Sum and differences of difference of two cubes and real number solutions

[Jenifer]

I have been working on these problems for 2 days I want to see if this one is correct and ask about a couple others. a³-8 is this true that (a-2)(a²+2a+4) the common is 2 and 2³is 8-8=0 and (2-2) is 0 and 2+2+4=8 and when you multiply 0*8 it is 0 and I want to see how to work 5x³=20x in my book the example subtracts from each side. I would also like to know if you can explain how to find real number solutions. I do not know how these problems are worked I want to make sure that I would subtract 20x which would zero it out, then when I subtract 20x from 5x it would it be 15x³ or is it -15x³ this must not be correct I wonder is it 50⁰⁼200 that is if the x is a real number of zero~dont know. Thanks for your help!!!

 

[Deleted member 4993]

I have been working on these problems for 2 days I want to see if this one is correct and ask about a couple others. a³-8 is this true that (a-2)(a²+2a+4) the common is 2 and 2³is 8-8=0 and (2-2) is 0 and 2+2+4=8 and when you multiply 0*8 it is 0 and I want to see how to work 5x³=20x in my book the example subtracts from each side. I would also like to know if you can explain how to find real number solutions. I do not know how these problems are worked I want to make sure that I would subtract 20x which would zero it out, then when I subtract 20x from 5x it would it be 15x³ or is it -15x³ this must not be correct I wonder is it 50⁰⁼200 that is if the x is a real number of zero~dont know. Thanks for your help!!!

5x³ = 20 x

5x³ - 20 x = 0

5x(x² - 4) = 0

5x(x² - 2²) = 0

5x(x-2)(x+2) = 0 so the solutions are:

x = 0

x - 2 = 0 → x = 2 and

x + 2 = 0 → x = -2

 

[JeffM]

I have been working on these problems for 2 days I want to see if this one is correct and ask about a couple others. a³-8 is this true that (a-2)(a²+2a+4) the common is 2 and 2³is 8-8=0 and (2-2) is 0 and 2+2+4=8 and when you multiply 0*8 it is 0 and I want to see how to work 5x³=20x in my book the example subtracts from each side. I would also like to know if you can explain how to find real number solutions. I do not know how these problems are worked I want to make sure that I would subtract 20x which would zero it out, then when I subtract 20x from 5x it would it be 15x³ or is it -15x³ this must not be correct I wonder is it 50⁰⁼200 that is if the x is a real number of zero~dont know. Thanks for your help!!!

This may be more information than you want.

A polynomial in one variable of degree n looks like this

\(\displaystyle a_1x^n + a_2x^{(n-1)} +\ ...\ a_{(n-1)}x^1 + a_n,\ where\ a_1 \ne 0.\)

So \(\displaystyle a^3 - 8 = 1 * a^3 + 0 * a^2 + 0 * a^1 + (- 8)\) is a polynomial in a of degree 3. Got the lingo?

In general, a polynomial of degree n has at most n real zeroes. If n is odd, it has at least one zero. If it is even, it may have no zeroes. This is rather unsatisfactory so it is frequently said that a polynomial of degree n has n zeroes, but not all of them are necessarily distinct and not all of them are necessarily real. If you are looking for the real zeroes, however, it helps to know that there are at most n of them, and if n is odd, there is at least one of them. None of this tells you how to find them. It just helps to know what you are looking for.

A polynomial of degree n with real coefficients can always be factored into (n/2) quadratics if n is even and can always be factored into a linear factor and (n - 2)/2 quadratics if n is odd. Of course some or all of the quadratics can then be factored into linear terms themselves. Every linear factor with real coefficients identifies a real zero.

\(\displaystyle x^3 - 8 = (x - 2)(x^2 + 2x + 4).\)

The quadratic term cannot be factored in real numbers as you can tell by applying the quadratic formula. So there is only one real zero.

\(\displaystyle x^3 - 8 = 0 \implies (x - 2)(x^2 + 2x + 4) = 0.\)

Now the only way that can happen is that the linear term is zero or the quadratic term is zero. If the linear term is 0,
then x - 2 = 0 or x = 2. And you have your answer.

So one way to find the real zeroes of a polynomial is to factor the polynomial into as many linear factors with real coefficients as possible. Each such factor will identify a real zero.

\(\displaystyle 5x^3 = 20x \implies 5x^3 - 20x = 0.\)

Polynomial of degree three with real coefficients. There is AT LEAST one real zero and AT MOST three real zeroes. Good so far.

One way to find them is to factor

\(\displaystyle 5x^3 - 20x = 5(x)(x^2 - 4) = 0 \implies x(x^2 - 4) = 0.\) OK we have one linear factor. Can we factor \(\displaystyle x^2 - 4\)?

\(\displaystyle x(x^2 - 4) = 0 \implies x(x - 2)(x + 2) = 0.\)

But that means either x = 0 or (x - 2) = 0 (meaning x = 2) or (x + 2) = 0 (meaning x = - 2) because the only way a product can be 0 is if one of the factors is 0.

So the zeroes are x = 0 or x = 2 or x = -2. <<< fixed a typo

Was this too much?

 

[HallsofIvy]

Another way of thinking about \(\displaystyle 5x^3= 20x\) is this:
x= 0 is obviously a solution since 0 times anything is 0 so 5(0)= 20(0). If x is NOT 0, we can divide both sides by 5x to get \(\displaystyle x^2= 4\) and then just take the square root or both sides: x= 2 and x= -2 so the three solutions to the original equation are x=0, -2, and 2.