# Factoring a quadratic: strange issues

• ducmod
In summary: The first one can be eliminated because it's a constant, so f-g= 0. The second one can be eliminated because it's also a constant, so g-f= 0. That leaves only the two terms:(x - f)(x - g) = x^2+ (f-g)x+ fg.Now take the square of each term:x^2+ (f-g)x+ fg= 4x^2+ (f-g)2+ fg.The two terms are equal, so the two roots are also equal:f=2 and g=3.
ducmod

## Homework Statement

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

## Homework 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.

## 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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ducmod said:

## Homework Statement

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

## Homework 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.

## 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!

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. ?

epenguin said:
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. ?
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 )))

ducmod said:
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 )))

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?

ducmod said:
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 )))
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.

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 $x^2- x- 6= 0$.

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

Factoring
Completing the square
Graphing

If a problem requires factoring and you can't figure it out that way, you can always use another method to find the roots and work backwards.

Suppose you use the quadratic formula to find roots f and g. The factors are then:

(x - f) and (x - g).

## 1. What is factoring a quadratic?

Factoring a quadratic is the process of rewriting a quadratic equation in the form of two binomials multiplied together. It is a useful technique for solving quadratic equations and understanding their properties.

## 2. Why is factoring a quadratic important?

Factoring a quadratic is important because it allows us to find the roots of a quadratic equation, which are the values of x that make the equation equal to 0. It also helps us to understand the behavior and properties of quadratic equations.

## 3. What are some common mistakes when factoring a quadratic?

Some common mistakes when factoring a quadratic include forgetting to factor out the greatest common factor, using incorrect signs when finding the factors, and forgetting to include both positive and negative solutions.

## 4. How do I factor a quadratic with strange coefficients?

To factor a quadratic with strange coefficients, you can use the AC method, which involves finding two numbers that multiply to give the coefficient of the x^2 term and add to give the coefficient of the x term. These numbers will then be used to factor the quadratic equation.

## 5. Are there any shortcuts for factoring a quadratic?

Yes, there are a few shortcuts for factoring a quadratic, such as using the difference of squares formula for perfect square trinomials or using the quadratic formula for quadratic equations that cannot be easily factored. However, it is important to understand the process of factoring and not rely solely on shortcuts.

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