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.

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 $a=\frac{r}{s},r\in F$₁$,s\in F$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 $y$.

∴$P(x)=(-3x+4y+2)(x+2y-3)$