Remainder Theorem Mathematics Notes | EduRev

Algebra for IIT JAM Mathematics

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Mathematics : Remainder Theorem Mathematics Notes | EduRev

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Introduction

Remainder Theorem Mathematics Notes | EduRev

Divide 15 by 6. What answer do you get? By using the simple division process, we find that the quotient is 2 and the remainder is 3. Hence, we write 15 = (2 x 6) + 3. Note: The remainder ‘3’ is less than the divisor ‘6’. On the other hand, when we divide 12 by 6, we get a quotient of 2 and remainder 0. In this case, we say that 6 is a factor of 12 OR 12 is a multiple of 6.

Dividing One Polynomial by Another
Let’s start by dividing a polynomial with a monomial as follows:
Dividend Polynomial: 2x3 + x2 + x
Divisor Monomial: x
We have, (2x3 + x2 + x) / x = (2x3)/x + x2/x + x/x = 2x2 + x + 1
Observe that ‘x’ is common to each term of the dividend
polynomial. We can also write the dividend as, x(2x2 + x + 1). Hence, ‘x’ and ‘2x2 + x + 1’ are factors of ‘2x3 + x2 + x’.

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Let’s look at another example.
Dividend polynomial: 3x2 + x + 1
Divisor Monomial: x
We have, (3x2 + x + 1) / x = (3x2)/x + x/x + 1/x = 3x + 1 + 1/x
Here, 1 is not divisible by x. So, we stop the division here and note that 1 is the remainder. Hence, we have 3x2 + x + 1 = {x(3x + 1)} + 1. So, the result of the division is,
Quotient: 3x + 1
Remainder: 1
Note that since the remainder is not zero, x is not a factor of 3x2 + x + 1.
Dividing a Polynomial by Any Non-zero Polynomial
Let’s do the following division: Dividend: x + 3x2 – 1 and Divisor: 1 + x
Step 1. We arrange the terms on the descending order of their degrees. Hence, we have dividend: 3x2 + x – 1 and divisor: x + 1
Step 2. Divide the first term of the dividend by the first term of the divisor: 3x2/x = 3x. This is the first term of the quotient.
Step 3. Multiply the divisor by the first term of the quotient and subtract this product from the dividend,
{3x2 + x – 1} – {3x(x + 1)}
= {3x2 + x – 1} – {3x2 + 3x}
= – 2x – 1
‘– 2x – 1’ is the remainder.
Step 4. Now, the new dividend is ‘– 2x – 1’ and the divisor is still the same. Repeat step 2 to get the next term of the quotient. Divide the first term of the new dividend by the first term of the divisor: (– 2x)/x = – 2 = the second term of the quotient
Step 5. Multiply the divisor by the second term of the quotient and subtract this product from the new dividend,
(– 2x – 1) – {– 2(x + 1)} = (– 2x – 1) – (– 2x – 2)
= – 2x – 1 + 2x + 2 = 1 = Remainder
Remember: This process continues until the degree of the new dividend is less than the degree of the divisor.
Step 6. Hence, we have quotient: 3x – 2 and remainder: 1. It is important to note here that, 3x2 + x – 1 = (x + 1) (3x – 2) + 1 or Dividend = (Divisor × Quotient) + Remainder
Therefore, we can conclude that, if p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that:
p(x) = g(x)q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and r(x) as remainder.
Next, let’s try to find a link between the remainder and the dividend. Let’s find the value of the dividend polynomial at x
= -1
p(x) = 3x2 + x – 1
p(-1) = 3(- 1)2 + (- 1) – 1
= 3 – 1 – 1
= 1 … which is the remainder!
So, the remainder of (3×2 + x – 1) / (x + 1) = Value of (3×2 + x – 1) at x = – 1 (or the zero of the divisor [x + 1]). In other words, the remainder obtained on dividing a polynomial by another is the same as the value of the dividend polynomial at the zero of the divisor polynomial. This brings us to the first theorem of this article.

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).
Proof: p(x) is a polynomial with degree greater than or equal to one. It is divided by a polynomial (x – a), where ‘a’ is a real number. Let’s assume that the quotient is q(x) and the remainder is r(x). So, we can write,
p(x) = (x – a)q(x) + r(x)
Now, the degree of (x – a) is 1. Also, 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. In other words, r(x) is a constant. Let’s call the constant ‘r’. Hence, for all values of ‘x’, r(x) = r. Therefore,
p(x) = (x – a) q(x) + r
Now, let’s find p(a) or the value of p(x) at x = a.
p(a) = (a – a) q(a) + r
= (0)q(a) + r
= r
We see that when a polynomial p(x) of a degree greater than or equal to one is divided by a 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.
For example, check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is a multiple of 2t+1.
Solution: q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t) with remainder zero. Let’s find the zero of the divisor polynomial:
2t + 1 = 0 Or  t = – ½
Next, let’s find the value of q(t) at t= – ½
q(- ½) = 4(- ½)3 + 4(- ½)2 – (- ½) – 1
= 4(- 1/8) + 4(- ¼) + ½ – 1
= – ½ + 1 + ½ – 1 = 0.
Hence, we can conclude that the remainder obtained on
dividing q(t) by 2t + 1 is 0. And, (2t + 1) is a factor of ‘4t3 +
4t2 – t – 1’.

Solved Examples for You
Question: Find the remainder when x+ x3 – 2x2 + x + 1 is divided by (x – 1).
Solution: Dividend Polynomial = p(x) = x4 + x3 – 2x2 + x + 1 and Divisor Polynomial = x – 1
Zero of the divisor polynomial is x – 1 = 0 or, x = 1.
Therefore, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1 = 1 + 1 – 2 + 1 + 1 = 2.
So, by Remainder Theorem, the remainder is 2.
Question: Find the remainder when (x3 – ax2 + 6x – a) is divided by (x – a).
Solution: Dividend Polynomial = p(x) = x3 – ax+ 6x – a and Divisor Polynomial = x – a
Zero of the divisor polynomial is x – a = 0 or, x = a.
Therefore, p(a) = (a)3 – a(a)2 + 6a – a = a3 – a3 + 6a – a = 5a
So, by Remainder Theorem, the remainder is 5a.

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