 In this video, we are going to look at application of remainder theorem. And this is also known as factor theorem in certain special cases. I personally love this theorem because the way it is presented, it's very elegant. It's like carving a very good statue out of a stone. So whenever you get to the result, you feel like, oh, it was right in front of me. So what exactly is remainder theorem? So first of all, let us try to understand the concept. I'll just take any number, say 16. And if I want to divide 16 by 3, this is going to be a divisor and 16 is a dividend. I can usually write 16 as 3 times 5 plus 1. And I know that this one here is a remainder. And because this is a remainder which is not 0, we say that 3 is not a factor of 16, right? This is about numbers. The same thing can be done with polynomials. Now, what if we had polynomials instead of 16 and 3? The same logic applies. There will be a divisor. Here 5 was quotient. So there will be some quotient and there will be a remainder. So let's see. So if I had a polynomial x square minus 5x plus 8, let's say this was the dividend. And if I wanted to divide this by x minus 2, I would find that I can write this as x minus 2 times x minus 3 plus 2. Don't worry about this. So right now this is my divisor. This is my quotient and this is the remainder. If I generalize this, I can write any given polynomial with a divisor of the form x minus small a and then there will be some q of x which is quotient and then there will be some remainder r of x. And if I wanted to find what is the remainder whenever I divided any given polynomial with a factor like this, I can put x equal to a and get the value r of x. How? If I put x equal to 2 here, this goes away. The divisor and quotient multiplication goes to 0 and I'm only left with 2 which is remainder and that's what the remainder theorem says. If you want to find the remainder of f of x when it is divided by x minus a, you simply have to put x equal to a and in that case I'm only left with remainder which is nothing but equal to f of a because we are putting x equal to a. So let me just summarize quickly. If x minus small a divides any polynomial f of x, then the remainder is given by f of a. How do we apply this theorem to find factors of the given polynomial? That is the question and then there comes a special case when a remainder is 0 which is when we call it a factor theorem. So when this remainder f of a becomes 0, that's a factor theorem and when remainder is 0, x minus a becomes factor of the given polynomial f of x. Usually the question is posed like this whenever we have to deal with the application of remainder theorem or the factor theorem, factorize a given polynomial for example x square minus 8x plus 12 but how do we use factor theorem? We already know if we knew certain factor x minus a to be the factor of this polynomial, then putting x equal to a in here would return 0. So we will first have to identify some x minus a factor and to do that either we could use trial and error but we also know this 12 here and this is a second degree polynomial or a quadratic expression which can be written as some x minus a times x minus b and once the constant terms multiply this gives me x square minus a plus b plus a b. So this 12 is nothing but a b. This 12 could be written as 1 times 12 or 2 times 6 or 3 times 4 etc. So we could use the small a to be among 1, 2, 3, 4, 6, 12 something out of these. So our search gets limited and now we can start putting either of these values and check whether these return 0 as a remainder. So let us just quickly name this as f of x. So what would be f of 1? f of 1 is when x equal to 1, so we have 1 square minus 8 times 1 plus 12 and this gives us 5 and so this is not a 0 remainder and that is why x minus 1 is not a factor. Now what about f of 2, let us see. So square of 2 minus 8 times 2 plus 12 and this gives me 0 and that means x minus 2 is a factor of f of x. Usually it suffices when we got at least one factor. Now we can use long division or other methods to break the expression down into its factors. So since we have confirmed that x minus 2 is a factor we can write f of x to be x minus 2 times some x minus b. If we do long division and divide the given polynomial by x minus 2 after the long division we realize that another factor is x minus 6 because we got the remainder 0 and so it is possible to write down all the factors once we find at least one factor using the factor theorem and we can use any of the available methods to us. Something that we just saw a long division and then we can write the other factor as x minus 6, sorry this b does look like 6 so we can write x minus 2 times x minus 6 and this is how we have completely factorized the given polynomial and that basically is the application of remainder theorem to find out rest of the factors. Let's see another example. Now let's say we have to factorize x cube plus 3x square plus 5x plus 3. We will look at the constant term again. So 3 can be broken down into 1 times 3 or minus 1 times minus 3. Now we can check that whether x plus 1 or x minus 1 or x plus 3 or x minus 3 are the factors of the given polynomial. Let's again name this as f of x and so we have f of x to be equal to x cube plus 3x square plus 5x plus 3. Now if x minus 1 is a factor of this then we will put x to be equal to 1 because what we are essentially doing is equating x minus 1 to be equal to 0 and then we get x equal to 1. So remember that. So putting x equal to 1 means x minus 1 is a factor that's what we are considering and let's see what is f of 1 whether do we get remainder equal to 0. So f of 1 is 1 plus 3 plus 5 plus 3 and this is equal to 12 which is not 0 and so x minus 1 is not a factor. Now let's see if x plus 1 is a factor that means we will have to see whether f of minus 1 gives us 0 remainder because x plus 1 equal to 0 and x would be equal to minus 1. So we write cube of minus 1 plus 3 times square of minus 1 plus 5 times minus 1 plus 3 and this gives us minus 1 plus 3 minus 5 plus 3 which is equal to 0 and therefore x plus 1 is a factor of f of x and so now we know one of the factors and now we can use long division to get the other factor. So to put long division in we will write x cube plus 3 x square plus 5 x plus 3 and we won't go into detail but we are going to divide this by x plus 1 let's see what do we get and so after the long division we get another factor which is quotient here as x square plus 2 x plus 3. And so one of the factors of the given polynomial is x plus 1 and the another is this quotient here and so we can write x cube plus 3 x square plus 5 x plus 3 to be equal to x plus 1 which is one of the factors times another factor that we just found x square plus 2 x plus 3 and this is how we can apply remainder theorem to factorize the given expression.