# Factor Theorem

We are already familiar with the content of the Factor Theorem, without explicitly having encountered its name till now. This theorem says that given a polynomial $$a\left( x \right)$$, if its value at x equal to some real number k is equal to 0, that is, if $$a\left( k \right) = 0,$$then the linear polynomial $$\left( {x - k} \right)$$ is a factor of $$a\left( x \right)$$, and we can write

$a\left( x \right) = \left( {x - k} \right)b\left( x \right),$

where $$b\left( x \right)$$ is a polynomial of degree one less than $$a\left( x \right)$$. Let’s take an example. Suppose that

$a\left( x \right) = {x^3} - 6{x^2} + 11x - 6$

We make the following observation: $$a\left( 1 \right) = 0$$. Thus,

$$a\left( x \right) = \left( {x - 1} \right)b\left( x \right),$$

where $$b\left( x \right)$$ is a quadratic polynomial. We make further observations (verify them):

$$a\left( 2 \right) = 0,\,\,a\left( 3 \right) = 0$$

Thus, $$\left( {x - 2} \right),\,\,\left( {x - 3} \right)$$ will also be factors of $$a\left( x \right)$$. Since a cubic polynomial can have at the most three linear polynomial factors, we can immediately conclude that

$a\left( x \right) = \lambda \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)$

where $$\lambda$$ is some real constant. But a simple comparison of coefficients of $$a\left( x \right)$$and the right side above tells us that $$\lambda$$ must be equal to 1, and so

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

We have thus succeeded in factorizing$$a\left( x \right)$$ by inferring the factors using the Factor Theorem.

Polynomials
Polynomials
Grade 10 | Questions Set 1
Polynomials