how do you factor an equation??

Question

lets say for example......x^2-6x-7=0 step by step would be nice so i can understand it thanx

Answer 1

you reduce the equation as ap product of irreducible factors
x^2-6x-7=0
split the middle term as-7x and x
x^2-7x+x-7=0
x(x-7+1(x-7)=0
(x-7)(x+1)=0

Answer 2

compare the equation with ax^2+bx+c = 0, where a = 1, b=-6 and c = -7

Now try to find out the possibilities of getting b, in terms of a and c. Remember, axc = -7 here, so we get -6 = -7 + 1, till now u have to just do it in your mind
so we do the following,
x^2 - 6x - 7 = 0
or, x^2 - 7x+x-7 = 0
or, x(x-7) + (x-7) = 0
or, (x-7)(x-1) = 0

let see another example,
2x^2 -5x -3 = 0

axc = 2 x -3 = -6
b = -5 = -6 + 1
Not possible, -5 = 2-3 (always remember to use the numbers with their sign)

So we get
2x^2 -5x -3 = 0
or, 2x^2 - 6x + x - 3 = 0
or, 2x(x-3)+(x-3) = 0
or, (x-3)(2x+1) = 0

Hope this helps.

Answer 3

given an equation, ax^2 + bx + c = 0
I would first look at the "c" term and factor it.

in your case, c = -7, if I were to factor that, I get

-7 = (-1)*(7) or (-7)*(1)

then, i see if any of the factored forms woud add up to the b term.

in your case, b = -6
from factoring the c term, we see that:
-1 + 7 = 6
-7 + 1 = -6

since the factored form does add up to the b term, that is your root. Keep in mind, this method is useful if a = 1. I would do practice with problems where a = 1 so you get a feel of how to do these problems.

Down the line, you will be able to use the quadratic formula and that will be a useful tool to check your answers.

I hope this helps. good luck.

Answer 4

practice, for one.

for a simple 2nd order polynomial, I look at the coefficient on the 0th degree (in this case -7), and think of its factors. then, what sum of those factors, multiplied by x, gives me the coefficient on the 1st degree term.

so, in your example, the factors of -7 are 1, -7, and -1, and 7. Which pairs add up to -6? 1, -7. Then you write

x^2-6x-7 = (x+1)(x-7)


rfamily shows the other way to go about factoring this. Many methods often work, but not neccessarily at the same time. Trial and error...trial and error.

Anything much beyond cubes is quite tricky.

Answer 5

(x-7)(x+1)=0

Answer 6

first split the middle term '6x' so tht it equals the 3rd term '7' you can take factors of 6 as 6 and 1. then follow the formula (a+b) (a-b)

Answer 7

1) find two numbers a,b such that a*b = -7 AND a+b=-6

a*b=-7 : -1,7 or -7,1
from these only -7 + 1 = -6

So that (x-7)(x+1) is the factoriaation.

NOTE that this only works when there is nothing before the x^2

Answer 8

x^2-6x-7=0
1. calculate using abc formula
a x^2 + b x + c = 0; for your eq. a = 1 b= -6 and c = -7
answer (x1, x2):
x1 = (-b+(b^2-4ac)^0.5)/2a
x2 = (-b-(b^2-4ac)^0.5)/2a

2. trial error.
factorize -7 = (-1,7),(-7,1)
pick one where the sum of the factors equal to -6, in this case (-7,1)
x-7 = 0 and x+1=1
henced
x=7 and x = -1

there you go

Answer 9

Let ax^2+bx+c=0 be the eqn
Now first check D=b^2-4ac for the sign of it
Then use the formula
ax^2+bx+c=[x-{(-b+sqrt(D))/2}]*[x-{(-b-sqrt(D))/2}]
These are the factors

Answer 10

x^2 - 6x - 7 = (x - 7)(x + 1)

use x = (-b ± sqrt(b^2 - 4ac))/(2a)

to find out what x is.