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### Course: Math for fun and glory > Unit 4

Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)

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# Sum of polynomial roots (proof)

Sum of Polynomial Roots. Created by Sal Khan.

## Want to join the conversation?

- Why, at1:27, does (x-r1)(x-r2) = 0?(2 votes)
- A polynomial of degree N can be factored into N factors of the form (x-Rn). These factors multiply to get the original polynomial. Since the polynomial is given as equal to 0, you can see how if any of the factors is 0 then the entire product will be zero. So the entire product will be zero and the equation given will be satisfied if X equals any of R1, R2, ... Rn. That's why these are "roots". These are the values of X where the polynomial evaluates to 0. If you think of the function Y = f(x) where f(x) is the polynomials, then graphically, the roots (R1, R2, ..., Rn), are the X values where the function evaluates to 0 or crosses the X axis.(14 votes)

- At00:47, why did Sal divide each side by A? Can anyone explain?(2 votes)
- to get the leading coefficient to be 1. it doesn't change the polynomial at all, but puts it in a nicer format for us.(6 votes)

- Does anyone know how to prove this so-called "Chiu's Method" here:

http://www2.sd38.bc.ca/~fharwood/FactoringNotes

It works, but I don't know how...(2 votes)- What he does is like using a new variable. 4x^2+5x-6, multiply the polynomial with the leading coefficent '4' and you get: 4*4x^2+4*5x-4*6=(4x)^2+5*(4x)-24. If you think of z=4x you have z^2+5z-24, which factorizes as (z-8)(z+3), and then replace the z with the original 4x: (4x-8)(4x+3). Since you multiplied the polynomial with 4, you have to divide it now by 4, so you get to the original polynomial and you get: (x-2)(4x+3). Dividing only by 4 the first bracket(5 votes)

- x(x^2 + 4) (x^2 - x - 6) = 0 has how many real roots?(2 votes)
- x(x^2 + 4) (x^2 - x - 6) = 0 this means that either x = 0, (x^2 + 4) = 0 or (x^2 - x - 6) = 0

The first: X=0 is a real root so that´s one.

The second: (x^2 + 4)= 0 gives x^2 = -4 which gives x = +/- sqrt(-4) which gives two imaginary roots

The third: (x^2 - x - 6) = 0 which gives x = 0,5 +/-sqrt(6,25) which gives two different real roots.

x(x^2 + 4) (x^2 - x - 6) = 0 has THREE real roots!(4 votes)

- Is in it the "Simon’s Favorite Factoring Trick" formula.(3 votes)
- I'm not sure what you mean, Simon's Favorite Factoring Trick is mainly used for polynomials with multiple variables, such as ab+a+b+1. We can apply it here to get (a+1)(b+1). It's sometimes known as "completing the rectangle".(2 votes)

- If you had to find the sum of the roots of f(x)=x^3-2x+2011, could you be able to find them without the x^2 term?(1 vote)
- it would be 0, because you have to remember that the x^2 term is still there, but its coefficient is just 0.(1 vote)

- what is 6x - 5y + 3x + 2y(0 votes)
- The answer is 9x-3y

6x - 5y + 3x + 2y

combine like terms 6x + 3x = 9x and -5y + 2y= -3y

Therefore the sum of the polynomial is 9x - 3y(5 votes)

- when they ask me to let a & b be the solutions of
**x^2+x-1=0**. find**a^2+b^2**.

how can I solve it?

i know that ab= -1 & a+b=1.. how can i continue(1 vote)- a^2+b^2= (a+b)(a+b)-2ab ----( It's true, you can try to expand it, you should remember to apply this when solving this kind of question)

a^2+b^2= (a+b)(a+b)-2ab

=( 1)(1)-2(-1) -----(substitution)

=1+2

=3 **(3 votes)

- At6:54, Sal says "it'll be a proof by induction..."

What is proof by induction?(1 vote)- A proof by induction is a proof where it is hard to prove a statement for an infinite number of examples, but instead proving one example leads to the proof of the next, and it acts like a domino effect. A good example is the proof that any number x^2 is the sum of the first x odd natural numbers.

https://en.wikipedia.org/wiki/Mathematical_induction(2 votes)

- Where will I find integral root theorem and rational root theorem for factorization of polynomials?(1 vote)

## Video transcript

What I want to do
in this video, is figure out if
there's any fast way to figure out the sum of
the roots of any polynomial. And actually, there actually is. And so that's why
I'm doing this video. So let's start with a
second degree polynomial. So let's say it's x squared
plus a1x, plus a2 is equal to 0. So this is just a standard set
quadratic equation, right here, second degree equation. And you might be saying, hey,
but you put the coefficient on the x term being equal to 1. And in general, you can
always convert any polynomial. And I'll do it with
a second degree, but if you have ax squared
plus bx plus c is equal to 0, you can just divide both
sides of this equation by a. And you're going to
have something that's in this form, where the
coefficient on the x squared term is going to be equal to 1. And you can do that
with any polynomial. You can do that
with any polynomial that's set equal to 0. So with that out of
the way, let's think about what the sum of the
roots of this are going to be. So this is a second
degree polynomial. It's a quadratic equation. So it'll have two roots. They could be real or complex. So let's call the
roots r1 and r2. And that tells us,
since these are roots, that x minus r1 times x minus
r2 is going to be equal to 0. And if we multiply this out,
we get x times x is x squared. x times negative
r2 is negative r2x. And then we have negative
r1 times x, so negative r1x. And then we have negative
r1 times negative r2, which is plus r1r2
is equal to 0. Then we can simplify this
middle term a little bit. It becomes x squared minus
r1 plus r2 x plus r1r2 is equal to 0. So let's think about what
the sum of the roots are. So based on what we're
looking at right over here, what is r1 plus r2? Well, we see it
right over here when we multiplied out
these two expressions. r1 plus r2 is the negative
of this second coefficient right over here. Or it's the negative of the
coefficient on the first degree term. So if we look at the first
degree term over here, a1 must be the same thing
as negative r1 plus r2. Or another way to
think about it is, r1 plus r2 must be
equal to negative a1. So that wasn't too bad for
the second degree case. Let's try the third degree case. Let's see, if we have x to
the third plus a1x squared plus a2x, plus a3 is equal to 0. Let's think about what the
sum of its roots might be. Well, this guy is now going
to have one more root. I'm not saying it's the
exact same equation, but let's say we're
keeping everything general. So now we could say we
have roots r1, r2, and r3. Or we could say that x minus
r1 times x minus r2 times x minus r3 is equal to 0. Now we could multiply
all of this out, but we already figured out what
x minus r1 times x minus r2 is. It's this business over here. So we just have to multiply
this times x minus r3. And actually, I'm not even
going to do the full expansion, because we have a hunch
here that it always deals, it seems, with the
coefficient on the term that's one degree lower than the
degree of the polynomial. This was a second degree,
we looked at the coefficient on the first degree term. Maybe to find the
sum, we only have to look at this
coefficient over here. So I'm only going to figure out
this product up to this point, and then we can just
ignore the rest of it and see if get something
that is useful. So let's do the expansion. Let's multiply this right over
here times this over here. And what do we get? To get the x to the
third term, the only way to get that is to multiply
this x times this x squared. So we're going to
get x to the third. And that's the only way to
get the x to the third term. That's the only way to
get the third degree term. Now how do we get
this term over here? How do we get-- let me
do this a new color-- how do we get the x
squared term over here? Well, we could multiply
this x times this over here, because
the x times the x is going to give us x squared. So it'll be negative r1 plus
r2 x squared times that term. And then if you multiply
this x times that, you'll get r1r2 x,
and all of that. But I'll just write,
so on and so forth. And then, what other ways can
we get an x squared term here? There's no other way to
get x to the third term, but what other ways could
we get an x squared term? Well, when we multiply
this times this, you have the negative r3. Let me do this in another color. You have the negative
r3 three times this. Your obviously going
to take this term multiplied by everything. And then add it to this term
multiplied by everything. But we only care
about the things-- I did this just to show you that
it's there-- but I only care, because we have a
hunch that what matters is the coefficient on
the x squared term. So we just want
to see, how do you build the x squared
term in our, I guess, our expanded polynomial? The other way to
get an x squared is to multiply the negative r3
times this x squared over here. So this is going to be
negative r3 times x squared. And then when you
multiply this times this, you're going to
get something else. Then when you multiply
this times this, you're just going to get
the negative product of all the roots. So that's just something else. And so when we add
everything together you get x to the third. And then what's the sum
of these two things? It's negative r1 plus
r2 plus r3 x squared. And then you're just going to
have a bunch of other stuff that we didn't take
the time to add. But it looks like
our hunch paid off, because what is now
the sum of the roots? r1 plus r2 plus r3. What is that equal to? Well, that's sitting
right over here. That's equal to the negative
of this coefficient right over here. That's once again
equal to negative a1. It's equal to the negative of
the coefficient on the degree term that's one less than
the degree of our polynomial. So we already see a pattern. Let's say if we can
prove it generally. So we've already proven,
essentially, two base cases. We proven it for a degree
of two, a degree of three. Let's assume that it's
true for degree of n, and then we can prove it
for degree of n plus 1. And essentially,
this will be a proof by induction for
polynomials of any degree. So let's just say we have
an nth degree polynomial. So let's just assume this. So we're going to
assume this step, and this gives a little proof
by mathematical induction practice. So let's just assume that we
have a polynomial x to the n-- it's nth degrees-- so x to the
n, plus a1x to the n minus 1, and it just goes on and
on and on all the way down to the 0th degree term. It has roots, obviously, r1,
r2, all the way to r sub n. And we're going to assume
that r1 plus r2 plus all the way to r sub n is
equal to negative a1. Or another way to
think about it, if we were to multiply x
minus r1 times x minus r2, and just keep multiplying all
the way to x minus r sub n is equal to 0, this should
give us x to the n minus r1, plus r2, plus all the way
to rn x to the n minus 1, and then plus a
bunch of other stuff that we're not
going to calculate. So this tells us this. That this coefficient
right over here is going to be equal to
this right over here. So that is what we
are going to assume. So now let's think
about-- and we're not assuming this part-- given that,
let's think about the situation where we have a polynomial
x to the n plus 1, plus a1x to the n, plus--
and it just keeps going. You get all the
degree terms there. So now this is going
to have the roots r1, r2, all the way to rn. And then it'll have rn plus 1. It'll have these n roots, and
then it'll have an rn plus 1. So essentially, that tells
us that we're going to have, essentially, this product
times x minus rn plus 1 is equal to 0. Let me write it out. I don't want to skip steps. So this tells us, if
these are all the roots, that x minus r1 times x
minus r2, blah, blah, blah, all the way to x
minus rn times x minus rn plus 1 is
going to be equal to 0. It has one more root than
the previous example, than the one where we
made the assumption. So if we were to expand this
thing out, how can we do that? Well, this thing, if we
were to multiply it out, it's just this thing over
here times this new binomial, because this thing
over here is-- let me set it up, let me use
color coded-- this thing, let me do a different color. This thing over here
we already established. We assumed is this
thing over here, which is this thing over here. So if we want this
entire expansion, we just have to multiply
this times this. So how can we get an x
to the n plus 1 term? How can we get to the
x to the n plus 1 term? Well, there's only one way to
get the x to the n plus 1 term. And that's by multiplying
this x times this x to the n. So that will give us
x to the n plus 1. That's the only way to do it. That's the only way this
highest degree term. Now how can we get--
let me do it in purple-- how can we get at
the x to the n term? Well, we can multiply this
x times this second term over here. So if you multiply this x
times this business over here, you're going to get negative
r1 plus r2, plus all the way to rn. x times x to the n minus 1, is
just going to be x to the n. And then obviously,
you're going to multiply this x times all
the other terms, so you're just going to
get a bunch of other stuff. So that's how we can get
these two terms using the x. And then can we get
either of these degree terms using the r sub n plus 1? Well, we can multiply
r sub n plus 1, or negative r sub n
plus 1 times x to the n. And then we'll also get
a similar degree term. So we'll have negative r sub
n plus 1 times x to the n. And then obviously,
we're going to multiply this times all of the other
characters in this polynomial. So you're going to have all
this plus so on and so forth. But this will suit our
purposes, because when you add these two
things, what do we get? We get x to the n
plus 1, and then we get minus-- well,
this and this are the coefficients on
the x to the n term. So this is going to be minus
r1 plus r2 plus all the way to plus rn. And then this is
minus rn plus 1. We have the minus out front. So this is plus r sub
n plus 1 x to the n. And then we're going to have
a bunch of other lower degree terms that don't matter. But we've just proven our case. What is r1 plus r2 plus all
the way to plus r sub n plus 1? What is this equal to? Well, in our expansion we
have it right over here. That's equal to that over there. That is equal to the
negative of this coefficient. It's equal to the
negative of a1. Now the reason why this
works out no matter what-- you say, hey, if I
have complex roots, how does it always end
up being a real number? And that's because the
imaginary parts cancel out when you take their sum. So let's just apply it. We've proven it. We've proven, actually,
two base cases. We've proven it for a
degree 2, for degree 3. And then we showed that if
it's true for any degree n, then it's definitely true
for any degree n plus 1. So we know if it's
true for degree 3, then it's definitely
true for degree 4, which means it's definitely
true for degree 5. So A mathematical induction
is kind of this domino proof that if it's true for 3,
which makes it true for 4, then assuming it's true
for 4 makes it true for 5. And it just keeps
going to any n. So in general, if
someone gives you a polynomial-- let me think
of some crazy polynomial. If I were to give you x
to the seventh, minus pi x to the sixth plus ex to the
fifth, minus square root of 2 x to the fourth, minus-- I
won't write out all the terms. So minus 3 is equal to zero. If someone were to give you this
polynomial, and they were say, what do the seven roots of
this polynomial add up to? You say, oh, well,
the sum of these roots is going to be the negative
of this coefficient right over here. So r1 plus r2 plus all the
way to r7, all seven roots are going to add up to the
negative of this coefficient. They're going to be equal to pi. And one other thing, you
might be saying, hey, wait. Every polynomial you
did had a 1 coefficient in front of the
highest degree term. You know, not all
polynomials have that. And the answer there
is-- let's say, if I were to give you a
polynomial, I don't know, 7x to the fifth, minus 6x to
the fourth, plus, I don't know, pi x to the third, plus a
bunch of stuff other stuff is equal to 0-- and
you wanted to find the sum of this
polynomials roots. You'd first divide
everything by 7. So it becomes x to
the fifth, minus 6 over 7x to the fourth, plus
pi over 7x to the third, plus so on and so forth, is
equal to 0/7 which is just 0. And now you look
at this polynomial, well the sum is going to be the
negative of this coefficient here. So it's going to be the
sum of all five roots. Or it's going to be 6/7. Anyway, hopefully
you enjoyed that.