Lec 33 - Algebra of polynomials: Multiplication
Summary
TLDRThis video tutorial teaches the method of multiplying polynomials of arbitrary degrees. It starts with the basics, explaining the multiplication of monomials and binomials using the FOIL method, then generalizes the process for polynomials. The instructor demonstrates multiplying a trinomial by a monomial, extending the concept to include quadratic expressions, and provides a formula for finding the coefficient of any term x^k in the product. The video concludes with an example of multiplying two quadratic polynomials, illustrating the systematic approach to polynomial multiplication.
Takeaways
- 📚 The video is a tutorial on how to multiply polynomials of arbitrary degrees, starting with the basics of binomial multiplication using the FOIL method.
- 🔢 It explains the multiplication of monomials and how to apply the law of exponents when multiplying terms with the same base.
- 📈 The script demonstrates the process of multiplying a quadratic polynomial by a cubic monomial, showing the application of the law of exponents and simplification of the result.
- 📝 The concept of extending the FOIL method for binomials to polynomials with more than two terms is introduced, emphasizing that FOIL only applies to binomials.
- 📚 The tutorial provides a step-by-step method for multiplying polynomials by converting them into monomials and multiplying each term individually.
- 🔑 A general formula is presented for finding the coefficient of a term \( x^k \) in the product of two polynomials, which involves summing the products of coefficients where the indices sum to \( k \).
- 📉 The script illustrates how to apply this formula to polynomials of arbitrary degree, showing that the degree of the resulting polynomial is the sum of the degrees of the original polynomials.
- 📝 The importance of recognizing that not all coefficients are listed and that missing coefficients are assumed to be zero is highlighted.
- 📑 An example is worked through to demonstrate the application of the formula, showing the computation of the coefficients for each power of \( x \) in the product of two quadratic polynomials.
- 🔍 The video concludes with a note on the difference between polynomial multiplication, which always results in a polynomial, and division, which may not.
- 👋 The script ends with a sign-off, indicating that the next video will cover polynomial division and its potential outcomes.
Q & A
What is the FOIL method mentioned in the video script?
-The FOIL method is a technique used to multiply two binomials. It stands for First, Outer, Inner, Last, which refers to the process of multiplying each term of the first binomial by each term of the second binomial.
How does the video script generalize the multiplication of polynomials beyond binomials?
-The script generalizes polynomial multiplication by considering monomials and extending the concept to polynomials of arbitrary degree. It involves multiplying each term of one polynomial by each term of the other polynomial and applying the law of exponents.
What is the standard rule of multiplication for monomials as described in the video?
-The standard rule for multiplying monomials involves multiplying the coefficients and adding the exponents of like bases. For example, multiplying 2x^3 by x^2 results in 2x^(3+2) = 2x^5.
How does the video script handle the multiplication of a trinomial and a monomial?
-The script suggests converting the trinomial into separate monomials and then multiplying each of these monomials by the monomial of the other polynomial. This is an extension of the method used for monomials.
What is the difference between the FOIL method and the method described for multiplying a quadratic and a linear polynomial?
-The FOIL method is specific to binomials, while the method described in the script for multiplying a quadratic and a linear polynomial involves treating each term of the quadratic as a separate monomial and multiplying it by the linear polynomial, which is also treated as a monomial.
How does the script suggest finding the coefficient of a specific term in the product of two polynomials?
-The script provides a general formula where the coefficient of x^k is the sum of products of coefficients from each polynomial, where the sum of the indices of the coefficients equals k.
What is the significance of the formula ∑(j=0 to k) a_j b_(k-j) in the context of the video script?
-This formula is used to find the coefficient of x^k in the product of two polynomials. It represents the sum of all possible products of coefficients from the two polynomials where the sum of the indices equals k.
How does the script approach the multiplication of two polynomials of arbitrary degrees?
-The script suggests using the general formula to find the coefficients of each term in the product polynomial. It involves summing over all possible combinations of coefficients from the two polynomials that result in the desired exponent.
What is the degree of the resulting polynomial when multiplying two polynomials of degrees m and n?
-The degree of the resulting polynomial is m + n, assuming m is not equal to n. If m is equal to n, the degree is 2n.
Can the multiplication of two polynomials always result in a polynomial, according to the video script?
-Yes, the multiplication of two polynomials will always result in another polynomial. However, the division of two polynomials may not always result in a polynomial, as mentioned towards the end of the script.
What is the next topic the video script suggests covering after polynomial multiplication?
-The next topic suggested by the script is the division of two polynomials, which may not always result in a polynomial.
Outlines
📚 Introduction to Polynomial Multiplication
The paragraph introduces the concept of multiplying polynomials, starting with a basic understanding of binomial multiplication using the FOIL method. It then extends the discussion to polynomials of arbitrary degrees, using examples of monomials and trinomials to illustrate the process. The video script explains how to multiply each term of one polynomial by every term of another, applying the law of exponents and simplifying the expression. It also hints at a more general approach for multiplying polynomials beyond the FOIL method, which is typically limited to binomials.
🔍 Extending the FOIL Method for Polynomial Multiplication
This section delves into extending the FOIL method to multiply polynomials that are not binomials. It uses a quadratic polynomial and a linear polynomial as examples to show how to convert them into monomials and then multiply term by term. The script discusses the process of adding the resulting polynomials by matching exponents. It also introduces the idea of finding a general formula for the coefficients of the resulting polynomial, especially focusing on the coefficient of x squared and how it can be derived from the given polynomials.
📘 General Formula for Polynomial Coefficients
The paragraph presents a general formula for determining the coefficients of the terms in the product of two polynomials. It explains how to find the coefficient of x raised to any power 'k' by summing the products of coefficients from the original polynomials where the exponents add up to 'k'. The explanation includes a detailed example with two quadratic polynomials, demonstrating how to apply the formula to find the coefficients for each term in the resulting polynomial. The script emphasizes the systematic approach to polynomial multiplication, which can be applied to polynomials of arbitrary degrees.
📘 Demonstrating the General Formula with an Example
This part of the script provides a step-by-step demonstration of the general formula using an example of multiplying two quadratic polynomials. It identifies the coefficients of the individual terms in the polynomials and then applies the formula to calculate the coefficients of the resulting polynomial for each degree of x. The explanation includes computing the constant term, the linear term, the quadratic term, and higher degree terms, ensuring to account for all possible combinations of coefficients that contribute to each term. The paragraph concludes with the final polynomial obtained from the multiplication.
🔚 Conclusion and Transition to Polynomial Division
The final paragraph wraps up the discussion on polynomial multiplication, summarizing the process and emphasizing that the result is always another polynomial. It contrasts this with polynomial division, which will be covered in a subsequent video and may not always result in a polynomial. The script ends with a sign-off, thanking the viewers for watching and indicating that the next topic will be polynomial division.
Mindmap
Keywords
💡Polynomials
💡Multiplication of Polynomials
💡Monomial
💡FOIL Method
💡Exponents
💡Coefficient
💡Quadratic Functions
💡General Formula
💡Term-by-Term Multiplication
💡Degree of a Polynomial
💡Summation
Highlights
Introduction to the multiplication of polynomials beyond binomials.
Explanation of the FOIL method for binomials and its limitations for higher degree polynomials.
Demonstration of multiplying monomials using the law of exponents.
General rule for multiplying polynomials of arbitrary degree through term-by-term multiplication.
Illustration of multiplying a cubic monomial by a quadratic polynomial.
Conversion of polynomials into monomials for multiplication purposes.
Multiplication of a binomial with a quadratic polynomial using extended FOIL method.
Addition of polynomials by matching exponents to combine like terms.
Presentation of a systematic approach to multiply polynomials by term-by-term method.
Explanation of how to find the coefficient of a specific term in the product of two polynomials.
General formula for the coefficient of x raised to the power of k in polynomial multiplication.
Application of the general formula to find coefficients in the product of two quadratic polynomials.
Identification of coefficients for a polynomial of degree n and m using the summation formula.
Demonstration of the process to compute the multiplication of two quadratic polynomials step-by-step.
Explanation of how to handle coefficients not listed in the polynomials as zeros.
Final computation of the product of two quadratic polynomials using the systematic method.
Conclusion emphasizing that polynomial multiplication always results in a polynomial, unlike division.
Teaser for the next video on polynomial division and its potential non-polynomial results.
Transcripts
In this video, we will learn how to multiply two polynomials. Let us start with basics
of multiplication of polynomials. We already know how to multiply two binomials. For example,
if you have been given two binomials of the form a x plus b into c x plus d, then you
know how to multiply these two binomials that is we will use the foil method. However, in
this context, we want to generalize the settings for multiplication of polynomials of arbitrary
degree. So, let us see, let us start with some simple
monomials with through examples. So, here is a polynomial given to you p x is x square
plus x plus 1 and q x is 2 x cube. The question is do I know how to multiply these two polynomials?
Remember this one is called monomial, it has only one term. So, a standard rule of multiplication
will mean we have seen this in our quadratic functions that I will consider the product
in this manner. Once I consider the product in this manner,
what we will do is we will try to multiply each term of this 2 x cube with each term
of this polynomial. So, there are three terms. And for each term this 2 x cube will be multiplied.
So, if I do that the law of exponents will apply.
For example, x raised to 2 plus into x raised to 3 will mean x raised to 2 plus 3. So, once
I applywe apply the law of exponents and add the exponents, obviously, 2 was a constant
coefficient of x cube which will be multiplied throughout the expression. And therefore,
the resultant is this which we can simplify as 2 x raised to 5 plus 2 x raised to 4 plus
2 x cube. This is how we will multiply a monomial. Now, as you can see the this polynomial has
three terms 1, 2 and 3. So, it is not a binomial; it is a trinomial. So, my foil method will
not work here. So, foil method will work only for these kind of expressions which are binomials.
So, let us go ahead and try to consider a similar expression that is a quadratic expression
and another binomial, and try to see how can I extend the basis of foil method right.
So, here is a binomial 2 x plus 1. Andhere is a general polynomial quadratic polynomial
which is x square plus x plus 1 same. Now, what will you do? So, naturally you will consider
p x into q x which will be written in this form. Now, if I want to extend the basis whatever
I did for monomial, that means, I need to convert this into two monomials.
So, what are those two monomials? One monomial is 2 x; another monomial is 1. So, if I treat
them separately that is if I write them in this manner, let me erase this, that is I
have written them in this manner. Then what can I do about it, that means, now
this this turned out to be a same expression instead of x cube, here it is x that is all
is the difference right. So, whatever I did here, I can do it here. And the last term
is actually multiplied with 1 which it suppressed because multiplication with 1 will not change
anything. So, I do not have to worry about the last term.
Now, I will multiply this 2 x with all the terms in for of p x x square plus x plus 1
which is similar to this particular thing. So, I will get 2 x raised to 1 plus 2 x raised
to 1 plus 1 2 x x square plus x plus 1. Now, the job is very simple. You can treat
this as one polynomial, and this one as a second polynomial, and then we have to add.
How we add polynomials? We will add polynomials by matching the exponents, matching the exponents
of x. So, if I want to add these two polynomials, what will I do, I will simply match the exponents
and I will add them which is given here. So, in this case 2 x raised to 3, there is
no competing term for x raised to 3. So, it remains 2; x square comes here and here, therefore,
I added the two which gives me 2 plus 1, in a similar manner the terms containing x are
these two. So, I have added these two, so 2 plus 1 x plus 1 which is similar to what
we have seen in the last video of addition of polynomials. And therefore, we get the
answer to be equal to 2 x cube plus 3 x square plus 3 x plus 1.
So, effectively what we have done is we know how to multiply the terms term by term. And
finally, if at all I want to seek an extension of a of a foil method, it will be a term by
term multiplication of polynomials, that means, you take the polynomial of least degree and
multiply it with the polynomial of highest degree term by term, add those term match
the powers and then write your answer. So, this is one prototype that we can follow
for finding multiplication of polynomials or result of the multiplication of polynomials.
Now, the next question is can I generalize this method or can I answer it programmatically,
that means, can I give a simple formula for what the coefficient of one part x raised
to m will be? For example, in this case can I give a general formula what will be the
coefficient of 3 x square provided I know polynomials p x and q x, p x and q x. So,
to answer that, let us go ahead and try to find a general formulation of this formof
this formula. Let us go ahead. And if you are asked given
one quadratic polynomial and one linear polynomial, you are asked to compute p x into q x, how
will you go about this? This is what our task is. Now, so naturally I will write p x into
q x, and then I will convert each of them into monomials that isone monomial will be
b 1 x, and second monomial will be b naught. In this case, what will happen is we will
simply multiply them as a separate term by term multiplication. So, in earlier case our
b naught was 1 when we studied one example. But here we are considering a general expression,
and none of the expressions are 0 that is what we are assuming none of the coefficients
at a 2, a 1, a naught, b 1 and b naught none of them are 0.
For example, if you consider b naught to be equal to 0, then this term itself will vanish
the second term itself will vanish; you will not have the second term. So, we are assuming
that all terms remain in the loop ok. So, now it simple, the job is multiplying these
two polynomials, and you will get some answers that is ok, but now our main worry is to find
a pattern in these answers ok. So, now, when I multiplied this, if you look
at this particular expression that is a 2 b 1 x raised to 2 plus 1, a 1 b 1 x raised
to 1 plus 1, a naught b 1 x, a 2 b naught x square, a 1 b naught x, a naught b naught.
Here you take a pause and examine the terms. For example, this term contains the coefficient
of x raised to 3, this is 2 plus 1. So, x raised to 3. So, in that case, what is happening
here is if you look at the suffixes of the coefficients this is a 2, this is b 1, so
together they will sum to 3. In a similar manner, you look at this term which contains
x square. And you look at the suffixes of the coefficients
that is a 1 b 1, together they will sum to the exponent that is a 1 plus 1 is 2. So,
this should be a coefficient of x square. Then if this logic is correct, what should
bethe coefficient of a constant? The coefficient of the constant that is x raised to 0.
So, the coefficient of the constant must be a naught b naught. In a similar manner you
can ask the question what is a coefficient of x? If you asked that question, you will
naturally get the answer you collect all the in all the coefficients such that their suffixes
will sum to 1 that is a 1 b naught plus b naughtb 1 a naught. So, is there anything
called b 1 a naught? Yes, it is here. So, this what we have actually done is we
have figured out a pattern; that means, if I want to find the coefficient of x raised
to k, then better the sum should be some a j and b k minus j, so that they both will
sum, they both will sum to it is not equal to the this is I I am saying x raised to coefficient
of x raised to k will be equal to of the will be of the form a j plus b k minus j. So, with
this understanding, let us go further and try to rewrite this sum ok.
So, once I have rewritten this sum, my analogy is further amplified. For example, if you
look at the coefficient of x square, yes, it was it is a 1 b 1 and a 2 b naught which
is the coefficient of x square, so that also means this means if I can sum over this j
from 0 to what point to a point where I want the sum the exponent is raisedd to k, then
I will get all possible combinations where sum is actually k.
um In a similar manner, you canpause this video and verify whether you are getting the
same expression for x raised to 1 and all others right. So, with this understanding,
I I am ready to generalize this demonstration or thistheory for a polynomial of an arbitrary
order. Let us consider polynomials of degree n and
m, and try to find the general answer for them, and that answer will be in this form.
So, if you are given a polynomial of degree n n p x, and if you are given another polynomial
of degree m q x, let us say m not equal to n.
Even if m is equal to n it does not matter, but for our purposes let us take m not equal
to n, then what will be the coefficient of each of the x raised to k's? The coefficient
is actually given here, summation over j is equal to 0 to k a j b k minus j this is what
we have figured out in this expression is the coefficient of x raised to k.
Then the question is how far the degree will go? The degree will go till m plus n m is
not equal to n; if m is equal to n then the degree will go to 2 n that is ok. So,k is
equal to 0 to m plus n, and each of the coefficient of x raised to k will be j is equal to 0 to
k a j b k minus j. Now, let us demonstrate this idea with one example. Let us go ahead
and see one example of this idea. So, now, you have been given two polynomials
two quadratic polynomials and you are asked to compute the multiplication of these two
polynomials. One way is very simple you will go with term by term multiplication, and it
simply means you have to multiply the terms of second polynomial with the first polynomial
in a term by term fashion, or you can actually use the formula that I have given you in the
previous slide. So, you can pause this video, and try to compute by yourself or you can
go along with me. So, let us recall that formula again that
is p x is equal to sum a, so my polynomial is a polynomial of degree n, and q x is a
polynomial of degree m. In this case, in this particular example, the polynomial the first
polynomial is of degree 2 as well as the second polynomial is of degree 2.
So, in order to find the product of these two polynomials, what do we need to find is
we simply need to find the coefficients of x raised to k. So, let us first identify what
are a k's and what are b k's, hm,j is a dummy index. So, it does not matter.
So, let us first identify what are a k's and b k's. So, a naught as you can see is 1, b
naught is 1, a 1 is 1 again, b 1 is 2, hm, correct, this is correct, and thena 2 and
b 2 both are 1 yeah. So, I have enlisted all the coefficients of this particular expression,
p x and expressions p x and q x. Now, we need to use this formula, then this formula which
gives me the sum. So, let us use this formula and figure out.
Remember, all the coefficients that are not listed here. For example, what will be a 4,
if at all, I will write a 4, what will be a 4 in this expression? It will be 0. What
will be a 3 in this expression? It will be 0. So, all the coefficients that
are not listed here are 0s. Keep this in mind and try to answer the question. So, now, computation
of coefficient; it is very easy. So, let us start with 0th degree term that is constant
term. So, here k will be equal to 0. So, the summation will actually go from j is equal
to 0 to 0, that means, it will have only one term which is a naught b naught.
What is a naught b naught? Look here 1 into 1, so it will give you 1 ok. Let us go for
a degree 1 term. hm So, j is equal to 0 to 1, j is equal to 0 to 1, so it will have a
naught b 1 a naught b 1 plus a 1 b naught, a naught b 1 plus a 1 b naught these two terms
are there. So, let us compute them through this table a 1 is 1, b naught is 1, so this
will retain 1. a naught is 1; b 1 is 2, so it will give you 2. So, together it is 1 plus
2 which is equal to 3. Let us go for a second order term that is
the monomial with degree 2. So, in this case, j will run from 0 to 2. So, I will have a
naught b 2, a 1 b 1, a 2 b naught, a naught b 2, a 1 b 1, a 2 b naught, this is correct.
Just go ahead and compute these terms, a naught is 1 b 2 is 1, so you will get 1. a 1 is 1
b 1 is 2, so you will get 2. And a 2 b naught that is a 2 is 1 b naught is 1, so you will
get another 1. So, you will get the sum to be 4.
Let us go for a third term x cube term, and just simplysubstitute this. So,we need to
find all possible combinations. So, if it is a degree 3 term and we start with a naught,
it will be a naught b 3, a 1 b 2, a 2 b 1, a 3 b naught, these are the terms. And thenyou
simply compute them. Remember here now we came up with b 3.
What is what is b 3? b 3 is not listed here, that means, b 3 must be 0. In a similar mannerhere
a 3 must be 0 correct. So, these 2 terms are chopped off right away they are 0. hm So,
let us focus on the other 2 terms the first term you can easily verify because b 2 is
1, and a 1 is 1 is 1. And a 2 b 1, b 1 is 2, a 2 is 1, so it will be 2. So, 1 plus 2
3; this is correct. Now, the final term - the final term is a
degree 4 term, correct. If you do a term wise multiplication, what you will come up with
is because the degree 4 will be contributed by the highest order terms. So, you will simply
multiply x square into x square, and you will get only 1 term. But in this formulation what
we are doing here is we are taking all possible terms of degree 4. So, even though they are
0, we will first list them, and we will put them as 0s.
So, now, when we consider degree 4 term, I will get a naught b 4, a 1 b 3, a 2 b 2, a
3 b 1, and a 4 b naught. So, all these terms are here. And most of the terms will obviously,
be 0 only 1 term is a contributor. For example, a naught b 4 is 0, a 4 b 4 will be 0, a 1
b3 is 0, a 3 b 1 is 0. Why? Because b 4, b 3, a 3, a 4 all are 0 only term that will
contribute is a 2 b 2 which will be 1 into 1, so 1. So, this gives us a clear cut answer,
and this is a systematic way to multiply two polynomials.
Therefore, the resultant polynomial p x into q x simply write the terms from this table,
so this is a coefficient of x raised to 0 is 1, so the constant term 1 is here coefficient
of x raised to 1 is 3, so so 3 x is here. So, in a similar manner x square coefficient
of x square is 4. So, you will get 4 x square here ok; x cube
is 3, so 3 x cube correct. So, this is also done. And then x raised to 4 has only 1 term
as 1, so x raised to 4. Therefore, you got the resultant polynomial to be equal to this.
Now, remember one side note the multiplication of two polynomials will always fetch you a
polynomial again ok. Next operation is division which we will see
in the next video, but the division of two polynomials will not always lead to a polynomial.
We will see that in the next video. Bye for now.
Thank you.
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