How do I factor 4x^2+9x+5=0? + others?

Question

2x^2-3x-2=0
12x^2+9x-3=0
9x^2-16=0

Answer 1

The quadratic formula is an absurd way to factor. Factoring is supposed to help you find the roots, you can't solve for the roots using a big stupid formula and then use that to factor.

Start with the easiest: 9x² - 16 = 0
This is a difference of two squares. See:
http://en.wikipedia.org/wiki/Difference_of_squares

9x² - 16 = 0
(3x - 4)(3x + 4) = 0


Now let's try another: 2x² - 3x - 2 = 0
We know it has to start like this:
(2x + a) (x + b)
for some a and b. Do it out:
(2x + a) (x + b) =
2x² + (a+2b)x + ab
Thus:
ab = -2
2b+a = -3
You should be able to tell now that b = -2, a = 1.
(2x + 1) (x - 2) = 0


Now on to the next one:
4x²+ 9x + 5 = 0
The first possibility is that we have:
(4x + a)(x + b)
The other possibility would be:
(2x + a)(2x + b)
But we'll try the first one first:
(4x + a)(x + b) =
4x² + (a+4b)x + ab
This means that:
a+4b = 9
ab = 5
Pretty clear that a=5,b=1.
Thus:
(4x + 5)(x + 1)


Finally:
12x²+ 9x - 3 = 0
First factor out a 3. This is important. The first answerer didn't do this, which means his answer is wrong (in addition to having a totally ridiculous explanation).
12x²+ 9x - 3 =
3 (4x² + 3x - 1)
We want to factor this as:
3(4x + a)(x + b) =
3(4x² + (a+4b)x + ab)
meaning:
ab = -1
a+4b = 3
So clearly b = 1, a = -1, and you get:
3(4x - 1)(x + 1) = 0

Notice that in each one, we follow these steps:
1) factor out a common term (like 3 in the last one)
2) come up with a "form" for it, like (4x+a)(x+b)
3) expand the form, like 4x² + (a+4b)x + ab
4) match up a+4b with the x coefficient and ab with the constant
5) figure out a and b, or if no such integers exist, repeat using a different "form" of the equation, like (2x+a)(2x+b)

The "form" depends on the factors of the x² coefficient, and you might notice that you always get ab=constant, for any form you use.

Never use the quadratic formula for these. It's absurd. Once you follow this explanation a few times, you'll get the hang of it and it will become easier. You should never be taking square roots and dividing and by 2a and all that junk. It's completely unnecessary.

Answer 2

The trick is noticing the degree of the polynomial, the prime factorization on the x^2 (or on an nth degree polynomial, x^n) coefficient, and the prime factorization of the last coefficient (the one without an x, or with lowest exponent on x)...

4x^2+9x+5=0...
(4x+5)(x+1)=0

2x^2-3x-2=0
(2x+1)(x-2)=0

12x^2+9x-3=0
(3x+3)(4x-1)=0

9x^2-16=0
(3x-4)(3x+4)=0

Answer 3

4x² + 9x + 5 - 0

4x² + 4x + 5x + 5 = 0

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

*x + 5(x + 1) = 0

- - - - - - - - -

2x² - 3x - 2 = 0

2x² - 4x + 1x - 2) = 0

2x(x - 2) + 1(x - 2) = 0

(2x + 1)(x - 2)

- - - - - - - -

12x² + 9x - 3 = 0

12x² + 12x - 3x - 3 = 0

12x (x + 1) - 3(x + 1) = 0

(12x - 3)(x + 1) = 0

- - - - - - s-

Answer 4

Use the quadratic formula...

-b +or - Sqrt b^2 - 4(a)(c) all over 2a

These are the answers...

1. (4x+5) (x+1)
2. (2x-1) (x+2)
3. (4x-1) (3x+3)
4. (3x+4) (3x-4)

You can find the value of x's for each by making each value in the parenthesis = 0 and then solving it...

These are all factorable w/o using the qradratic formula, but the formula works for all quadratic equations...