Interesting Polynomial Coefficient Problem Khan Academy English Spanish

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I've been sent is pretty interesting algebra problem,

Me enviaron este bonito e interesante problema de álgebra,

and I think the question isn't properly formed, or it might be

y creo que la pregunta no está correctamente formulada, o podría

missing some piece of information.

faltar una parte de la información

But it's an interesting question nonetheless so I

No obstante, es una pregunta interesante, así que

thought I'd work it out, and we can talk about why

pensé que la trabajaría, y podemos hablar el por qué

it's not properly formed.

no está correctamente formulada

So they give us a function, f of x.

Entonces, tenemos una función, f de x

They say it's a third degree polynomial of the form

Se dice que es un polinomio de tercer grado de la forma

ax to the third plus bx squared plus cx plus d.

ax³ + bx² + cx + d

And they tell us a couple of the 0's of this polynomial.

y nos dicen un par de 0's del polinomio

They say that we have 0's at the point minus 1, 0 and

Dicen que tenemos 0's al punto -1 , 0 y

the point 2 comma 0. And they tell us we have a y-intercept, y-intercept,

at the point 0, minus 2.

y nos dicen que tenemos una intersección con el eje Y en el punto 0 , -2

And what they ask us is, what is a plus b plus c plus d?

So let's see if we can make some headway on this.

So the first thing is to think about, is what would a third

degree polynomial look like and what are we even talking

about when we say 0's.

So let me just draw a little bit of a graph.

And I don't know exactly what this third degree polynomial

looks like just yet.

Let me draw my axis.

Now a third degree polynomial can have as many as three 0's.

And 0's are just the points where the

function is equal to 0.

So, we have a 0 at minus 1, 0.

So that's minus 1, 0, that's right there.

We have a zero 2, 0.

So those are the two 0's that they've given us.

They also told us the y-intercept at 0, minus 2.

So it intersects the y-axis right there.

But this guy can have as many as three 0's.

Now some of them might be complex.

But complex 0's come in pairs of two, the two

complex conjugates.

I won't go into too much detail here.

So this one must have a third real 0, because if the third

one was complex, you'd need another complex 0, and you

can't have four 0's for a third degree polynomial.

So that third root, it could be sitting some place out

here, maybe over here.

Or it could even be a repeat of one of these

two roots over here.

But let's just assume that there's some third root, we

don't know what it looks like.

Let's say it looks something like-- let's say that third

root is sitting out here some place.

And then a potential graph, and I could be completely wrong,

I'm just guessing, but just to get a sense of what third

degree polynomials look like, and how can a graph intersect

the x-axis three times, a potential graph might look

something like this.

Curves and bam, hits the other 0, hits the y-intercept, and

then goes back up like that.

It might go like that, it might go the other way, go like that.

Up, and then down like that.

Well, there's many ways you could draw something that

essentially curves twice to intersect these three points

and also that y-intercept, but I'm not going to go into

all of those right now.

But let's see if we can figure out the coefficients here.

So the key point is me telling you that there

must be a third 0 here.

Let's call that third 0, it's at the point, it's at the

point-- let's call it --r3 comma 0.

And I'm using the letter r for roots.

Roots are the x values of the 0's.

So if we say f of minus 1 is equal to 0, we say that x is

equal to minus 1 is a root.

Similarly, we know that f of 2, f of 2, is equal to 0.

Or we could say x equals 2 is a root, these are roots.

So when someone tells you a 0 they're also essentially

telling you the roots, and then we know there must be a third

root at x is equal to r3.

So x is equal to r3.

Now, if we have a third degree polynomial where these three x

values make it 0, we can rewrite this third degree

polynomial as-- we can rewrite it as-- I'll do it in a

slightly different color --f of x is equal to x plus 1-- you'll

see why I'm doing x plus 1 instead of x minus 1 in a

second --plus 1 times x minus 2 times x minus r3.

And we want to put an a out front, and I'll tell you in

a second why I'm putting this a out front.

Because if you were to multiply just the x terms here, when

actually do your distributed property and multiply out these

binomials, you'll just get x to the third, but we had

an ax to the third.

So you just needed a constant a there to make this

an ax to the third.

Now, why did I write out like this?

Well, if you put a minus 1 in here what's going to happen?

This term is going to be minus 1 plus 1.

It's going to be 0.

Who cares what these are or what a is, the whole

thing's going to be 0.

If we put x is equal to 2, this term right

here is going to be 0.

Who cares what the other terms or the a is, f

of 2 is going to be 0.

Similarly, when x is equal to r3, this last term is 0, who

cares about anything over here.

So we know that this guy up here can be rewritten

in this form over here.

And so if we can solve for r3 we can just multiply this thing

out and try to do a little bit of pattern matching to figure

out what these coefficients are going to be.

Now, the other big clue they gave us is a y-intercept.

That point right there, the point 0, minus 2.

They're telling us that f of 0 is equal to minus 2.

Well what's f of 0?

f of 0-- I'll write it here --f of 0-- If you put 0 in here,

this term becomes 0, this term becomes 0, that term becomes 0,

you are just left with the d.

So f of 0 is equal to d.

Or we could say that d must be equal to minus 2.

d is equal to minus 2, so we solved at least one

of the call efficient.

The constant term. d is equal to minus 2.

Now, if d is equal to minus 2, that means all of these

constant terms, when you were to multiply out this

expression, and I would include the a, because the a is scaling

everything, that must be equal to minus 2.

Let me be clear here.

So let me rewrite this expression.

This is equal to, I'll just multiply this a times

this last term.

We do it in any order you want, so this is the same thing as x

plus 1 times x minus 2 to times a a minus r, let me

write, minus ar3.

This is the exact same thing as this thing up here, I'll just

write this is with f of x is equal to, and this is the exact

same thing is that up there.

Now, to solve for an r3, or attempt to solve for an r3, you

just have to realize that when you multiply this thing out,

the way to get the constant term over here, or the d over

here, that's going to be a product of the constant

terms in our expressions.

Because if I introduced any products with any x terms,

you're going to get one of these other terms over here.

The ax cubed term is generated from that term, that term,

and that term, being multiplied together.

and then the constant term is multiplied by the

constant terms to be multiplied together.

And then the two in the middle are multiplied by different

combinations of constant terms.

And you'll see that if you actually want to

multiply this out.

But if you take my word for it, that times that times that

times, actually let me write it this way.

That times that times that, have got to be equal to date we

can then attempt to find some relationship between a and d,

which is mine is minus 2.

So we could say 1 times minus 2 times minus ar3, these two

minuses will cancel out.

It is going to be equal to our d terms.

It's going to be equal to minus 2, and so if we simplify that a

little bit we get 2ar3 is equal to minus 2, divide both sides

by 2a and you get r3 is equal to minus 1/a.

So we haven't solved for the third root, we've just gotten

it in terms of, in terms of our coefficient a.

But letls see if we can still use that in some useful way.

So if r3 is minus 1/a, this term right here becomes what?

Let me rewrite it.

Let me rewrite it.

All right, I'll do it in blue.

f of x is going to be equal to x plus 1 times x minus 2 times

ax-- now minus a times r3.

Let me do it over here.

Minus a times r3 three is minus 1/a.

So that's a minus times a minus.

The a's cancel out.

You just get a 1.

So this term now if we substitute r3 with this will

just turn into a this will just turn into a plus 1.

Minus a times minus 1/a that's plus 1.

We just substituted that for that to get this right here.

Now let's multiply this thing out.

Let's actually do the algebraic multiplication

and see what we get.

So first I'm going to multiply these two terms.

I'm going to do it in a different color.

Let me multiply those two terms right there.

So this is going to be equal to x squared, x squared plus 1x

minus 2x So that's minus x, right?

1 times x is x, minus 2 times x is x minus 2x.

So it's minus x, and then minus 2, just multiplied these two

binomials out right here.

And then I'm going to have to multiply that times ax plus 1,

so I'm going to write the ax plus 1 down here

just to save space.

And now we just do a little bit of algebraic multiplication.

1 times minus 2 is-- I'll do it in magenta --minus 2.

1 times minus x is minus x.

1 times x squared is x squared.

Now ax times minus 2 is minus 2ax.

ax times minus x is minus ax squared, right?

Minus ax squared and then ax times x squared

is ax to the third.

ax to the third.

Then we add everything together, and we get our f of x

is going to be equal to ax to the third-- let me scroll to

the left a little bit --ax to the third.

x squared minus ax squared.

So that's plus 1 minus a, 1 minus a x squared.

And then we have these two added together so this is minus

1 minus 2a or we could call this equal to-- so minus 1

minus 2a --or we could say minus 1 plus 2ax and then we

have our minus 2, which makes sense because that is our d.

So this is really the solution.

This is the form that are equation is going to take and

the reason why I said in the beginning of the problem that

the equation or the problem wasn't well defined is that I

can pick any real number a and this equation will

satisfy these conditions that we were given.

So I suspect that I wasn't given all of the constraints

on this problem.

Maybe there was a third constraint that a is equal to

1, or b is equal to something else, or a plus b is

equal to something else.

But just using these constraints, just using these

constraints, the equation will take this form.

So, if it was in this form, of course this is a, this is b,

which would be 1 minus a, that's b, that's a, and then

this right here is equal to c.

So we could pick, you know, we could pick one particular

choice if we just want to get an answer to their problem, but

it might not be the answer they were looking for because they

might have picked a different a.

But if you just pick a simple a is equal to 1, than

what's b equal to?

Then b is equal to 1 minus a, which is equal to 0.

Then b is equal to 0, and then c is equal-- actually c would

include this minus sign right there --c would be one plus 2a.

So 2a is 2, 1 plus 2 is 3.

Put a minus sign.

So c would be equal to minus 3.

And of course no matter what we do, d is going

to be equal to minus 2.

So if we pick a is equal to 1, the answer here to what is a

plus b plus c plus d, would be equal to 1 plus minus 3.

1 plus minus 3 is minus 2.

Minus another minus 2 is minus 4.

That's another possibility, but we just as easily could have

picked, you know a could have been something, it could be a

fraction, it could be anything.

It could be, you know a could be minus 1.

a could be equal to minus 1, if a is equal to minus 1, then b

is equal to 2, then b is equal to 2, right?

1 minus minus 1.

And then c is equal to 1 minus 2, which is minus 1, but then

you have a minus out front.

So then c is equal to 1, and then d is always going

to be equal to minus 2.

d is equal to minus 2.

And then you have a plus b plus c plus d would be equal to.,

well these would cancel out, you'd get them equal to 0.

So, we don't know exactly what answer the problem writer

was looking for, but it is a pretty interesting problem.

And then you know, just out of curiosity, it-- let's take this

first example that we took --where a is equal to 1, we

could then look at what our root must be.

If a is equal to 1, then our third root is that r3 would be

equal-- that's just for this situation right here --than r3

would be equal to negative 1 divided by 1 would be equal to

minus 1, which means that we would have a repeated root, we

would have a repeated root, right there.

We wouldn't necessarily have a distinct root out there.

And if we're curious about what that would look like, let's

look at the situation where a is equal to 1.

Actually maybe we wil look at both of them.

When a is equal to 1, our equation is x to the third--

where is the x button --x to the third, there's no x

squared term, minus 3x minus 3x, minus 3x, minus 2.

So let's grab both of them simultaneously.

So that's this one, where we picked this choice

of coefficients.

And then let's do this choice of coefficients, if we have, if

we have minus x to the third, right? a is minus 1 plus 2 x

squared, and then we have, plus x, c is 1, plus x minus 2.

So those are 2 graps.

Let's see what they look like.

Let's graph them.

So our first one, there we go, we hit that point,

that's our double root.

Notice, both of these two graphs meet our

two constraints.

They both have a root there at minus 1.

They both have a y-intercept at y is equal to minus 2.

And they both have a root at x is equal to

positive 2 right there.

Now the second one right here had a separate distinct third

root, while the first one, here, has a double root

right over there.

Anyway, I actually think this problem was somewhat more

interesting by the fact we had to actually look at the

different solutions to it.

And there's actually an infinite number of solutions

to this based on what you choose for your a.

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