Using the Quadratic Formula on a difference of cubes to factor x^3 - 1

[DustinC]

I've been asked to factor x^3 - 1 and then use the quadratic formula on it.

So x^3 - 1 turns into (x - 1)(x^2 + x + 1)

Question is, how do I use both expressions.

If I only had to use the second expression (x^2 + x + 1) I'd know what to do. But what do I do with the (x - 1)?

Do I turn (x - 1) into (0x^2 + x - 1) and then put both that expression and the (x^2 + x +1) into the quadratic formula separately? use the quadratic formula on both of them?

 

[JeffM]

I've been asked to factor x^3 - 1 and then use the quadratic formula on it.

So x^3 - 1 turns into (x - 1)(x^2 + x + 1)

Question is, how do I use both expressions.

If I only had to use the second expression (x^2 + x + 1) I'd know what to do. But what do I do with the (x - 1)?

Do I turn (x - 1) into (0x^2 + x - 1) and then put both that expression and the (x^2 + x +1) into the quadratic formula separately? use the quadratic formula on both of them?

If I understand the question (and I am not sure I do FULLY understand it), you have merely been asked to factor the expression; no equation is involved. Are you to restrict yourself to real coefficients.

i am guessing that you have not been taught the misnamed Fundamental Theorem of Algebra. It says that any polynomial of even degree with real coefficients can always be factored into quadratics with real coefficients and that any polynomial of odd degree greater than one and with real coefficients can always be factored into at least one linear factor with real coefficients and quadratic factors with real coefficients. And you should certainly know that some but not all quadratics with real coefficients can be factored into 2 linear factors with real coefficients.

So in this case, you started with a third degree polynomial. And you have found a linear factor with real coefficients and a quadratic factor with real coefficients. Perfect so far. If the quadratic factor can be broken down into two linear factors, then you can factor further. So how do you use the quadratic formula to see whether a quadratic can be factored.

 

[DustinC]

So the math book wants me to take x^3 - 1 and use the quadratic formula to solve it. To do that, I first need to factor x^3 - 1 (which is a difference of cubes)

You get (x - 1)(x^2 + x + 1)

How do you use the quadratic formula on 2 espressions such as those.

I already know how to factor a difference of cubes. I need to know how to input the factored parts of a difference of cubes into the Quadratic Formula.

I can stick the variables from the x^2 + x +1 into the formula just fine. But what do I do with the x - 1 part in the beginning. How do i use the Quadratic Formula on both of them.

 

[mmm4444bot]

… (x - 1)(x^2 + x + 1)

How do you use the quadratic formula on 2 espressions such as those.

You don't. The first factor (x-1) is a linear polynomial. We don't use the Quadratic Formula with linear polynomials.

If you want to find the values of x for which the other factor (x^2+x+1) equals zero, you would use the Quadratic Formula to do that. In other words, the Quadratic Formula is used to solve quadratic equations, like this one:

x^2 + x + 1 = 0

For the linear factor, you would solve a linear equation:

x - 1 = 0

If I only had to use the second expression (x^2 + x + 1) I'd know what to do.

Show us what you would do. :cool:

… factor x^3 - 1 and then use the quadratic formula on it …

That's not the wording in your assignment. Please type the complete exercise statement, worded exactly as you received it, including all instructions. Thanks.