Polynomial basics
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Terms coefficients and exponents in a polynomial
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Interesting Polynomial Coefficient Problem
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Polynomials1
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Polynomials 2
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Evaluating a polynomial at a given value
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Simplify a polynomial
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Adding Polynomials
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Example: Adding polynomials with multiple variables
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Addition and Subtraction of Polynomials
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Adding and Subtracting Polynomials 1
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Adding and Subtracting Polynomials 2
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Adding and Subtracting Polynomials 3
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Subtracting Polynomials
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Subtracting polynomials with multiple variables
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Adding and subtracting polynomials
Interesting Polynomial Coefficient Problem Finding the coefficients of a third degree polynomial given 2 roots and the y-intercept
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- I've been sent is pretty interesting algebra problem,
- and I think the question isn't properly formed, or it might be
- missing some piece of information.
- But it's an interesting question nonetheless so I
- thought I'd work it out, and we can talk about why
- it's not properly formed.
- So they give us a function, f of x.
- They say it's a third degree polynomial of the form
- ax to the third plus bx squared plus cx plus d.
- And they tell us a couple of the 0's of this polynomial.
- They say that we have 0's at the point minus 1, 0 and
- the point 2 comma 0.
- And they tell us we have a y-intercept, y-intercept,
- at the point 0, minus 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.
- 1, 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.
- That's a plus.
- 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.
- 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.
- ax plus 1.
- 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.
- 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.
- That's 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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