He **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. A polynomial is said to be divisible among others when its remainder or remainder is zero.

A polynomial is an expression of the form:

P(x) = an xn + an-1 xn-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 an, an-1 , …… + a1 , a0 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 in the form:

(x–ri)

Where ri is the ith root of P(x):

P(x) = an (x – r1) (x – r2) …..(x – rn)

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

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**examples**

**– Example 1**

Consider for example the polynomial:

P(x) = 3⋅x2 – 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 not. So we 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 a root of P(x), so according to the factor theorem, the binomial (x – 2) is indeed a factor of P(x).

Let’s move on to direct verification by performing the division. The detail of how the division is made 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⋅x2 – 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) times the quotient (3⋅x – 1) plus the remainder (zero in this case):

(3⋅x2 – 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 the original polynomial as a product of polynomials:

(3⋅x2 – 7⋅x + 2) = (x -2)(3⋅x – 1)

### – Example 2

Let Q(x) = x3 – x + 2 be the polynomial. 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, we simply have to verify 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) it’s not divisible by (x + 1), since Q(-1) ≠ 0.

Now we will proceed to divide Q(x) by the binomial (x + 1) as a verification method 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 the highest degree to zero degree all the coefficients of the polynomial, including the missing ones, since these have a 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 since it is placed in the first column with a changed sign, that is, -1.

The following figure illustrates how synthetic division is performed:

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 by the factor theorem. Note also that when substituting x = -1 in Q(x) what is obtained is precisely the residue or remainder of the division of polynomials, since Q(-1) = residue = 2.

Of course, the division provides the additional information of the quotient C(x) = x2 – 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 are left with the expansion of the polynomial Q(x) as follows:

Q(x) = (x + 1) (x2 – 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.

## Exercises

### – Exercise 1

Find the factors of the polynomial

P(x) = x3 – 5 x2 + 2x + 8

And also write its factorization.

**Solution**

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

Since 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⋅ 12 + 2⋅1 + 8 = 1 – 5 + 2 + 8 = 6 which is different from 0, so +1 is not a root.

We explore -1:

P(-1) = (-1)3 – 5⋅ (-1)2 + 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 try the following which is +2:

P(+2) = (+2)3 – 5⋅ (+2)2 + 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 see if it cancels the polynomial:

P(+4) = (+4)3 – 5⋅ (+4)2 + 2⋅(+4) + 8 = 64 – 80 + 8 + 8 = 0.

That is to say that +4 is the root of P(x) and therefore the binomial ( x – 4 ) is another of its factors.

We no longer have to keep looking, because it is a polynomial of degree 3 that has three roots at most. In this exercise all the roots turned out to be real and integer.

Therefore the polynomial P(x) factors like this:

P(x) = x3 – 5 x2 + 2 x + 8 = ( x + 1 ) ( x – 2 ) ( x – 4 ).

### – Exercise 2

Let be the polynomial 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 nullifies the polynomial then (x – (-2)) is a factor of that polynomial.

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

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

Now the value of p is solved so that the equality to zero is fulfilled:

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

Baldor Aurelius. Algebra. Homeland Publishing Group.

Demana, W. Precalculus: Graphical, Numerical, Algebraic 7th Ed. Pearson Education.

Jiménez, R. 2008. Algebra. Prentice Hall. Stewart, J. 2006. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning. Zill, D. 1984. Algebra and Trigonometry. McGraw Hill.