For a long time I wanted to write about ** remainder theorem**, as most of my students used to ask me frequently about this and I thought probably you also want to read about this topic. So, here it is…

Remainder theorem

As I’ve have discussed about Polynomial in my earlier post and also about Long Division method to solve any fractions of polynomials, Now this time is to explain the most favourable method to find remainders of fractions.

*With the help of long division method remainder and quotient can be evaluated but this turn to be lengthy method, while we are only tend to find, whether any fractions are divisible or not.* So, when someone is looking to find whether the Polynomial is perfectly divisible by another polynomial or not i.e.

*remainder theorem comes in existence to find remainders direct without any hassle and struggle with only in*__two steps__.### • So here is step by step method of remainder theorem:-

Let any quadratic, cubic or any Polynomial P(x) is divided by another polynomial say g (x) then remainder can be evaluated by following method of two easy steps

Say P(x) = x³ + 3x² – 5x +6

& g(x) = x + 1

**STEP 1:**by g(x) = 0,

x+1= 0

x = -1

**STEP 2:**P (-1) = (-1)³ + 3 (-1)² – 5(-1) + 6

= -1 + 3 ×1 + 5 + 6

= -1 + 3 + 11

P (-1) = 13

hence, If P(x) = x³ + 3x² – 5x +6 when divided by g(x) = x + 1 left with remainder 13.

the same remainder can be obtained from long Division method and can verify this.

let’s have an another example:

P(x) = x³ + 3x² + 3x + 1

& g(x) = 5+2x

**STEP 1:**5+2x = 0

2x = -5

x = -5/2

now,

**STEP 2:**

**P(**-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1

= -125/8 + 75/4 – 15/2 + 1

= -27/8

hence, I hope you’ve got to know about remainder theorem and it’s most utility in the field of algebra.

**I personally consider this method is a shortcut to verify whether any Solution of Long Division is correct or not by observing remainders.**

*now over to you, what you have found about this tutorial special and what else you want to add here do let me know by commenting.*