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### Course: Class 10 (Foundation)>Unit 3

Lesson 1: Zeroes of a polynomial

# Zeros of polynomials & their graphs

Sal uses the zeros of y=x^3+3x^2+x+3 to determine its corresponding graph. Created by Sal Khan.

## Want to join the conversation?

• Why do complex solutions come in pairs?
• A polynomial of degree n has n solutions. So let's look at this in two ways, when n is even and when n is odd.
1. n=2k for some integer k. This means that the number of roots of the polynomial is even. Since the graph of the polynomial necessarily intersects the x axis an even number of times. If the graph intercepts the axis but doesn't change sign this counts as two roots, eg: x^2+2x+1 intersects the x axis at x=-1, this counts as two intersections because x^2+2x+1=(x+1)*(x+1), which means that x=-1 satisfies the equation twice. This means that for an even n we have an even number of real solutions.
2. n=2k+1 for some integer k. This means that the graph of the polynomial intersects the x axis an odd amount of times, but at least once. So there is at least one real solution to the polynomial, meaning that we have an even amount of unspecified solutions, which brings us back to the n=2k train of thought. This means that for an odd n we have an odd number of solutions.
Putting together 1. and 2. we get that there always has to be an even number of complex solutions. I hope this answers your question :)
• Why did Sal neglect to evaluate the situation where x=0? With much less computation, that would have simplified the equation to y=3 and the y-intercept of (0,3) is only found on graph A.
• I think because he wanted to mention the roots, etc. Without that information, we the viewers would be less capable of solving similar problems in the future. Sal is always good about trying to impart some insight, rather than just showing the 'easy' way to solve a particular problem.
• So what are the two complex solutions for the ` x^2 = -1 `? I understand that the square root of -1 is i. But why is it also -i ? Is it because ` -i * -i = -1 `, and also ` i*i = -1 `?
Basically the same reason that ` 1*1 = 1 ` and ` -1*-1 = 1 ` ?
• Yes, you are correct. Remember: when taking the square root of both sides, if a side is not yet squared, you have to consider +- the root.
For example:
x^2 = 256
x = +-√256
x = 16 or x = -16
• At he says that `C` only has two real roots so it's excluded. But how can we say that for sure? Couldn't the graph curve back down off to the left or the right side somewhere that we don't see?
• As always, there are some unspoken assumptions. In this type of problem, the implication is that all of the interesting features of the graph are visible. In practice, if you saw a curve of this nature and did not know the formula, you could say that it is at least locally approximated by a curve with only two zeroes.
• What would happen if an imaginary number is a zero too? Then how would we graph it?
• Imaginary numbers and complex numbers with imaginary parts can be graphed on what we call an "Argand diagram" which is just a coordinate plane but with the y-axis being imaginary and the x-axis being real.
• What is the difference between complex and imaginary?
• Imaginary numbers are just a multiple of i;( For example: 3i). Complex numbers are imaginary numbers and another number; (For example: 3i-8.)
• It seems to me that the complex numbers Sal found at the end caused the curved line from points (-1, 5) to (0, 3), but how do we know where the effect of the complex numbers will take place on our graph?
• If you think of it as (x^2+1)*(x+3) you have the product of a parabola (x^2 translated up 1 ) and a straight line x, slope 1, translated up 3).
Or, you can picture it as (this is more helpful) the sum of x^3, the parabola 3x^2, the straight line x, and horizontal line 3.
As for the complex factors that "x^2+1" can be factored into, I can't visualize them at all.
• x^2=-1 is analogous to i, yes? x=sqrt(-1)?
• How do you know how many zeros a polynomial function has?
• The Fundamental Theorem of Algebra tells us that every n-degree polynomial has exactly n complex roots. Keep in mind, that this theorem does not account for multiplicity. In other words, some of the roots may not be unique. For example, the quadratic:
𝑥² + 6𝑥 + 9 = 0
Has only one unique solution: 𝑥 = -3. However the Fundamental Theorem of Algebra says that since the degree of the polynomial was 2, then there must have been 2 solutions. We call 3 a DOUBLE ROOT of the polynomial, because it accounts for both roots of the polynomial. Furthermore, that quadratic could have been factored as:
(𝑥 + 3)(𝑥 + 3) = 0
Here, two binomial factors both yield 𝑥 = -3 so again we can see that -3 must be a double root. We can also say that the root -3 has a multiplicity of 2. A triple root would have a multiplicity of 3 etc. Therefore, the Fundamental Theorem of Algebra can be used to find the number of zeros a polynomial function and if there is an apparent "contradiction" then it is because of one or more of the roots may have a multiplicity greater than 1. Comment if you have questions.