Timeless Theorems of Mathematics/Polynomial Factor Theorem

The Polynomial Factor Theorem is a theorem linking factors and zeros of a polynomial. It is an application of the Polynomial Remainder Theorem. It states that a polynomial $$f(x)$$ has a factor $$(x - a)$$ if and only if $$f(a) = 0$$. Here, $$a$$ is also called the root of the polynomial.

Statement
If $$P(x)$$ is a polynomial of a positive degree and if $$P(a) = 0$$ so $$(x - a)$$ is a factor of $$P(x)$$.

Proof
According to the Polynomial Remainder Theorem, the remainder of the division of $$P(x)$$ by $$(x - a)$$ is equal to $$P(a)$$. As $$P(a) = 0$$, so the polynomial $$P(a)$$ is divisible by $$(x - a)$$

&there4; $$(x - a)$$ is a factor of $$P(x)$$. [Proved]

Converse of Factor Theorem
Proposition :  If $$(x - a)$$ is a factor of the polynomial $$P(x),$$ then $$P(a) = 0$$

Example 1
Problem : Resolve the polynomial $$P(x) = 18x^3 + 15x^2 - x - 2$$ into factors.

Solution : Here, the constant term of $$P(x)$$ is $$-2$$ and the set of the factors of $$-2$$ is $$F$$1$$ = $${&plusmn;1, &plusmn;2}

Here, the leading coefficient of $$P(x)$$ is $$18$$ and the set of the factors of $$18$$ is $$F$$2$$ = $${&#xB1;1, &#xB1;2, &#xB1;3, &#xB1;6, &#xB1;9, &#xB1;18}

Now consider $$P(a)$$, where $$a = \frac {r}{s}, r\isin F$$1$$, s\isin F$$2

When,

$$a = 1, P(1) = 18+15-1-2 \ne 0$$

$$a = -1, P(-1) = -18+15+1-2 \ne 0$$

$$a = -\frac{1}{2}, P(-\frac{1}{2}) = 18(-\frac{1}{8})+15(-\frac{1}{4})+\frac{1}{2}-2 = 0$$

Therefore, $$(x + \frac{1}{2}) = \frac{1}{2}(2x+1)$$ is a factor of $$P(x)$$

Now, $$P(x) = 18x^3 + 15x^2 - x - 2$$ $$= 18x^3 + 9x^2 + 6x^2 + 3x - 4x - 2$$ $$= 9x^2(2x + 1) + 3x(2x + 1) - 2(2x + 1)$$ $$= (2x + 1)(9x^2 + 3x - 2)$$ $$= (2x + 1)(9x^2 + 6x - 3x - 2)$$ $$= (2x + 1)(3x(3x + 2) - 1(3x + 2))$$ $$= (2x + 1)(3x + 2)(3x - 1)$$

&there4;$$P(x) = (2x + 1)(3x + 2)(3x - 1)$$

Example 2
Problem : Resolve the polynomial $$P(x) = -3x^2 -2xy + 8y^2 + 11x - 8y - 6$$ into factors.

Solution : Considering only the terms of $$x$$ and constant, we get $$-3x^2+ 11x - 6$$.

$$-3x^2+ 11x - 6 \equiv (-3x+2)(x-3)..(i)$$

In the same way, considering only the terms of $$y$$ and constant, we get $$8y^2 - 8y - 6$$.

$$8y^2 - 8y - 6 \equiv (4y+2)(2y-3)..(ii)$$

Combining factors of above (i) and (ii), the factors of the given polynomial can be found. But the constants $$+2, -3$$ must remain same in both equations just like the coefficients of $$x$$ and $$y$$.

&there4;$$P(x) = (-3x+4y+2)(x+2y-3)$$