# What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$?

What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$?

I know its answer: 0, -2, -1+i, -1-i.

But I dont know how to find? Please show me this. I know this is so trivial, but important for me. Thank you.

• It's a 4'th degree polynomial so it can't have more than 4 roots... Oct 29 '13 at 21:49
• I think there might be a bit of a typo: maybe $-1,+i$ should have been $-1+i$ etc.? Oct 29 '13 at 22:10

Add 1 to both side to get:

$$z^4+4z^3+6z^2+4z+1 = 1$$

i.e.

$$(z+1)^4 = 1$$

can you finish from there?

• Well, in general how to find such equations'root?
– 1190
Oct 29 '13 at 21:49
• Sorry, I asked too trivial. I realized. But explain the last question please, I Will be happy.
– 1190
Oct 29 '13 at 21:50
• In general, 4th order equations are much harder - this one has been specially chosen to allow the trick we have used. Oct 29 '13 at 21:50
• One "easy" way I saw this trick is by noticing that the coefficients are the fourth row of Pascal's triangle with the last term missing. The row is 1,2,6,4,1 so in order to make it a perfect 4th power, you add and subtract the last term which should be 1. Oct 29 '13 at 22:16
• Yes, that is pretty much what went on in my head - but I think you have a typo in your comment? (1.4.6.4,1 not 1,2,6,4,1 ?) Oct 29 '13 at 22:18

Hint: Add $1$, and use the binomial theorem, for

$$z^4 + 4z^3 + 6z^2 + 4z + 1 = (z + 1)^4$$

So the problem is equivalent to solving $$( z+1)^4 = 1$$

• Expanding a little: the way to find this is by looking at the polynomial and recognising the coefficients 1,4,6,4 as something kind of familiar — noticing that they’re almost what you’d get from the binomial expansion of some fourth power. The general lessons to learn from this are (1) when factorising polynomials, if you don’t have a general method that applies, look for a way to relate them to something familiar you understand; and (2) know the binomial theorem like you know your alphabet! Oct 29 '13 at 22:28

$(z+1)^4=z^4+4z^3+6z^2+4z+1$.${}{}{}{}{}$

• Remark: Suppose that someone out of cruelty replaces $6z^2$ by $6.1 z^2$. No problem!. Divide through by $z^2$, and make the substitution $w=z+\frac{1}{z}$. We get a quadratic in $w$. Oct 30 '13 at 0:10

First, find the real rational roots. One of them is $0$ because $z$ divides the polynomial. Divide by $z-0 = z$. For the rest, look at the remaining polynomial $z^3 + 4z^2 + 6z + 4$. Its rational roots have numerator dividing $4$, the integer coefficient, and denominator dividing $1$, the coefficient of the highest term. The sign is not restricted. So, the possibilities are $\pm 1, \pm 2,\ \text{and}\ \pm 4$. Trial and error gives $-2$ only. Divide now by $z - (-2) = z+2$. Now you have a quadratic.

put z4+4z3+6z2+4z = 0
z(z3 + 4z2+6z + 4) = 0
z = 0 (first root)

By hit n trial, -2 is also a root,
So divide z3 + 4z2+6z + 4 by (z+2) : You will get another equation which can be solved using quadratic formula.