Factor theorem and factorization of polynomials. Exercises resolved.

Next I’m going to explain the factor theorem, which will help us factor polynomials and we’ll see that it also has a lot to do with the roots of a polynomial. Also, we’ll see how to factor polynomials.

Factor theorem

The theorem of the factor says:

If a polynomial P(x) equals zero for x=a, then the polynomial P(x) can be broken down into form factors P(x)=(x-a).Q(x)
.

What does this mean?

In a division of polynomials, we know that the dividend can be expressed as the divider by the quotient plus the rest:

factorization of polynomials exercises solved step by step

If the division is exact, i.e. if the rest is 0 and we are left:

factorization exercises solved step by step

That is, we can express the polynomial as the product of two polynomials, that is, as the product of two factors.

Let’s see it with an example.

We have the following polynomial:

factor theorem exercises resolved

We know that x=2 is a root of the polynomial, since the value of polynomial is 0 for x=2:

factorize polynomials solved exercises

Therefore, if we divide the polynomial into x-2, that is, between a binomial of the form (x-a), where a is a root of the polynomial, we would have an exact division and can express that polynomial as the product of two factors.

We divide it by (x-2):

factorization of polynomials resolved exercises

We use the Ruffini rule to perform the division:

factorization of polynomials exercises

We express the polynomial using the following formula:

factorization of polynomials

Where the dividend is:

factorization of grade 3 polynomials exercises resolved

The divisor:

factorización ejercicios resueltos

And the quotient:

factor theorem

And we have left:

how to factor polynomials step by step

So finally, we have expressed the polynomial as a product of two factors.

This is only possible when the result of the division is exact.[/box]

You need to know the root properties of a polynomial, which will be very helpful in expressing polynomials as a product of factors.

Resolved exercises of factorization of a polynomial

Let’s see now with a solved exercise step by step, how we can factor a polynomial with the Ruffini rule, taking into account everything learned so far.

Let’s break down the next polynomial:

exercise factor theorem

To decompose it, we must divide it by a binomial (x-a), so that the division is exact and we can express it as the product of two polynomials of lesser degree.

Then, we must find what must be that coefficient a of the binomial (x-a) by which we will begin to divide the polynomial with the Ruffini rule.

For the division to be exact, that coefficient a, must be a root of the polynomial, for the division to be exact.

Since we don’t know what the roots of the polynomial are, we must test with different coefficients in the Ruffini rule until the rest is zero, or in other words, until the element of the last row is 0.

And what numbers should we try with? All of them?

Fortunately not.

According to the first property the roots of the polynomial are divisors of its independent term.

The independent term of the polynomial is 4 and has as divisors: 1, -1, 2, -2, 4 and -4.

Those divisors are the candidates to be the roots of the polynomial and they are the ones we must try with the Ruffini rule. In this case, if we tried with 3 or with 5, we would not have as rest 0.

Let’s start testing with 1:

polynomial factor theorem

It has given us 0 as the rest, therefore, on the one hand, 1 is a root of the polynomial and on the other hand, the polynomial can be expressed as a product of the quotient by the divisor:

polynomial factorization exercises

Now, we have two factors. The first is grade 1 and can no longer be broken down, but the second is grade 2 and we will continue to break it down by Ruffini.

The second polynomial is:

ejercicios teorema del factor

The independent term is -4, therefore, the coefficients with which we must test in Ruffini are: 1, -1, 2, -2, 4 and -4 .

This time, with 1 and -1 we do not have a rest equal to 0, but we do have it with 2:

polynomial factorization exercises

We express this last grade 2 polynomial as the product of the divisor by the quotient:

factor theorem exercises

And finally, if we replace the grade 2 polynomial by the multiplication of its two factors, the original polynomial can be expressed as:

polynomial factorization exercises

We have already factored it.

The original polynomial, which was grade 3, has been factored as the product of three grade 1 binomials.

The coefficients a, of each binomial (x-a) correspond to the roots of the polynomial: 1, 2 and -2 are the roots of the polynomial.

If after testing with all the divisors of the independent term, we do not get the rest to be 0, it means that the polynomial is irreducible and therefore cannot be decomposed.

Let’s solve this exercise:

Factor the following polynomials using the Ruffini rule:

Solution:

Paragraph a

The independent term is 18, therefore, for the rest to be 0, we must test with its divisors: 1, -1, 2. -2, 3 y -3.

The first of those divisors with which it gives us 0 of rest is 2:

Therefore, we express the polynomial as a product of the divisor by the quotient:

The second factor, which is grade 2, can be broken down further:

Its independent term is -9. The divisors to be tested with are: 1, -1, 2, -2, 3 and -3.

The first divisor with which we have rest 0 is 3:

We write the polynomial as the multiplication of the divisor by the quotient:

We replace the polynomial of degree 2 with the multiplication of its two factors. The original polynomial can be expressed as:

Paragraph b

The independent term is -6, therefore, for the rest to be 0, we must test with its divisors: 1, -1, 2. -2, 3 y -3.

The first of those divisors with which it gives us 0 of rest is 1:

Therefore, we express the polynomial as a product of the divisor by the quotient:

The second factor, which is grade 2, can be broken down further:

The independent term is -6, therefore, for the rest to be 0, we must test with its divisors: 1, -1, 2. -2, 3 y -3.

The first of those divisors with which it gives us 0 of rest is 2:

We write the polynomial as the multiplication of the divisor by the quotient:

We replace the polynomial of degree 2 with the multiplication of its two factors. The original polynomial can be expressed as:

Paragraph c

In this case, the polynomial has no independent term. Similarly, let’s try 1, -1, 2 and -2 until the rest is 0.

The first of those divisors with which it gives us 0 of rest is 2:

Therefore, we express the polynomial as a product of the divisor by the quotient:

The second factor, which is grade 3, can be broken down further:

In this case, the polynomial has no independent term. Similarly, let’s try 1, -1, 2 and -2 until the rest is 0.

The first of those divisors with which it gives us 0 of rest is -2:

We write the polynomial as the multiplication of the divisor by the quotient:

We replace the polynomial of grade 3 with the multiplication of its two factors. The original polynomial can be expressed as:

We can no longer decompose, because x², even if it is grade 2, is another factor.

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