Timeless Theorems of Mathematics/Polynomial Factor Theorem

The Polynomial Factor Theorem is a theorem linking factors and zeros of a polynomial.^[1] 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.^[2]

Proof

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 $(x - a)$

∴ $(x - a)$ is a factor of $P (x)$ . [Proved]

Converse of Factor Theorem

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

Factorization

Example 1

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

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

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

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

When,

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

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

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

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

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

∴ $P (x) = (2 x + 1) (3 x + 2) (3 x - 1)$

Example 2

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

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

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

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

$8 y^{2} - 8 y - 6 \equiv (4 y + 2) (2 y - 3) .. (i i)$

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 $y$ .

∴ $P (x) = (- 3 x + 4 y + 2) (x + 2 y - 3)$

References

↑ [1] Byjus.com, Maths, Factor Theorem
↑ [2] Byjus.com, Maths, Roots of Polynomials

[1] [1] Byjus.com, Maths, Factor Theorem

[2] [2] Byjus.com, Maths, Roots of Polynomials

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