Remainder Theorem is used to calculate the remainder when a polynomial is divided by another linear polynomial. It is an application of the Euclidean Division of Polynomials. It states that “If p(x) is any polynomial of degree greater than or equal to one and is divided by the linear polynomial (x – a) where ‘a’ is any real number, then the remainder is p(a).”
What is Remainder Theorem?
Remainder Theorem states that “If p(x) is any polynomial of degree greater than or equal to one and is divided by the linear polynomial (x – a) where ‘a’ is any real number, then the remainder is p(a).”
Remainder Theorem: Formula
Remainder Theorem formula is given by the expression: p(x) = (x-a) q(x) + r(x)
When p(x) is divided by (x-a), r = p(a),
and when p(x) is divided by (ax+b), r = p(-b/a)
The video below explains this:
Polynomials Detailed Video Explanation:
Remainder Theorem: Proof
Let us look at the statement to prove the Remainder Theorem.
We have a polynomial p(x) with a degree greater than or equal to 1.
Let us say, it is divided by another polynomial (x-a), where ‘a’ is a real number.
Assuming that q(x) and r(x) are quotient and remainder, we can write
p(x) = (x-a)q(x) + r(x)
Degree of the divisor (x-a) is 1. Since r(x) is the remainder, its degree is less than the degree of the divisor (x – a). Therefore, the degree of r(x) = 0. This means r(x) is a constant.
Therefore,
p(x) = (x-a)q(x) + r
The value of p(x) at x=a can be calculated as follows-
p(a) = (a – a) q(a) + r
= (0)q(a) + r
= r
With this we conclude that when a polynomial p(x) of a degree greater than or equal to one is divided by another linear polynomial (x – a), where a is a real number, then the remainder is r which is also equal to p(a). This proves the Remainder Theorem.
Example:
Ques. Check if p(x) = x4 + x3 – 2x2 + x + 1 is a multiple of (x – 1).
Ans. Zero of divisor (x-1) is 1
Value of p(x) at x=1 is
p(1) = 1+1-2+1+1= 2
This means 2 is the remainder when p(x) is divided by (x-1).
Hence, p(x) is not a multiple of (x-1).
Important Topics for JEE MainAs per JEE Main 2024 Session 1, important topics included in the chapter remainder theorem are as follows:
Some memory based important questions asked in JEE Main 2024 Session 1 include:
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Points to Remember
- Remainder Theorem states that “If p(x) is any polynomial of degree greater than or equal to one and is divided by the linear polynomial (x – a) where ‘a’ is any real number, then the remainder is p(a).”
- Remainder Theorem formula is given by the expression: p(x) = (x-a) q(x) + r(x)
Sample Questions
Ques 1. State and prove the Remainder Theorem. (3m)
Ans. Remainder Theorem states that “If p(x) is any polynomial of degree greater than or equal to one and is divided by the linear polynomial (x – a) where ‘a’ is any real number, then the remainder is p(a).”
Proof:
We have a polynomial p(x) with a degree greater than or equal to 1.
Let us say, it is divided by another polynomial (x-a), where ‘a’ is a real number.
Assuming that q(x) and r(x) are quotient and remainder, we can write
p(x) = (x-a)q(x) + r(x)
Degree of the divisor (x-a) is 1. Since r(x) is the remainder, its degree is less than the degree of the divisor (x – a). Therefore, the degree of r(x) = 0. This means r(x) is a constant.
Therefore,
p(x) = (x-a)q(x) + r
The value of p(x) at x=a can be calculated as follows-
p(a) = (a – a) q(a) + r
= (0)q(a) + r
= r
With this we conclude that when a polynomial p(x) of a degree greater than or equal to one is divided by another linear polynomial (x – a), where a is a real number, then the remainder is r which is also equal to p(a). This proves the Remainder Theorem.
Ques 2. Find the remainder when g(x) = x4 – x3 + x2 – 2x + 1 is divided by x – 2. (1m)
Ans. We know that zero of x-2 is 2.
As stated in the Remainder Theorem, the remainder of g(x) will be g(2).
Hence, g(2) = 24-23-222(2)+1 = 17
Ques 3. Find the root of the polynomial x2 – 5x + 4 (1m)
Ans. Let f(x)=x2 – 5x + 4
By hit and trial method, we need to find x as that number which gives zero remainder.
Here, we have
f(4) = 42-5(4)+4 = 0
Hence,(x-4)is a factor of the given polynomial f(x).
Ques 4. Find the remainder when t3 – 2t2 + 4t + 5 is divided by t – 1. (2m)
Ans. According to the remainder theorem, if p(t) is any polynomial of degree greater than or equal to one and is divided by the linear polynomial (t – a) where ‘a’ is any real number, then the remainder is p(a).
Here,
p(t)= t3 – 2t2 + 4t + 5
(t-a) is (t-1)
⇒ p(1) is the remainder
⇒ p(1) = 1-2(1)+4(1)+5 = 8
8 is the remainder when t3 – 2t2 + 4t + 5 is divided by t – 1.
Ques 5. Find the r ( d ) of the polynomial d4 - 2d3 + 4d2 - 5 if it is divided by d - 2 (1m)
Ans. We have d=2.
Hence, r(d) = r(2) = 24 - 2(2)3 + 4(2)2 - 5 = 16 - 16 + 8 - 5 = 3
Ques 6. Determine that x = 1 is a root of P(x).
Ans. Assuming that x = 1 may be a root of P(x) we can say that (x – 1) may be a factor of P(x).
If we divide P(x) by (x – 1), we will get a new smaller polynomial and a remainder of zero.
Ques 7. Example: Find the root of the polynomial x2-3x -4
Ans. x2-3x -4
f(4) = 42-3(4) -4
f(4) = 16-12-4 = 0
This gives (x-4) as the factor of x2-3x -4.
Ques 8. Find the remainder (without division) when 8x2+5x + 1 is divisible by x-10
Ans. Here, f(x)= 8x2+5x + 1
By the remainder theorem, when f(x) is divided by (x-10), the remainder is f(10).
Put x=10 in f(x)
f(10)= 8(10)2+5(10)+ 1
= 800+50+1
= 851
Ques 9. Find the remainder when x3-ax2+6x-a is divisible by x-a.
Ans. Here f(x)=x3-ax2+6x-a and divisor is (x-a)
Remainder will be f(a).
Put x=a in f(x), we get
f(a)=a3-a(a)2+6a-a = 5a is the remainder
Ques 10. Find the remainder (without division) when 4x3 -3x2+2x-4 is divisible by x+2.
Ans. Here, f(x)= 4x3 -3x2+2x-4 and divisor (x+2) gives x= -2
By remainder theorem, f(-2) will be the remainder when f(x) is divided by (x+2)
Hence,
Putting x=-2 in f(x), we get,
f(-2)= 4(-2)3 -3(-2)2+2(-2)-4 = -52
Ques 11. Check whether the polynomial: f(x) = 4x3+4x2-x-1 is a multiple of 2x + 1.
Ans. f(x)= 4x3+4x2-x-1 and divisor (2x+1) gives, x=-12
Hence, remainder = f(-1/2) = 4(-1/2)3+4(-1/2)2-(-1/2)-1 = 0
We get the remainder as 0, therefore, f(x) is a multiple of (2x+1)
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