Find the remainder – Vedic Algebra

Vedic Algebra

I have got a mail from some QuickerMaths follower, to illustrate usage of Vedic Mathematics in branches of mathematics other than arithmetic. This post is for that purpose only. Here I am highlighting the usage of Vedic Mathematics in finding out the remainder when an algebraic expression is divided by another.

Finding out the remainder becomes extremely easy using Vedic Maths.

So lets begin with a simple example –


Find the remainder when

x3 + 4x2 + 6x – 7 is divided by (x + 5)

Solution:

Step I: Put divisor equal to 0 .i.e.

x + 5 = 0

x = -5

Step II: The remainder will be f(x).

f (-5) = (-5)3 + 4(-5)2 + 6(-5) – 7

= -125 + 100 – 30 – 7

= -62

Example 2: (Mx3 + 3x2 -3) and (2x3 – 5x +M) leaves the same remainder when divided by (x -4) Find the value of M.

Solution:

Let R1 and R2 be remainder for 1st and 2nd equation simultaneously

Rl = f (4) = M (4)3 + 3(4)2 – 3 = 64M+ 45

R2 = f (4) = 2(4)3 - 5(4) + M = M + 108

They leave the same remainder. So,

Since, Rl = R2. We have

64 M + 45= M +108

Or, 63 M = 63

M = 1

Example 4: (Mx3 + x2 – 2x – N) is exactly divisible by (x – 1) and (x + 1). Find the value of M and N.

Soln: When the expression is exactly divisible by any divisor, the remainder will be zero.

Now, the remainder, when the divisor is x-1, is

f (l) = M + 1 – 2 – N = 0          .

\M – N = 1                 ………….(1)

And the remainder, when the divisor is x + 1, is

f( -1) = M( -1)3 + (-1)2 – 2(-1) – N = 0

-M + 1 + 2 – N = 0

M + N = 3                     …..(2)

Solving (1) & (2), we have,

M = 2, N = 1

Thanks to Nehul from Nagpur for asking this question. If you have any similar question, go to Contact Page and post your queries/suggestions.

Take Care. God Bless!!

Vineet Patawari

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