# Solving for Zeros in Polynomials of Higher Degree

• B
Gold Member

## Main Question or Discussion Point

Please see the attached image which is presented in my text. This is the end step after using the Rational Zeros Theorem to find possible rational zeros, testing by synthetic division, and then factoring.

What I don't understand here is that we have the term
$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$

This the 27 is broken up into 3⋅3⋅3 and distributed to three of the four other factors to give us $\left(x+1\right)\left(3x-4\right)\left(3x-i\right)\left(3x+i\right)$

The zero's are then found to be as listed.

$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$ gives us the exact same zeros as is. So what was the purpose of this step?

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mathman
It is easier to see the zeros (roots) when the factors are of the form x - r, and any constants are taken out.

Gold Member
But in the first case, the factors are exactly the same except for the leading constant, so cant we just disregard the leading constant? Or even distribute it into a single factor?

fresh_42
Mentor
Please see the attached image which is presented in my text. This is the end step after using the Rational Zeros Theorem to find possible rational zeros, testing by synthetic division, and then factoring.

View attachment 227519

What I don't understand here is that we have the term
$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$

This the 27 is broken up into 3⋅3⋅3 and distributed to three of the four other factors to give us $\left(x+1\right)\left(3x-4\right)\left(3x-i\right)\left(3x+i\right)$

The zero's are then found to be as listed.

$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$ gives us the exact same zeros as is. So what was the purpose of this step?
How do you factor $g(x)\,?$ It seems that $g(x)$ comes from somewhere with a factor $27x^2+3$ given. Now this has to be split, and the quadratic formula for it is a way to do it. Of course you can also guess solutions.

Mark44
Mentor
What I don't understand here is that we have the term
$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$
In the above, I think you meant to write $(x + 1)(3x - 4)(3x + i)(3x - i)$. As already stated, writing the factors in the form (x - r) makes it clearer exactly what the zeroes are.

As a side note, you can save yourself a lot of typing by using ( in place of \left and ) in place of \right. The LaTeX versions are useful if they enclose a "tall" expression, say with complicated fractions, but aren't needed for simpler expressions.
opus said:
This the 27 is broken up into 3⋅3⋅3 and distributed to three of the four other factors to give us $\left(x+1\right)\left(3x-4\right)\left(3x-i\right)\left(3x+i\right)$

The zero's are then found to be as listed.

$27\left(x+1\right)\left(x-\frac{4}{3}\right)\left(x-\frac{i}{3}\right)\left(x+\frac{i}{3}\right)$ gives us the exact same zeros as is. So what was the purpose of this step?

Gold Member
How do you factor $g(x)\,?$ It seems that $g(x)$ comes from somewhere with a factor $27x^2+3$ given. Now this has to be split, and the quadratic formula for it is a way to do it. Of course you can also guess solutions.
Yes, so the $27x^2+3$ is from what was left after guessing possible zeros. Of course since it's now in quadratic from, we can go on to factor. Now the $27x^2+3$ factors into the $27\left(x+\frac{i}{3}\right)\left(x-\frac{i}{3}\right)$
And now, the factors are stated, and the zeros are clear. So why continue? Can't we just disregard the 27 when we're talking about the zeros now?

In the above, I think you meant to write $(x + 1)(3x - 4)(3x + i)(3x - i)$. As already stated, writing the factors in the form (x - r) makes it clearer exactly what the zeroes are.

As a side note, you can save yourself a lot of typing by using ( in place of \left and ) in place of \right. The LaTeX versions are useful if they enclose a "tall" expression, say with complicated fractions, but aren't needed for simpler expressions.
That statement was just copied from the second line from the top in the picture. The 27 is leading the four factors that correspond to the zeros.
Thank you for the parentheses tip! I always mess something up with the LaTeX so I have to sift through and find it. That'll make things much easier.

Mark44
Mentor
And now, the factors are stated, and the zeros are clear. So why continue?
Because the problem is asking for the factorization.
We can then write the factorization using the result from synthetic division.
In your image the two lines above the part with "Actual zeroes" are the factorizations -- the first with all factors in the form (x - r) and the second with all factors using integer coefficients in all factors.

fresh_42
Mentor
Yes, so the $27x^2+3$ is from what was left after guessing possible zeros. Of course since it's now in quadratic from, we can go on to factor. Now the $27x^2+3$ factors into the $27\left(x+\frac{i}{3}\right)\left(x-\frac{i}{3}\right)$
And now, the factors are stated, and the zeros are clear. So why continue? Can't we just disregard the 27 when we're talking about the zeros now?
It's a matter of taste. I first misunderstood your question, as I thought you didn't see what the small calculation on the right is for. I personally favor the expression $g(x)=27 \ldots$ so that the zeroes can directly be seen, but this is as if we debated whether $g'(x)$ has to be preferred or $\frac{d}{dx}g(x)$. It depends on what it is used for, or what an author favors. There is no deeper reason behind that.

Gold Member
So is it fair to say that in both cases, we can easily find the zeros, but we need to distribute that 27 into the rest of the expression for the expression to be factored?

fresh_42
Mentor
So is it fair to say that in both cases, we can easily find the zeros, but we need to distribute that 27 into the rest of the expression for the expression to be factored?
Not really. Both expressions are factored. You can always only factor up to units and since we are over the complex numbers, up to any number from $\mathbb{C}-\{0\}$. $g(x)=36 \cdot i \cdot \ldots$ would also be a valid decomposition. This in mind it makes sense, and which is why I prefer this notation, to write $g(x)=27\cdot h(x)$ with a monic polynomial $h(x)$.

mathman
General opinion: this issue is a quibble - for practical purposes, it really doesn't matter.

Gold Member
Thanks guys.
General opinion: this issue is a quibble - for practical purposes, it really doesn't matter.
I dont see anything quibble about discussing two different forms of the same solution. Especially since the text deliberately made note of distributing the coefficient throughout the other factors.

StoneTemplePython
You would be wise to put in the effort to try and write every polynomial as $g(x) = k \dot h(x)$ as Fresh suggested -- focusing on the monic polynomial is key. The fact that $g(x)$ and $h(x)$ have the same roots is important -- if you don't understand why, it's worth dwelling on it and asking more questions. This idea of "distributing the coefficients" to other factors seems odd and a bit like a waste of time. It's ok to follow the text but you should know that sometime authors make mountains out of molehills. Understanding and embracing monic polynomials would be the takeaway from the thread here.