fast way to factor a quadratic equation?

Question

i remember back in junior high learning some sneaky snake fast way to factor a quadratic... aside from using the LONG drawn out quadratic formula... it was something like, if the middle sign is negative then you know the signs are different in the two factors (as in:

( __ + __ )( __ - __ )

and if the last sign is negative it means the second sign is negative? GAAAAH i know i am butchering this, but if you kno0w what i'm talking about PUHHHHLEASE inform me!

i jsut realized how absolutely confusing that sounds if you don't know what i'm talking about... haha

Answer 1

This is easiest to understand through example.

For instance, if you have: x^2 - 8x + 7.

First off, the coefficient in front of the x^2 is 1, simplifying the work. Because there is a '+ 7', we know that the two numbers that multiply to equal 8 must be either both positive, or both negative. Since the coefficient in front of the x is '- 8', then we know the two numbers must add up to a negative number.
In this case, we know that adding two positive numbers will not result in a negative number, so the two numbers we are looking for must be negative, thus giving us
( x - __ )( x - __ )
Now we think, "What numbers add up to -8 and multiply to give us +7?" Those numbers are -1 and -7 so we have:
( x -1 )( x - 7 ).

This also works for something like x^2 +3x - 4

Once again, the coefficient in from of x^2 is 1, so we can start off with:
( x +/- __ )( x +/- __ )
We notice the '+ 3' and the '- 4' so we think, "What two numbers will add up to 3 and multiply to give -4. The numbers +4 and -1 should jump into your head so you can complete the factoring:
( x + 4 )( x - 1 ).

The key is to see whether the last "sign" is positive or negative. If it is positive, then the numbers will either be both positive or both negative. If it is negative, then the numbers will have opposite signs.

Answer 2

Example

2x² + 8x - 10 = 0

Common factor first

2(x² + 4x - 5) = 0

The middle term inside the parenthesis is + 4x

Find the sum of the middle term

x = 1

Multiply the first term 1 times the last term 5 equals 5 and factor

The factors of 5 =

1 x 5

- 1 and + 5 satisfy the sum of the middle term

1 = x

insert = x and + 5x into the equation

2(x² + 4x - 5) = 0

Factor by grouping

2(x² - x + 5x - 5) = 0

2[x(x - 1) + 5(x - 1) = 0

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

- - - - - - - - - -

Completing the square and the quadratic formula also works.

- - - - - - -s-

Answer 3

The b moves the vertex along the parabola c - ax², and just like other translations f(x - a) is f(x) moved 'a' units right and f(x + a) is moved 'a' units left. So when b is negative it is similar to the translation f(x-a). Let's do an example. Say we have the parabola y = x² - 4x + 5 Now it's vertex form is y = (x - 2)² + 1 So the vertex is at the point (2,1) which lies on the parabola f(x) = 5 - x², specifically f(2) = 1 as expected. y = (x - 2)² + 1 is the parabola g(x) = x² + 1 translated 2 units to the right. So you see, when b is negative (and a ia positive) the graph moves right and when b is positive (and a is positive) the vertex moves left. If a is negative then this is reversed as you can factor the negative out then apply the conditions above.

Answer 4

Well, there are a couple ways i know.
1) you guess
2) you take the product of a and c, then find two factors of that that when added, equals b. Then, separate b into those two numbers, then simplify.
3) this is my very own equation, and it only works when there are two solutions and a b and c are not zero. Simply, you take the solutions (using the quadratic formula). Then you take one solution, and take its reciprocal. If the solution is 4/7, then take 7/4, and make 7x=4. Then do it with the other solution, and tada, you have the two factors.

Answer 5

the square minus square theorem
factor a^2-b^2=(a+b)(a-b)
this only works if there is no
this has nothing to do with quadratics.
the only way to properly solve a quadratic is to use the formula.
the other factored form is just using random guess work.
there is a third way, which is to put the equation in to
a(x-h)^2+y, where h is the x of the vertex, and y is the y of the vertex. this method is called completing the square. after doing this, the solution is x=sqrt(-y/a)+h.
this way is very long, and it is the internal method of the quadratic formula.

Answer 6

I believe you are referring to FOIL - First, Outside, Inside, Last
It's more for "un-factoring"

Ok I think I know what you mean-
x^2-4 = (x+2)(x-2)

so it ends up being x^2-y^2 = (x+y)(x-y)
where y is whatever the constant is.

Completing the square is what it's called. It can be done for other numbers too - but is great if you already have a square constant.!


Hope I helped

Answer 7

My college math teacher said that the best way was to have a sharp pencil and a good eraser.

You are right though, if the middle sign is negative then one must be negative and one must be poitive.

Middle sign is positive then either both are poitive or both are negative.