# Factoring a quadratic: strange issues

1. Jul 6, 2015

### ducmod

1. The problem statement, all variables and given/known data
Hello!

One of the easiest rules (when possible to apply) to factor a quadratic is to find both x-s by

x1 + x2 = b
x1 * x2 = c

2. Relevant equations
Please, take a look at what is written in the book. I can't grasp why x1 = -2 and x2 = 3, and not, as I thought, x1 = -3 and x2 = 2.

3. The attempt at a solution
What is wrong in my understanding?

x1 + x2 = -1
x1 * x2 = -6
Therefore, x1 = -3 and x2 = 2; and the equation (x + 3 )(x - 2) =0

I would be grateful for the help.
Thank you!

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2. Jul 6, 2015

### epenguin

Can see only two lines of your quote so not sure what your problem is, but could it be this
x1 + x2 = b

Which should be -b. ?

3. Jul 6, 2015

### ducmod

Thank you for your reply. There is a screenshot attached, and my equations.
Do you say that x1 + x2 equal -b and not b? Mathisfun lied to me )))

4. Jul 6, 2015

### epenguin

Check that again, but better and faster would be to look see if roots are a and b so that they satisfy

(x - a)(x - b) = 0

what is the coefficient of x in the quadratic you get expanding that?

5. Jul 6, 2015

### SteamKing

Staff Emeritus
This page from Math is Fun gives the correct relationship for the sum and product of the roots of a quadratic:

http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

You can always check with Wikipedia for quadratic equations.

BTW, these formulas are called Vieta's formulas, for the famous French mathematician.

6. Jul 7, 2015

### HallsofIvy

You can multiply the left side to see that $(x+ a)(x+ b)= x^2+ (a+ b)x+ ab$. In your given example, $x^2- x- 6$, so we must have a+ b= -1. And ab= -6. Obviously, 3 and -2 multiply to give -6. But -2+ 3= 1, not -1. We must have a=- 2 and b= 3. Once we have that we have the two factors (x+ (-2)(x+ 3)= (x- 2)(x+3)= 0, we must have either x- 2= 0 or x+ 3= 0. From those, x= 2 and x= -3 are the roots of the equation $x^2- x- 6= 0$.

7. Jul 7, 2015

### Dr. Courtney

There are several ways to find the roots of a quadratic:

Factoring