The factoring method is quite commonly used to easily find the real roots of a quadratic equation

2 things to know before doing this is
Firstly, when we have
(x−a)(x−b)
, we set the other side to 0, because we want the x values where y is 0, so we have (x-a)(x-b) =0. And because of this, we separate it into two equations, where x-a=0 and x-b=0, so at the end, positive a and positive b are the x-intercepts.
Secondly, the FOIL method used to check.
FOIL method: forward, outside, inside, last
(a+b)(c+d) = ac +ad + bc + bd (usually we are able to combine ad and bc)

And how do we get to factored form from standard form?
Well, let’s first take an example:
x2+7x+6

In this, from seeing x^2, we know that we can set it to (x +/- _)(x +/- _)
And now we want to set the signs, we can set this one right when we see the equation in this example, because both b and c are positive integers (in ax^2 + bx + c)
So we can write (x+_)(x+_)
Nowe we have to fill in the two blanks. We look at the c value, and see that it is 6, from this, we can list out all the factors that multiply to 6, and then find which pair adds up to 7
So we would have:
1 & 6 - 1+6 = 7
2 & 3 - 2+3=5
So, from above, we can see that the two blanks we are about to fit in are 1 and 6. So we have
(x+6)(x+1)
and you can also check this by using the FOIL method

Check: (x+6)(x+1) = x^2 + x + 6x + 6 = x^2 + 6x + 6
So we are correct

And from (x+6)(x+1) we can determine that -6 and -1 are the roots, and so the coordinates are (-6,0) and (-1,0)

And so now, we can move on to the next scenario, if we had, for example
x2−7x+6
As you can see, I’m using the same numbers, just changing b into a negative integer
From this, we know that we need to integers that multiply up to 6, but add up to -7, in these kind of cases, it is obvious that the two signs in the the expression are both negative, because negative and negative makes positive, which gets our c value to be a positive integer, but also gets our b value to a negative number since -x+(-y) = -x-y, so it is negative
So we can set this up to (x-_)(x-_)
From here, we can see that again, 1 times 6 is six, and adds up to 7, so we have
(x−6)(x−1)
And again, we can check using the FOIL method

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

And from (x-6)(x-1) we can determine that 6 and 1 are the roots, so the coordinates are (6,0) and (1,0)

The next scenario is where one sign is positive, the other is negative
For example, if we had
x2+2x−15

Question

The factoring method is quite commonly used to easily find the real roots of a quadratic equation

2 things to know before doing this is 
Firstly, when we have 
(x−a)(x−b)
, we set the other side to 0, because we want the x values where y is 0, so we have (x-a)(x-b) =0. And because of this, we separate it into two equations, where x-a=0 and x-b=0, so at the end, positive a and positive b are the x-intercepts.
Secondly, the FOIL method used to check.
FOIL method: forward, outside, inside, last
(a+b)(c+d) = ac +ad + bc + bd (usually we are able to combine ad and bc)

And how do we get to factored form from standard form?
Well, let’s first take an example: 
x2+7x+6

In this, from seeing x^2, we know that we can set it to (x +/- _)(x +/- _)
And now we want to set the signs, we can set this one right when we see the equation in this example, because both b and c are positive integers (in ax^2 + bx + c)
So we can write (x+_)(x+_)
Nowe we have to fill in the two blanks. We look at the c value, and see that it is 6, from this, we can list out all the factors that multiply to 6, and then find which pair adds up to 7
So we would have:
1 & 6   -> 1+6 = 7
2 & 3  -> 2+3=5
So, from above, we can see that the two blanks we are about to fit in are 1 and 6. So we have 
(x+6)(x+1)
 and you can also check this by using the FOIL method

Check: (x+6)(x+1) = x^2 + x + 6x + 6 = x^2 + 6x + 6
So we are correct

And from (x+6)(x+1) we can determine that -6 and -1 are the roots, and so the coordinates are (-6,0) and (-1,0)

And so now, we can move on to the next scenario, if we had, for example 
x2−7x+6
 As you can see, I’m using the same numbers, just changing b into a negative integer
From this, we know that we need to integers that multiply up to 6, but add up to -7, in these kind of cases, it is obvious that the two signs in the the expression are both negative, because negative and negative makes positive, which gets our c value to be a positive integer, but also gets our b value to a negative number since -x+(-y) = -x-y, so it is negative
So we can set this up to (x-_)(x-_)
From here, we can see that again, 1 times 6 is six, and adds up to 7, so we have 
(x−6)(x−1)
 And again, we can check using the FOIL method

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

And from (x-6)(x-1) we can determine that 6 and 1 are the roots, so the coordinates are (6,0) and (1,0)

The next scenario is where one sign is positive, the other is negative
For example, if we had 
x2+2x−15

marskixmourn · Answer

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snowflake0531 · Answer

I absolutely love how you copied and pasted what i wrote