 Hello friends, welcome to another session on problem-solving related to factor theorem. The question says without using the long division method prove that x square plus 2x minus 3 is a factor of fx which is x to power 4 plus 2x cube minus 2x square plus 2x minus 3. So as you can see the question clearly says that you cannot use the long division method if we could use that what we could have done is we could have divided fx by the given divisor and showed that remainder is 0 and hence we could have established that x square plus 2x minus 3 is factor of fx. Okay, so if the long division method is not allowed then what else or what other approach is left to us? So let us go by the fundamental equation we have been writing so far and that is dividend is equal to divisor into quotient plus the remainder. Now this particular relationship holds for numbers as well as polynomials we have established that in previous sessions so we can use it for our advantage to find out or establish this given fact or whatever it has been the demand of the question. Now how do we prove that something is a factor of something else right? So we have discussed that as well that if somehow you show that remainder is 0 then automatically the divisor or the quotient becomes the factor of dividend so 8 is a factor of 24 because it doesn't leave any remainder when it divides 24. So all of you know 8 threes are 24 so this is 0 remainder is 0 so in a division process if remainder is 0 then the divisor is considered to be factor of the dividend and not only with the numbers it also holds with polynomials we have seen this in the previous sessions. Now let us say the given dividend is fx clearly now fx is equal to what is the divisor? Divisor here is x square plus 2x minus 3 and let us say during this division process you get a quotient qx and a remainder rx correct you know so you can get a polynomial quotient and a polynomial remainder right now the same equation can be written as fx is equal to can I factorize this itself and why will that help we will see in the subsequent steps so I can split the middle term we have learned how to factorize such expressions so 3x minus x minus 3 times whatever qx is I don't know let it be like that correct now fx will be equal to if you factorize it further x common x plus x plus 3 and minus x plus 3 correct into qx plus rx okay so again this can be written as fx is equal to x plus 3 times x minus 1 times qx plus rx correct now this is a very interesting step why because this itself can establish whether the given divisor that is x square plus 2x minus 3 is a factor of fx now let it will be advisable now to write the full fx as well so if you see this is x to the power 4 plus 2x cubed minus 2x squared plus 2x minus 3 this was fx and this is equal to x plus 3 and x minus 1 qx plus rx now we have to find out let us what basically we have to do is if we somehow again prove by any means that rx is 0 then clearly the given divisor would be factor of fx is it so we have done nothing we have expressed fx like this we have expressed divisor like that and this is the qx which you will obtain and this is the remainder so if somehow we prove that rx is 0 come what may then our job is done now if you see I can I can make this entire term 0 by two values of x and you could have figure it by figure it by now that if you put x is equal to minus 3 or x is equal to 1 then you can clearly see x plus 3 and x minus 1 will become 0 isn't it so how does it help let's see so let us first take x equals to minus 3 okay so if I take x equals to minus 3 so deploy x as minus 3 so minus 3 to the power 4 plus 2 minus 3 cubed minus 2 minus 3 whole squared plus 2 minus 3 and then minus 3 right minus 3 to the power 4 plus 2 times minus 3 cubed minus 2 times minus 3 squared plus 2 times minus 3 minus 3 this is my fx and clearly rhs will become 0 into x minus 1 into qx into rx so don't you think that we are getting rx without division c this entire item becomes 0 so rx is whatever is left here isn't it now if somehow if it is it comes out to be 0 then we'll get rx as 0 and the moment remainder is 0 we can prove that the given factor is or the given divisor is factor of the polynomial fx so let's check the lhs what is this guys this is clearly 81 then this one be minus 54 is it it so 81 and minus 54 please be very very careful with the um solution and then if you see this is minus 2 times 9 and minus 6 and minus 3 right and this is equal to rx and if you calculate 18 and 54 is 72 and 9 is 81 so this is 0 right rx is coming out to be 0 so clearly remainder is equal to 0 correct remainder is 0 when x is equal to minus 3 or or that means by factor theorem what can we say we can say x plus 3 is a factor of fx correct x plus 3 is a factor of fx isn't it right similarly similarly similarly add x is equal to 1 what will happen let's deploy 1 instead of minus 3 now so 1 to the power 4 plus 2 to the power 2 into 1 to the power 3 minus 2 1 square plus 2 times 1 minus 3 is equal to again this item will become 0 in this case correct in the second case so it will become 0 into qx plus rx okay so if you see this is 1 plus 2 minus 2 plus 2 minus 3 which is 0 and this is equal to rx again remainder comes out to be 0 right so by factor theorem by factor theorem x minus 1 is a factor of fx right now since x minus 1 and x plus 3 are both factors of fx therefore their product also will be a factor of so x minus 1 and x plus 3 will be a factor of fx isn't it let's take an example just to make you understand 3 divides 48 correct and 4 divides 48 correct right 3 divides 48 and 4 divides 48 so their product that is 12 is also factor of 48 12 divides 48 right so hence if one is a factor of fx another one is a factor of fx their product will also be a factor of fx so this will be x square plus 2x minus 3 is a factor of fx hence proved right so this is how if there is a so in this case there was a quadratic divisor so hence you took out the independent or individual factors of that quadratic divisor and prove that each one of the linear factors of the divisor itself is a factor of polynomial right so you express this as x minus 1 and x plus 3 and you used factor theorem twice to prove that x minus 1 and x plus 3 both are factors of fx hence their product will also be a factor of fx that's how you have to solve these kind of problems