Is what I've done correct, and how proceed ?
x^4-5x^2+4=0
x(x^3-5x+4)=0
Did you check your factoring by expanding your second line? No, because that would give you 4x at the end not just 4.
So your factoring is incorrect.
Let a = x^2, and rewrite your first equation in terms of a. Can you see what the next step will be?
No, you've misunderstood what I said.So
x^4-5x^2+4=0
x(x^3-5x+4x)=0
???
Although we are happy to tell you whether you are correct or not, you should learn how to check your own work.Is what I've done correct, and how proceed ?
x^4-5x^2+4=0
x(x^3-5x+4)=0
Is what I've done correct, and how proceed ?
x^4-5x^2+4=0
x(x^3-5x+4)=0
To cut to the chase, look at the first line. Notice that there are three terms (expressions set apart by addition or subtraction). Whatever you try to factor out of the full expression must appear in ALL of the terms. The third term is just the constant 4. It does not contain a factor of x that can be factored out.
What you have is a trinomial. Look at the powers of the three terms. The first exponent is 4, the second is 2, and the third is 0. This is somewhat similar to if they were exponents of 2, 1, and 0, yes? How?
Well, if you multiply x times x, you get x^2. That is the basis of a quadratic equation. For example, if you FOIL out (x+1)(x+1) you get (x^2 + 2x + 1). But now imagine that instead of that, you had (x^2 + 1)(x^2 + 1). You would get (x^4 + 2x^2 + 1). What was traditionally your first order term in your factored binomials is, in this case, a second order term.
So if, in your trinomial, the exponent of your first term is twice the exponent of the second term, and the third term is constant, then it is just like a quadratic, except that what would normally be the first order middle term is of a higher power.
TLR
We often call this "W substitution." In this case, W is x^2. Substitute W in for every instance of x^2 and it becomes a straightforward quadratic. Factor in terms of W. Then, substitute x^2 back in. You can do this for any power of x if the first term is twice that power and the last term is constant.
Like (x^6 + 2x^3 +1) where W is x^3
or (x^16 + 2x^8 +1) where W is x^8
I agree that x(x^3-5x)+4=0 is equivalent to what you started with. The problem is when you factor you are supposed to have a product of things, like (3x+4)times(7x-9). What you have is x(x^3-5x) plus 4,which is not just a product. It has more than one term. The terms are in two different colors: x(x^3-5x)+4. Terms are separated by + and - symbols which are not in parenthesisx^4-5x^2+4=0
x(x^3-5x)+4=0
Like that ? It's what I understood from the fact that I have to factor in everything that's IF I understood corretly. Is that right?
x^4-5x^2+4=(x^2-4)(x^2-1) = 0. Continue from here.Is what I've done correct, and how proceed ?
x^4-5x^2+4=0
x(x^3-5x+4)=0
x^4-5x^2+4=0
x(x^3-5x)+4=0
Like that ? It's what I understood from the fact that I have to factor in everything that's IF I understood corretly. Is that right?
They are indeed equivalent expressions in that they denote the same value given any specific value of x.x^4-5x^2+4=0
x(x^3-5x)+4=0
Like that ? It's what I understood from the fact that I have to factor in everything that's IF I understood corretly. Is that right?
It helps to quote what you don't understand. Your previous post was #9. The post above is #15. In between were five different posts.I don't understand where -1 comes from in (x^2-1)
When x = -1, (x^2-1) = 0.I don't understand where -1 comes from in (x^2-1)
x^4-5x^2+4=0
x(x^3-5x)+4=0
Like that ? It's what I understood from the fact that I have to factor in everything that's IF I understood corretly. Is that right?
No no. if x2 = 4, it does NOT follow that x = sqrt(4), rather x = +/-sqrt(4)x4-5x2+4=0
what drudge was saying; substitute w=x2
w2-5w+4=0
(w-4)(w-1)=0
w=4, and 1
then, since w=x2
x=sqrt4, and sqrt1