Roots and factors of a polynomial

Let P(x) be a polynomial. A number a such that P(a)=0 is called a **root **or** zero** of P.

If x=a is a root of a polynomial, then x-a is a **factor** of that polynomial.

Consider the polynomial P(x)=x^{2}+2x-3.

Let's plug x=1 into the polynomial: p(1)=1^{2}+2·1-3=0.

Consequently x=1 is a root of the polynomial P(x)=x^{2}+2x-3.

Note that x-1 is a factor of the polynomial P(x)=x^{2}+2x-3.

**The fundamental theorem of algebra** stated that a polynomial P(x) of degree n has n roots, some of which may be degenerate.

What are the possible integer roots of P(x)=2x^{5} -3x^{3} +4x^{2} -9x + 6 ?

If there are integer roots, they will be factors of the constant term 6; namely 1, -1, 2, -1, 3, -2, 6, -6.

Now, is 1 a root? To answer, we will divide the polynomial by x-1 and hope for remainder 0. Yes! 1 is a root!

Now, is 1 a root again? To answer, we divide the resulting polynomial by x-1 and hope for remainder 0. Yes! 1 is a root again!.

Now, is -2 a root? Yes! -2 is a root!

We have: 2x^{5} -3x^{3} +4x^{2} -9x + 6 =(x-1)^{2}(x+2)(2x^{2} +3)where 2x^{2} +3 does not have integer roots.

Conclude that the integer roots of P(x)=2x^{5} -3x^{3} +4x^{2} -9x + 6 are a=1, b=1 and c=-2