9th class Maths _Polynomials_Remainder theorm

1. 9th Class Maths Polynomials-Remainder Theorem Dear students, we all are well versed with division process and know how we can divide a number with another number. We are now going to discuss an important topic for class 9Remainder Theorem. Study Online Classes available on Takshila Learning for better understanding of this topic. While learning division, we come across different terms used in the division Dividend- a number to be divided by another number. Divisor- a number by which another number is to be divided Remainder-a number that is left over Quotient- a result obtained by dividing one quantity by another. So, we know if we will divide 21 by 5 we will get quotient 4 and remainder as 1 21=(4*5)=1 If we will divide 50 by 10 we will get 5 as quotient and 0 as a remainder So now the question striking to class 9 students is can we divide one polynomial by another? Let’s say (3x²+x+1)÷x So now how we will solve this

2. (3x²÷x)+(x÷x)+(1÷x) {x(3x+1)}+1 In this case (3x+1) is the quotient and 1 is the remainder.Since the remainder is not 0, hence it is not the factor. Study more about this topic with online classes for 9th Class Maths. Remainder Theorem: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x-a then the remainder is p(a). Now, how to proof the above saying. so moving ahead as per the syllabus of class 9 we will now proof the above theorem. Proof: Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by(x-a), the quotient is q(x) and the remainder is r(x) P(x) =(x-a) (q(x) +r(x) Since the degree of (x-a) is 1 and the degree of r(x) is less than the degree of x-a, the degree of r(x) =0.This means that r(x) is constant say r. For every value of x, r(x) =r Therefore p(x) =(x-a) q(x) +r If x=a P (a) = (a-a) q (a) +r =r Hence proved the above theorem. Watch Animated video for better understanding of Remainder theorem of class 9th Maths, click here. NCERT Maths solutions class 9 Exercise 2.3 Q1-Find the remainder when x³+3x²+3x+1 is divided by A1 (i) x+1

3. P(x) =x³+3x²+3x+1 Q(x) =x+1 0=x+1 X=-1 P (-1) = (-1)³+3((-1)²+3(-1) +1 =-1+3-3+1 =2-2 =0 (ii)x-? ? Following the first steps as in first part X=? ? P(x) =x³+3x²+3x+1 = (1/2)³+3(1/2)²+3(1/2) +1 =???????? ? =?? ? (iii) X X=0 P (0) =0³+3(0²) +3(0) +1 =1 (iv) X+∏ X=-∏ P (∏) =-∏³+3(-∏²) +3(-∏) +1 --∏³+3∏²+-3∏+1 (V) 5+2x

4. X=-? ? ?+3(-5/2)² + 3(−5/2) + 1 P (-? ?) =−? ? = (-125/8) + (75/4)-(15/2) +1 = (25/8)-(15/2) + (2/2) = (25/8)-(13/2) = (25-52)/8 = (-27/8) So, with this we can conclude that polynomial can be divided by another polynomial using the remainder theorem easily. For more NCERT solutions class 9 mathsand practicing CBSE question paper for Class 9 to score good marks. Kindly visit Or Call us : 8800999280/8800999284/011-45639131