Question

Explain Remainder theorem with example​

Answer 1

Answer:

➡️Remainder Theorem is an approach of Euclidean division of polynomials. According to this theorem, if we divide a polynomial P(x) by a factor ( x – a); that isn't essentially an element of the polynomial; you will find a smaller polynomial along with a remainder.

➡️After dividing we get the answer 2x+1, but there is a remainder of 2. Say we divide by a polynomial of degree 1 (such as "x−3") the remainder will have degree 0 (in other words a constant, like "4").

➡️Example: 2x2−5x−1 divided by x−3

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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’.

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