Polynominal equation: x^4-5x^2+4=0 x(x^3-5x+4)=0
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Harry_the_cat
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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?
Harry_the_cat
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No, you've misunderstood what I said.
Initially you said:
x^4-5x^2+4=0
x(x^3-5x+4)=0
What I'm saying is, if you check your factoring by expanding your second line, you get x^4 -5x^2 +4x, which is NOT the same as your first line because you have 4x where it should be 4, as in the original equation.
Your second post is clearly not correct. Whenever you factorise you should always check it by expanding out and seeing if it is what it should be.
Starting again:
Note that x^4-5x^2+4 = (x^2)^2 - 5x^2 +4 = 0
Let a = x^2, and rewrite in terms of a. It should be clear what the next step is.
x(x^3-5x+4)=0
What I'm saying is, if you check your factoring by expanding your second line, you get x^4 -5x^2 +4x, which is NOT the same as your first line because you have 4x where it should be 4, as in the original equation.
Your second post is clearly not correct. Whenever you factorise you should always check it by expanding out and seeing if it is what it should be.
Starting again:
Note that x^4-5x^2+4 = (x^2)^2 - 5x^2 +4 = 0
Let a = x^2, and rewrite in terms of a. It should be clear what the next step is.
Although we are happy to tell you whether you are correct or not, you should learn how to check your own work.
Two expressions in x are equal if and only if their difference is zero for every possible value of x.
\(\displaystyle \text {Is } x(x^3 - 5x + 4x) \text { a correct factoring of } x^4 - 5x^2 + 4?\)
\(\displaystyle x(x^3 - 5x + 4x) = x(x^3) + x(-\ 5x) + x(4x) = x^4 - 5x^2 + 4x^2 = x^4 - x^2.\)
\(\displaystyle x^4 - 5x^2 + 4 - (x^4 - x^2) = 4 - 4x^2.\)
\(\displaystyle \text {BUT } 4 - 4(10^2) = 4 - 400 = -\ 396 \ne 0.\)
\(\displaystyle \text {Therefore, } x(x^3 - 5x + 4x) \text { is not a correct factoring of } x^4 - 5x^2 + 4?\)
As our revered feline has explained, it is easy to find the correct factoring with a clever substitution of variables.
mmm4444bot
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Harmless Drudge
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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.
TL
RWe 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
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.
TL
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
That was a good explanation, thanks ! It made a lot of sense, and you explained it so simple.
Sorry for the late replies I'm busy with finals currently.
Steven G
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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 parenthesis
Steven G
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x^4-5x^2+4=(x^2-4)(x^2-1) = 0. Continue from here.
Harmless Drudge
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Look at it in reverse. Multiplication is distributive across addition. As such, when you multiply one or more terms by a different collection of one or more terms, the result is a product. It contains both factors. When you factor (literally, when you unmultiply back into factors) you are trying to end up with only one term. That term may have several factors (which may have their own terms inside of them), but it is still only one term.
That is to say, when you factor something out of an expression, it has to be factored out of every term.
In your example, you had three terms. Sometimes that means you can factor a single term out of all three. Like (x^3 + x^2 +x). You can factor x out of all three, because all three contain x.
But in your example, the last term is a constant. Typically, a constant is the product of two or more constants. [The exception to this is when you multiply positive and negative powers of equal total magnitude, like x^3 * x^(-3), but polynomials are typically assumed to only have positive integer powers of the independent variable.]
So if a constant is a constant times a constant, you are looking for two or more factors that end in constants. Your equation is a trinomial quadratic, meaning that if it factors politely, it will be into two binomials (in a process that reverses FOIL).
Let's break it down:
x^4-5x^2+4=0 is what happened from FOILing two things.
Last times Last equals 4, and because it is restricted to polynomials, no negative powers of x are allowed. So it is two integers that times out to 4. Your choices are:
1*4, 2*2, (-1)(-4), and (-2)(-2).
Now look at the product of the firsts. First times First equaled x^4. So what could that be? This is a trinomial arising from FOIL. That means the Outside times Outside and the Inside times Inside were added together to form one term. That means that they were the same power of x, otherwise they would not add, and you would have four terms to factor, not three. So only x^2 * x^2 is allowed, unless you are allowed non-integer coefficients.
So that means you have two factors, each being x^2 plus or minus a factor of 4. So now we need to find the factors of 4 that also add to (-5) because that is the coefficient of your b term. (-1) and (-4) add to (-5).
So your factors are (x^2 - 1)(x^2 - 4)
But now check your work by distributing across the addition, timesing every term in the first factor by every term in the second:
x^4 - x^2 - 4x^2 +4
combine like terms and you are back to (x^4 -5x^2 +4).
They are indeed equivalent expressions in that they denote the same value given any specific value of x.
You can verify that by the method I told you earlier.
\(\displaystyle (x^4 - 5x^2 + 4) - \{x(x^3 - 5x) + 4\} = x^4 - 5x^2 + 4 - (x^4 - 5x^2 + 4) = 0\)
no matter what value x may have.
So it, and an infinity of other expressions, is a valid equivalent, but it is not a useful equivalent.
One way to solve a polynomial equation is to factor it into a product of linear and quadratic terms and then apply the zero-product property.
\(\displaystyle 0 = x^4 - 5x^2 + 4 = (x^2 - 4)(x^2 - 1) = (x - 2)(x + 2)(x - 1)(x + 1).\)
This gets us to a point where we can apply the zero-product property because we have a product of terms equal to zero.
If two or more numbers when multiplied together result in zero, then at least one of those numbers is zero.
Applying that to our problem, we know that
\(\displaystyle (x - 2),\ (x + 2),\ (x - 1), \text { or } (x + 1) \text { must equal } 0.\)
\(\displaystyle x - 2 = 0 \implies x = 2,\ x + 2 = 0 \implies x = -\ 2,\)
\(\displaystyle x - 1 = 0 \implies x = 1, \text { and } x + 1 = 0 \implies x = -\ 1.\)
We have four possible solutions. Let's check.
\(\displaystyle x = \pm 2 \implies x^2 = 4 \implies x^4 = 16 \implies\)
\(\displaystyle x^4 - 5x^2 + 4 = 16 - 5 * 4 + 4 = 20 - 20 = 0.\)
So both 2 and - 2 work.
\(\displaystyle x = \pm 1 \implies x^2 = 1 \implies x^4 = 1 \implies\)
\(\displaystyle x^4 - 5x^2 + 4 = 1 - 5 * 1 + 4 = 5 - 5 = 0.\)
And both 1 and - 1 work.
To sum up, every expression has an infinite number of equivalent expressions. The object is to find helpful equivalents. To use the zero-product property, you need a product, not a sum.
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Harry_the_cat
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This is how I'd set it out:
\(\displaystyle x^4-5x^2+4=0\) … I call this a "hidden" quadratic equation
\(\displaystyle (x^2)^2 - 5 (x^2) +4 = 0\)
Let \(\displaystyle x^2 =a \)(for now):
\(\displaystyle a^2 -5a +4=0\) … now it's a normal everyday quadratic equation
\(\displaystyle (a-1)(a-4)=0\)
So \(\displaystyle a=1\) or \(\displaystyle a=4 \)
But remember that \(\displaystyle a = x^2 \)
So \(\displaystyle x^2 =1\) or \(\displaystyle x^2 =4\)
This means that \(\displaystyle x = 1\) or
\(\displaystyle x^4-5x^2+4=0\) … I call this a "hidden" quadratic equation
\(\displaystyle (x^2)^2 - 5 (x^2) +4 = 0\)
Let \(\displaystyle x^2 =a \)(for now):
\(\displaystyle a^2 -5a +4=0\) … now it's a normal everyday quadratic equation
\(\displaystyle (a-1)(a-4)=0\)
So \(\displaystyle a=1\) or \(\displaystyle a=4 \)
But remember that \(\displaystyle a = x^2 \)
So \(\displaystyle x^2 =1\) or \(\displaystyle x^2 =4\)
This means that \(\displaystyle x = 1\) or
\(\displaystyle x = -1\)
or \(\displaystyle x = 2\)
or\(\displaystyle x = -2\)
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.
Steven G
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When x = -1, (x^2-1) = 0.
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
Steven G
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No no. if x2 = 4, it does NOT follow that x = sqrt(4), rather x = +/-sqrt(4)