Division of Polynomial by Linear Factor

Division of Polynomial by Linear Factor

Go back to  'Polynomials'

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?

Download Polynomials Worksheets
Polynomials
Grade 9 | Questions Set 1
Polynomials
Grade 9 | Answers Set 1
Polynomials
Grade 10 | Questions Set 1
Polynomials
Grade 10 | Answers Set 1
  
More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus