Explanation factor theorem, examples, exercises

The factor theorem states that a polynomial P (x) is divisible by a binomial of the form (x - a) if x = a is a root of P (x), that is, P (a) = 0. It is said that a polynomial is divisible between another when its residual or remainder is zero.

A polynomial is an expression of the form:

P (x) = a_n xⁿ + to_n-1 x^n-1 +… + A₁ x + a₀

Where:

-n is the degree of the polynomial, where n is the largest integer to which the independent variable x is raised,

-The values ​​to_n, to_n-1 ,… + A₁ , to₀ are the coefficients of the polynomial, which are generally real numbers, but could also be complex numbers.

A polynomial of degree n can be decomposed as the product of n binomials of the form:

(x - r_i)

Where r_i is the i-th root of P (x):

P (x) = a_n (x - r₁) (x - r_two)… (X - r_n)

Since the number of roots of a polynomial is equal to its degree.

Article index

• 1 Examples
  • 1.1 - Example 1
  • 1.2 - Example 2
• 2 Exercises
  • 2.1 - Exercise 1
  • 2.2 - Exercise 2
• 3 References

Examples

- Example 1

Let us consider by case the polynomial:

P (x) = 3⋅x^two - 7⋅x + 2

You want to know if this polynomial is divisible by the binomial (x - 2). If the factor theorem is used, then we must evaluate P (x = 2) to know if the value 2 is a root or it is not. We then proceed to evaluate the expression:

P (2) = 3⋅22 - 7⋅2 + 2 = 3⋅4 - 7⋅2 + 2 = 12 - 14 + 2 = 12 - 12 = 0.

It turns out that x = 2 is the root of P (x), so according to the factor theorem, the binomial (x - 2) is effectively a factor of P (x).

Let's move on to direct verification by dividing. The detail of how the division is carried out is shown in the following figure:

It is verified that the quotient between P (x) and (x-2) gives a polynomial of a lower degree called the quotient C (x) = 3⋅x - 1 with remainder 0.

We can summarize the result as follows:

(3⋅x^two - 7⋅x + 2) ÷ (x -2) = (3⋅x - 1) + 0

The previous expression can be written in another way, simply remembering that the dividend P (x) is equal to the product of the divisor (x -2) by the quotient (3⋅x - 1) plus the remainder (zero in this case):

(3⋅x^two - 7⋅x + 2) = (x -2) (3⋅x - 1) + 0

In this way it was possible to factor the polynomial P (x), that is, to write as a product of polynomials, the original polynomial:

(3⋅x^two - 7⋅x + 2) = (x -2) (3⋅x - 1)

- Example 2

Let the polynomial Q (x) = x³ - x + 2. We want to know if it is divisible by the binomial (x + 1).

The most direct way is to simply apply the factor theorem. In this case, you simply have to check if x = -1 cancels or not the polynomial Q (x).

We proceed by substituting:

Q (-1) = (-1)³ - (-1) + 2 = -1 + 1 + 2 = 2

The result is different from zero, therefore the factor theorem assures us that the polynomial Q (x) is not divisible by (x + 1), since Q (-1) ≠ 0.

Now we will proceed to carry out the division of Q (x) by the binomial (x + 1) as a method of verification of our conclusion.

On this occasion, the division will be carried out using the synthetic division method, which consists of placing in the first row ordered from highest degree to zero degree all the coefficients of the polynomial, including the missing ones, since they have zero coefficient.

Then in the first column the independent term of the divisor is placed but with the sign changed, in our case the divisor is (x + 1). Its independent term is 1, but as in the first column it is placed changed sign, that is, -1.

The following figure illustrates how the synthetic division is done:

With this result it is verified that (x + 1) is not a factor of the polynomial Q (x) = x³ - x + 2 since the remainder is not zero.

This conclusion is not surprising, because it had already been predicted with the factor theorem. Also note that when substituting x = -1 in Q (x) what is obtained is precisely the remainder or remainder of the division of polynomials, since Q (-1) = remainder = 2.

Of course, the division provides the additional information of the quotient C (x) = x^two - x.

Remembering that the dividend Q (x) is equal to the divisor (x + 1) by the quotient C (x) plus the remainder r = 2, we have the expansion of the polynomial Q (x) as follows:

Q (x) = (x + 1) (x^two - x) + 2 = x (x + 1) (x - 1) + 2

It should be noted that this expression is not the factorization of said polynomial, since there is a non-zero term adding, which is precisely the remainder of value 2.

Training

- Exercise 1

Find the factors of the polynomial

P (x) = x³ - 5 x^two + 2 x + 8

And also write your factorization.

Solution

The factor theorem tells us that we should look for the roots to to then find the factors (x - to), in this case as it is a polynomial of degree three, there must be three roots. 

As it is a polynomial with integer coefficients, the roots must be between the divisors of the independent term, which in this case is 8. These divisors are:

± 1, ± 2, ± 4, ± 8.

We start by exploring +1: P (+1) = 1³ - 5⋅ 1^two + 2⋅1 + 8 = 1 - 5 + 2 + 8 = 6 which is different from 0, therefore +1 is not a root.

We explore -1:

P (-1) = (-1)³ - 5⋅ (-1)^two + 2⋅ (-1) + 8 = -1 - 5 - 2 + 8 = 0

From the result it is concluded that -1 is the root of P (x) and (x - (-1)) = (x + 1) is a factor of the polynomial.

Two more factors remain to be found:

We prove the following which is +2:

P (+2) = (+2)³ - 5⋅ (+2)^two + 2⋅ (+2) + 8 = 8 + (-20) + 4 + 8 = 0

Again we get zero. So the other factor is (x - 2).

Since it is a polynomial of degree three, we only need to find a factor. Now we test the value +4 to find out if it cancels the polynomial:

P (+4) = (+4)³ - 5⋅ (+4)^two + 2⋅ (+4) + 8 = 64 - 80 + 8 + 8 = 0.

That is, +4 is the root of P (x) and therefore the binomial (x - 4) is another of its factors.

No more searching, because it is a polynomial of degree 3 that has at most three roots. In this exercise all the roots turned out to be real and integer.

Therefore the polynomial P (x) is factored like this:

P (x) = x³ - 5 x^two + 2 x + 8 = (x + 1) (x - 2) (x - 4).

- Exercise 2

Let the polynomial be p⋅x³ - x + 2p. Determine the value of p so that the polynomial is divisible by (x + 2).

Solution

We use the factor theorem, which states that if x = -2 cancels the polynomial then (x - (-2)) is a factor of said polynomial.

Then we substitute x for (-2) in the original polynomial, simplify it and set it equal to zero:

p⋅ (-2)³ - (-2) + 2p = 8p + 2 + 2p = 10p + 2 = 0

Now the value of p is cleared so that the equality to zero is satisfied:

p = -2 / 10 = -⅕ 

This means that the polynomial: 

-⅕⋅x³ - x - ⅖

It is divisible by (x + 2), or what is equivalent: (x + 2) is one of its factors.

References

1. Baldor Aurelio. Algebra. Patria Editorial Group.
2. Demana, W. Precalculus: Graphical, Numerical, Algebraic 7th Ed. Pearson Education.
3. Jiménez, R. 2008. Algebra. Prentice hall.
4. Stewart, J. 2006. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning.
5. Zill, D. 1984. Algebra and Trigonometry. Mcgraw hill.