# Division of Polynomial by Linear Factor

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Consider the following relation involving polynomials:

$$\left( {{x^3} + x + 1} \right) = \left( {x + 1} \right)\left( {{x^2} - x + 2} \right) + \left( { - 1} \right)$$

Let us denote the different polynomials involved in this relation as follows:

$$a\left( x \right):{x^3} + x + 1$$

$$b\left( x \right):x + 1$$

$$q\left( x \right):{x^2} - x$$

$$r: - 1$$

In terms of these, we can write the polynomial relation as follows:

$a\left( x \right) = b\left( x \right)q\left( x \right) + r$

We can look at this in the following manner: if we divide the polynomial $$a\left( x \right)$$ by the linear polynomial $$b\left( x \right)$$, the quotient polynomial is $$q\left( x \right)$$, while the remainder (which is a constant) is r. The polynomial $$a\left( x \right)$$ is the dividend, while the polynomial $$b\left( x \right)$$ is the divisor. Let’s look at another example:

$\left( {4{x^4} - {x^2} + 1} \right) = \left( {2x + 1} \right)\left( {2{x^3} - {x^2}} \right) + \left( 1 \right)$

The dividend in this case is $$a\left( x \right):4{x^4} - {x^2} + 1$$ while we are taking the divisor to be $$b\left( x \right):2x + 1$$. Thus, the quotient is $$q\left( x \right):2{x^3} - {x^2}$$ and the remainder is $$r:1$$.

We make the observation that whenever a polynomial is divided by a linear divisor, the remainder will always be a constant. Can you see why?