# Math Help - Help with Factorising of Polynomials in C

1. ## Help with Factorising of Polynomials in C

Can someone help me with getting the factors for the following problem? I can't seem to get the other set.

If z-3i is a factor of 2z^4 - 4z^3 + 21z^2 - 36z +27, find the remaining factors.

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Let P(z) = 2z^4 - 4z^3 + 21z^2 - 36z +27

If z[1] = z-3i
then z[2] = z+3i (Conjugate Root Theorem)
Let z[3], z[4] = third and forth zeros respectively
then P(z) = (z - z[1])(z - z[2])(z - z[3])(z - z[4])
= (z - (z-3i))(z - (z+3i))(z - z[3])(z - z[4])
= (3i)(-3i)(z - z[3])(z - z[4])
= (-9i^2)(z - z[3])(z - z[4])
= 9(z - z[3])(z - z[4])

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I'm pretty sure this is correct so far but I'm not sure how to get the remaining two factors. I'm not sure if you need to divide P(z) by 9 through long division or divide it by z[1]*z[2].

If someone would be so kind as to help me get the remaining factors which will likely be conjugates.

Note: z[1] is meant to be z with a subscript 1, etcetera. Not sure what the notation for that is though.

2. Originally Posted by joumas
Can someone help me with getting the factors for the following problem? I can't seem to get the other set.

If z-3i is a factor of 2z^4 - 4z^3 + 21z^2 - 36z +27, find the remaining factors.

---

Let P(z) = 2z^4 - 4z^3 + 21z^2 - 36z +27

If z[1] = z-3i
then z[2] = z+3i (Conjugate Root Theorem)
Let z[3], z[4] = third and forth zeros respectively
then P(z) = (z - z[1])(z - z[2])(z - z[3])(z - z[4])
= (z - (z-3i))(z - (z+3i))(z - z[3])(z - z[4])
= (3i)(-3i)(z - z[3])(z - z[4])
= (-9i^2)(z - z[3])(z - z[4])
= 9(z - z[3])(z - z[4])

---

I'm pretty sure this is correct so far but I'm not sure how to get the remaining two factors. I'm not sure if you need to divide P(z) by 9 through long division or divide it by z[1]*z[2].

If someone would be so kind as to help me get the remaining factors which will likely be conjugates.

Note: z[1] is meant to be z with a subscript 1, etcetera. Not sure what the notation for that is though.
We know that z - 3i and z + 3i are factors, so (z - 3i)(z + 3i) = z^2 + 9 is a factor. Thus z^2 + 9 divides 2z^4 - 4z^3 + 21z^2 - 36z +27. Doing the division gives:
2z^2 - 4z + 3

There are two ways to proceed from here. The first is by use of the quadratic formula (since the remaining polynomial doesn't factor over the reals.) So solving 2z^2 - 4z + 3 = 0 gives z = 1 + i(sqrt{2}/2) and z = 1 - i(sqrt{2}/2). So the remaining factors of the original polynomial are:
z - 1 - i(sqrt{2}/2)
z - 1 + i(sqrt{2}/2)
However, if you look carefully you will note that we are off by a factor of 2 somewhere. Multiplying these two factors together gives z^2 - 2z +3/2, not the required 2z^2 - 4z + 3. A simple way to fix this is to multiply one of our factors by 2, giving the factors of the original polynomial as:
z - 3i
z + 3i
2z - 2 - i*sqrt{2}
z - 1 + i(sqrt{2}/2)

The second way is to note that we have two more factors to find: z - z[3] and z - z[4]. So a(z - z[3])(z - z[4]) = az^2 - a(z[3] + z[4])z + a*z[3]*z[4] must be the same as the polynomial 2z^2 - 4z + 3. Thus
a = 2
a(z[3] + z[4]) = 4
a*z[3]*z[4] = 3
We may solve the system of equations for a, z[3], and z[4] and we get the same set of factors as we did in the first method.

I recommend the first method over the second as it is slightly simpler.

-Dan

3. Thanks for the help. The problem had me stumped for a while due to the 2 outside the brackets so I thought I had done something wrong.

4. My teacher has taught me a way of long division where you divide an equation by another. Its pretty hard to describe but here's a page that tell you how to do it Polynomial Long Division

Obviously when you get a remainder of zero you know you've found a factor of the original equation.

Hope this helps
Derg