Factors of a Polynomial

Factors of a Polynomial

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Observe the following:

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

We have split the polynomial on the left side into a product of two linear factors. In other words, we have factorized the polynomial. Here is another example of factorization:

\[{x^3} - 6{x^2} + 11x - 6 = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]

Can we always split a polynomial into a product of (simpler) factors? Note that at our level, we are only dealing with polynomials which have real numbers as coefficients – complex coefficients are outside the scope of on discussion. Given this constraint, we cannot always split a particular polynomial into simpler factors. For example, \(p\left( x \right):{x^2} + x + 1\) has no (two) linear factors (try it!).

Now, suppose that a polynomial \(p\left( x \right)\) has a linear factor, \(q\left( x \right):x - a\). The Remainder Theorem tells us that if we substitute x equal to a in \(p\left( x \right)\), we will obtain 0, that is, \(p\left( a \right) = 0\). This should be obvious to you by now. If \(q\left( x \right)\) is a factor of \(p\left( x \right)\), then we have

\[p\left( x \right) = \left( {x - a} \right)r\left( x \right)\]

where \(r\left( x \right)\) is a polynomial of degree one less than the degree of \(p\left( x \right)\). Now, substituting x equal to a on both sides, we immediately have \(p\left( a \right) = 0\).

A polynomial of degree n can have at the most n linear factors, but as we have already seen, it may not always have n linear factors. Consider

\[p\left( x \right):{x^4} - 1\]

When we factorize this, we have

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

This has three factors, two of which are linear and one is a quadratic. The quadratic factor \({x^2} + 1\) cannot be factorized further into (two) linear factors as long as we are restricted to considering real polynomials.

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