Zeros of a Polynomial
Polynomials are used to model some physical phenomena happening in real life, they are very useful in describing the situations mathematically. They are used in almost every field of Science, even outside of science like for example in Economics and other related areas. Zeros or roots of these polynomials are a very important aspect of their nature and can be very useful while describing them or plotting them on a graph. Let’s look at their definition and methods of finding out the roots in detail.
Zeros/Roots of a Polynomial
We say that x = a is the root of the polynomial if P(x) = 0 at that point. The process of finding zero is basically the process of finding out the solutions of any polynomial equation. Let’s look at some examples regarding finding zeros for a second-degree polynomial.
Question 1: Find out the zeros for P(x) = x2 + 2x – 15.
x2 + 2x – 15 = 0
⇒ x2 + 5x – 3x – 15 = 0
⇒ x(x + 5) – 3(x + 5) = 0
⇒ (x – 3) (x + 5) = 0
⇒ x = 3, -5
Question 2: Find the out zeros for P(x) = x2 – 16x + 64.
x2 – 16x + 64 = 0
⇒ x2 – 8x – 8x + 64 = 0
⇒ x(x – 8) – 8(x – 8) = 0
⇒ (x – 8) (x – 8) = 0
⇒ (x – 8)2 = 0
x = 8, 8
This is called a double root.
Suppose we have a polynomial P(x) = 0 which factorizes into,
P(x) = (x – r)k(x – a)m
If r is a zero of a polynomial and the exponent on its term that produced the root is k then we say that r has multiplicity k. Zeroes with a multiplicity of 1 are often called simple zeroes.
Question 3: P(x) is a degree-5 polynomial, that has been factorized for you. List the roots and their multiplicity.
P(x) = 5x5−20x4+5x3+50x2−20x−40=5(x+1)2(x−2)3
Given, P(x) = 5(x+1)2(x−2)3
Putting this polynomial equal to zero we get the root,
x = -1, -1, 2, 2, 2
Notice that -1 occurs two times as a root. So its multiplicity is 2 while the multiplicity of the root “2” is 3.
Fundamental Theorem of Linear Algebra
If P(x) is a polynomial of degree “n” then P(x) will have exactly n zeros, some of which may repeat.
This means that if we list out all the zeroes and listing each one k times when k is its multiplicity. We will have exactly n numbers in the list. This can be useful as it can give us an idea about how many zeros should be there in a polynomial. So we can stop looking for zeros once we reach our required number of zeros.
For the polynomial P(x),
- If r is a zero of P(x) then x−r will be a factor of P(x).
- If x−r is a factor of P(x) then r will be a zero of P(x).
This can be verified by looking at previous examples. This factor theorem can lead to some interesting results,
Result 1: If P(x) is a polynomial of degree “n”, and “r” is a zero of P(x) then P(x) can be written in the following form,
P(x) = (x – r) Q(x)
Where Q(x) is a polynomial of degree “n-1” and can be found out by dividing P(x) with (x – r).
Result 2: If P(x) = (x-r)Q(x) and x = t is a zero of Q(x) then x = t will also be zero of P(x).
To verify the above fact,
Let’s say “t” is root Q(x), that means Q(t) = 0.
We know that “r” is a root of polynomial P(x), where P(x) = (x – r) Q(x),
So we need to check if x = t is also a root of P(x), let’s put x = t in P(x)
P(t) = (t – r) Q(t) = 0
So, x = t is also a root P(x).
Question 1: Given that x = 2 is a zero of P(x) = x3+2x2−5x−6. Find the other two zeroes.
From the fundamental theorem we studied earlier, we can say that P(x) will have 3 roots because it is a three degree polynomial. One of them is x = 2.
So we can rewrite P(x),
P(x) = (x – 2) Q(x)
For finding the other two roots, we need to find out the Q(x).
Q(x) can be found out by dividing P(x) by (x-2).
After dividing, the Q(x) comes out to be,
Q(x) = x2 + 4x + 3
The remaining two roots can be found out from this,
Q(x) = x2 + 3x + x + 3
⇒ x(x + 3) + 1(x + 3)
⇒ (x + 1) (x + 3)
Q(x) = 0,
x = -1, -3
Thus, the other two roots are x = -1 and x = -3.
Question 2: Given that x = r is a root of a polynomial, find out the other roots of the polynomial.
P(x) = x3−6x2−16x; r = −2
We know that x = -2 is a root,
So, P(x) can be rewritten as, P(x) = (x + 2) Q(x).
Now to find Q(x), we do the same thing as we did in the previous question, we divide P(x) with (x + 2).
Q(x) = x2 – 8x
Now to find the other two roots, factorize Q(x)
Q(x) = x (x – 8) = 0
So, the roots are x = 0, 8.
Thus, we have three roots, x = -2, 0, 8.
SO, this polynomial can also be written in factored form,
P(x) = (x + 2) (x) (x – 8)
Question 3: Find the roots of the polynomial, 4x3-3x2-25x-6 = 0
Trick to solve polynomial equations with degree 3,
Find the smallest integer that can make the polynomial value 0, start with 1,-1,2, and so on…
Here we can see -2 can make the polynomial value 0.
Write (x+2) at 3 places and then write the coefficients accordingly to make the complete polynomial
4x2 (x+2) -11x(x+2) -3(x+2) =0
Now, notice carefully, the first coefficient is 4x2, because when it is multiplied with the x inside the bracket, it gives 4x3
When 4x2 is multiplied with 2, it gives 8x2, but the second term must be -3x2, hence the coefficient added next is -11x
Now, we know how to adjust the terms so that when we simplify it gives back the original polynomial.
We get a quadratic equation and a root is already there,
(4x2-11x-3)(x+2) = 0
Factorize the quadratic equation,
(4x2-12x+x-3)(x+2) = 0
(4x(x-3)+1(x-3))(x+2) = 0
(4x+1)(x-3)(x+2) = 0
x = -2, x = 3, x = -1/4
Question 4: Find the zeros of the polynomial, 4x6– 16x4= 0
The Polynomial has up to degree 6, hence, there exist 6 roots of the polynomial.
4x4(x2-4) = 0
4x4(x2-22) = 0
4x4[(x+2)(x-2)] = 0
Therefore, x= 0, 0, 0, 0, 2, -2