Bezout’s theorem: finding the remainder of dividing a polynomial by a binomial

In this post, we will look at Bezu’s theorem, with which you can find the remainder of dividing a polynomial by a binomial, and also, learn to apply it in practice to solve examples.

Statement of Bezout’s theorem

Remainder of dividing a polynomial P(x) of binomial (x-a) equals P(a).

P_n(x) = a₀xⁿ + a₁x^n-1 +… + A_n-1x + a_n

Corollary from the theorem:

Number a is the root of the polynomial P(x) only if the polynomial P(x) divides without remainder into a binomial (x-a).

From this corollary follows the following assertion: set of polynomial roots P(x) is identical to the set of roots of the corresponding equation P(x)=0.

Solution examples

Example 1

Find the remainder of a polynomial 5x² – 3x + 7 of binomial (x-2).

Solution

To find the remainder of the division, according to Bezout’s theorem, it is required to find the value of the polynomial at the point a (ie instead of x substitute the value a, which in our case is equal to the number 2).

5⋅2² – 3 ⋅ 2 + 7 = 21.

That is remainder equal 21.

Example 2

Using Bezout’s theorem, find out if a polynomial is divisible 3x⁴ + 15x – 11 of binomial (x+3) without a remainder.

Solution

In this case, a = -3. We substitute this number for x into a polynomial and get:

3 ⋅ (-3)⁴ + 15 ⋅ (-3) – 11 = 187.

This means that division without a remainder is impossible.

Example 3

Find out at what value y, a polynomial x²³ + yx + 16 divides without remainder into a binomial (x+1).

Solution

Applying the Bezout theorem, we find the zero remainder of the division:

(-1)²³ + and ⋅ (-1) + 16 = 0

-1 – and + 16 = 0

y = 15

Thus, at y, flat 15, the remainder will be 0.