When a polynomial is divided by a divisor and the remainder is zero, this means that the divisor is a factor.
Therefore, from the Remainder Theorem, we arrive at the Factor Theorem.
The Factor Theorem states:
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If f(a) = 0 then (x − a) is a factor of the polynomial f(x)
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Example
Show that (x − 3) is a factor of f(x) = x3 − 2x2 − 5x + 6
Using the Factor Theorem
f(3) = 33 − 2.32 − 5.3 + 6
= 27 − 18 − 15 + 6
= 0
Therefore (x − 3) is a factor of f(x) = x3 − 2x2 − 5x + 6
Factorising using the Factor Theorem
Quadratic expressions can often be factorised by inspection. It is more difficult with cubic expressions. The Factor Theorem makes the factorising of cubics and higher order polynomials a little easier.
Example
Factorise f(x) = x3 − 2x2 − 5x + 6
The factors of 6 are {-1, -2, -3, -6, 1, 2, 3, 6}
So select a couple of these to try with the Factor Theorem.
Try 2 and -2 first:
f(2) = (2)3 − 2(2)2 − 5.2 + 6 = -4
so (x − 2) is not a factor.
f(-2) = (-2)3 − 2(-2)2 − 5.-2 + 6 = 0
so (x + 2) is a factor.
If (x+2) is a factor, we should now try 3 and -3. Because 3 x 2 = 6
f(3) = (3)3 − 2(3)2 − 5.3 + 6 =0
so (x − 3) is a factor.
So far (x + 2) and (x − 3) are factors so, by inspection the final factor must be (x − 1).
Because 2 x -3 x -1 = 6
Therefore f(x) = (x + 2)(x − 3)(x − 1)
(Alternatively once a factor is found it can be divided out using long division and the resulting quotient factorised.)
