....) Your signs are different, so you want to subtract factors to get 1. Those factors are 5 and 6. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(6x - 6)(6x + 5)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 6 but the second parentheses does not reduce.

(6x - 6)(6x + 5)
----------
.....6
This reduces to your final factors of

(x - 1)(6x + 5) <==This is the answer to your problem.


I hope that helps!! :-)

6x² - x - 5 = 0
x² - 1/6x = 5/6
x² - 1/12x = 5/6 + (- 1/12)²
x² - 1/12x = 120/144 + 1/144
(x - 1/12)² = 121/144
x - 1/12 = +/- 11/12

= x - 1/12 - 11/12, = x - 12/12, = x - 1
= x - 1/12 + 11/12, = x + 10/12, = x + 5/6, = 6x + 5

Answer: (x - 1)(6x + 5)

Proof:
= (x - 1)(6x + 5)
= 6x² + 5x - 6x - 5
= 6x² - x - 5

34g

6x^2 - x - 5
= 6x^2 + 5x - 6x - 5
= (6x^2 + 5x) - (6x + 5)
= x(6x + 5) - 1(6x + 5)
= (6x + 5)(x - 1)

(6x+5)(x-1)

Howdy & I sure hope all is well. :)

As for your question, algebraic expressions are much like numbers. It's just that they're more cumbersome, since you can't reduce them as you do with numbers. For example, you can add 2 & 3 and get 5. But X + 2 + 2X + 3 can only be reduced to 3X + 5.

Now, just as you can rewrite numbers as products, 12 can be rewritten as 2*6 or 3*4, you can do the same with algebraic expressions. But before you work on that, you need to remember how it is that you get that kind of expression in the first place.

Remember when you learned how to multiply together numbers that have 2 or more digits, such as 2 times 47? You would go:

2 times 7 is 14, & would write down the 4 & carry the 1. Then you'd multiply 2 & 4, & get 8, then add the 1 to it, giving you 9. So you'd put the 9 to the left of 4, something like this:

1
47*
2
--
94

You do that because of the distribution of the product. Meaning that, since 47 is really 40 + 7, what you're doing is 2*40 + 2*7, except you do it right to left. (It's actually more efficient to do it from left to right, even in the method above). So 2*47 = 2*40 + 2*7. This relates to algebraic expressions, since you're multiply two expressions together, then adding up the common terms. For example:

(aX + b)*(cX + d) =

acX^2 + adX + bcX + b*d =

acX^2 + (ad + bc)X + bd

So when you're factoring, the very first step is to figure out what numbers multiply together to give ac & bd, & add up to give ad + bc. In your question

6X^2 -- X -- 5,

only 1 & 5 multiply to give 5. As for the 6, 1*6 = 6 & 2*3 = 6. But, with 2 & 3, you'll either get 2 -- 5 or 3 -- 10. So the pair you need is 1 & 6, since 5 -- 6 = -- 1. Now, you have to be careful where you put the 6. You're given enough clues, the -- 1 for the X & the -- 5. So you know that 5 & 6 must be together in the same expression, since you don't want to multiply them together. So the expression can now be re written as:

6X^2 -- X -- 5 =

(X -- 1)*(6X + 5)

You can confirm that this will yield the desired expression by multiplying the two brackets:

6X^2 + 5X -- 6X -- 5 =

6X^2 -- X -- 5

I sure hope this was clear & helpful enough. :) Good luck, take care, & have a great day. :)

Cheers! :)

6x² - x - 5

>> First, multiply 6 with -5 = -30
>> Find the factors of -30 =
(-1)(30), (-2)(15), (-3)(10), (-5)(6),
(-6)(5), (-10)(3), (-15)(2), (-30)(1)
>>
use -6 + 5 = -1 because thge coeffisient of -x = -1

>> Then:
6x² - 6x + 5x -5

Factorize:

6x(x - 1) + 5(x - 1)

(6x + 5)(x - 1)

x's=[1+ & - (1+120)^1/2]/12,then 6(x-x1)(x-x2).
6(x-1)(x+5/6)=
=(6x+5)(x-1)

It is suppose to be like this:

6x2 - x - 5 = 0

(6x + 5)(x - 1) = 0

Check:
6x times x = 6x2
6x times -1 = -6x
5 times x = 5x
5 times -1 = -5
so -6x plus 5x = -x

so you get back 6x2 - x - 5 = 0

6x^2 - x - 5 = 0

(6x + 5) (x - 1) = 0