# 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 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 $P(x) = 18x^3 + 15x^2 - x - 2$ into
factors.

**Solution :** Here, the constant term of $P(x)$ is $-2$ and the set of
the factors of $-2$ is $F$**~1~**$=${±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
$a = \frac {r}{s}, r\isin F$**~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, $P(x) = 18x^3 + 15x^2 - x - 2$
$= 18x^3 + 9x^2 + 6x^2 + 3x - 4x - 2$
$= 9x^2(2x + 1) + 3x(2x + 1) - 2(2x + 1)$ $= (2x + 1)(9x^2 + 3x - 2)$
$= (2x + 1)(9x^2 + 6x - 3x - 2)$ $= (2x + 1)(3x(3x + 2) - 1(3x + 2))$
$= (2x + 1)(3x + 2)(3x - 1)$

∴$P(x) = (2x + 1)(3x + 2)(3x - 1)$

### 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 $y$.

∴$P(x) = (-3x+4y+2)(x+2y-3)$

## References


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

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