# How to solve algebraic equations by long division method?

To Do :

We have to explain , how to solve algebraic equations by long division method.

Solution :

For example,

1. Divide the polynomial $6-t+$$t^5$ by the polynomial $t^2$$+5$.

Write both the dividend and the divisor in descending order of the powers of the variables.

Therefore, the dividend is $t^5$$-t+6$ and the divisor is $t^2$$+5$.

To make the division process simpler, represent the dividend as $t^5+0\times t^4+0\times t^3+0\times t^2-t+6$ because zero multiplied by anything will result in zero, and thus will not change the value of the expression.

The first step is to divide the first term of the dividend by the first term of the divisor. So, divide $t^5$ by $t^2$ to get $t^3$

The second step is to multiply the divisor by this $t^3$, that is, $t^3$ forms the first term of the quotient. Subtract the result, $t^5+0\times t^4+5\times t^3$gives the remainder as $-5t^3+0t^2-t+6$

The degree of the remainder is still greater than the degree of the divisor. Perform the above process till the degree of the remainder is less than the degree of the divisor.

$t^2+0t+5$ ) $\ t^5+0\times t^4+0\times t^3+0\times t^2-t+6$ ($t^3$                                  $\frac{t^5}{t^2}=t^3$

$t^5+0\times t^4+5t^3$

------------------

$\ t^2+0t+5$$\times t^2-t+6$

$\frac{-5t^3}{t^2}=-5t$

$-5t^3+0\times t^2-25t$

--------------------

$24t+6$

The degree of the new remainder 24t+6 is less than the degree of the divisor. So, stop the process here.

When the polynomial $6-t+t^5$ is divided by the polynomial $t^2+5$ , the quotient is  $t^3-5t$ and the remainder is 24t+6.