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**Introduction**

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 â€˜3â€™ is less than the divisor â€˜6â€™. 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**

Letâ€™s start by dividing a polynomial with a monomial as follows:

Dividend Polynomial: 2x^{3} + x^{2} + x

Divisor Monomial: x

We have, (2x^{3} + x^{2} + x) / x = (2x^{3})/x + x^{2}/x + x/x = 2x^{2} + x + 1

Observe that â€˜xâ€™ is common to each term of the dividend

polynomial. We can also write the dividend as, x(2x^{2} + x + 1). Hence, â€˜xâ€™ and â€˜2x^{2} + x + 1â€™ are factors of â€˜2x^{3} + x^{2} + xâ€™.

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Letâ€™s look at another example.

Dividend polynomial: 3x^{2} + x + 1

Divisor Monomial: x

We have, (3x^{2} + x + 1) / x = (3x^{2})/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 3x^{2} + 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 3x^{2} + x + 1.**Dividing a Polynomial by Any Non-zero Polynomial**

Letâ€™s do the following division: Dividend: x + 3x^{2} â€“ 1 and Divisor: 1 + x

Step 1. We arrange the terms on the descending order of their degrees. Hence, we have dividend: 3x^{2} + x â€“ 1 and divisor: x + 1

Step 2. Divide the first term of the dividend by the first term of the divisor: 3x^{2}/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,

{3x^{2} + x â€“ 1} â€“ {3x(x + 1)}

= {3x^{2} + x â€“ 1} â€“ {3x^{2} + 3x}

= â€“ 2x â€“ 1

â€˜â€“ 2x â€“ 1â€™ is the remainder.

Step 4. Now, the new dividend is â€˜â€“ 2x â€“ 1â€™ 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: (â€“ 2x)/x = â€“ 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,

(â€“ 2x â€“ 1) â€“ {â€“ 2(x + 1)} = (â€“ 2x â€“ 1) â€“ (â€“ 2x â€“ 2)

= â€“ 2x â€“ 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 â€“ 2 and remainder: 1. It is important to note here that, 3x^{2} + x â€“ 1 = (x + 1) (3x â€“ 2) + 1 or Dividend = (Divisor Ã— Quotient) + Remainder

Therefore, we can conclude that, if p(x) and g(x) are two polynomials such that degree of p(x) â‰¥ degree of g(x) and g(x) â‰ 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, letâ€™s try to find a link between the remainder and the dividend. Letâ€™s find the value of the dividend polynomial at x

= -1

p(x) = 3x^{2} + x â€“ 1

p(-1) = 3(- 1)^{2} + (- 1) â€“ 1

= 3 â€“ 1 â€“ 1

= 1 â€¦ which is the remainder!

So, the remainder of (3Ã—2 + x â€“ 1) / (x + 1) = Value of (3Ã—2 + x â€“ 1) at x = â€“ 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 â€˜aâ€™ be any real number. If p(x) is divided by the linear polynomial (x â€“ 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 â€“ a), where â€˜aâ€™ is a real number. Letâ€™s assume that the quotient is q(x) and the remainder is r(x). So, we can write,

p(x) = (x â€“ a)q(x) + r(x)

Now, the degree of (x â€“ a) is 1. Also, since r(x) is the remainder, its degree is less than the degree of the divisor: (x â€“ a). Therefore, the degree of r(x) = 0. In other words, r(x) is a constant. Letâ€™s call the constant â€˜râ€™. Hence, for all values of â€˜xâ€™, r(x) = r. Therefore,

p(x) = (x â€“ a) q(x) + r

Now, letâ€™s find p(a) or the value of p(x) at x = a.

p(a) = (a â€“ 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 â€“ 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) = 4t^{3} + 4t^{2} â€“ t â€“ 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. Letâ€™s find the zero of the divisor polynomial:

2t + 1 = 0 Or t = â€“ Â½

Next, letâ€™s find the value of q(t) at t= â€“ Â½

q(- Â½) = 4(- Â½)3 + 4(- Â½)2 â€“ (- Â½) â€“ 1

= 4(- 1/8) + 4(- Â¼) + Â½ â€“ 1

= â€“ Â½ + 1 + Â½ â€“ 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 â€˜4t^{3} +

4t^{2} â€“ t â€“ 1â€™.

**Solved Examples for You****Question: Find the remainder when x ^{4 }+ x^{3} â€“ 2x^{2} + x + 1 is divided by (x â€“ 1).**

Zero of the divisor polynomial is x â€“ 1 = 0 or, x = 1.

Therefore, p(1) = (1)

So, by Remainder Theorem, the remainder is 2.

Zero of the divisor polynomial is x â€“ a = 0 or, x = a.

Therefore, p(a) = (a)

So, by Remainder Theorem, the remainder is 5a.

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