4 FACTORING EXERCISES WITH SOLUTIONS - MATHS - 2025

The exercises factorization help understand this technique, much used in mathematics and is in the process of writing a sum as a product of certain terms.

The word factorization refers to factors, which are terms that multiply other terms. For example, in the prime factorization of a natural number, the prime numbers involved are called factors.

That is, 14 can be written as 2 * 7. In this case, the prime factors of 14 are 2 and 7. The same applies to polynomials of real variables.

That is, if you have a polynomial P (x), then factoring the polynomial consists of writing P (x) as the product of other polynomials of degree less than the degree of P (x).

Factoring

Various techniques are used to factor a polynomial, including notable products and calculating the roots of the polynomial.

If we have a second-degree polynomial P (x), and x1 and x2 are the real roots of P (x), then P (x) can be factored as "a (x-x1) (x-x2)", where "a" is the coefficient that accompanies the quadratic power.

How are the roots calculated?

If the polynomial is of degree 2, then the roots can be calculated with the formula called "the resolvent".

If the polynomial is of degree 3 or more, the Ruffini method is usually used to calculate the roots.

4 factoring exercises

First exercise

Factor the following polynomial: P (x) = x²-1.

Solution

It is not always necessary to use the resolvent. In this example you can use a remarkable product.

Rewriting the polynomial as follows we can see which notable product to use: P (x) = x² - 1².

Using the remarkable product 1, difference of squares, we have that the polynomial P (x) can be factored as follows: P (x) = (x + 1) (x-1).

This further indicates that the roots of P (x) are x1 = -1 and x2 = 1.

Second exercise

Factor the following polynomial: Q (x) = x³ - 8.

Solution

There is a remarkable product that says the following: a³-b³ = (ab) (a² + ab + b²).

Knowing this, the polynomial Q (x) can be rewritten as follows: Q (x) = x³-8 = x³ - 2³.

Now, using the remarkable product described, we have that the factorization of the polynomial Q (x) is Q (x) = x³-2³ = (x-2) (x² + 2x + 2²) = (x-2) (x² + 2x + 4).

The quadratic polynomial that arose in the previous step remains to be factorized. But if you look at it, Remarkable Product # 2 can help; therefore, the final factorization of Q (x) is given by Q (x) = (x-2) (x + 2) ².

This says that one root of Q (x) is x1 = 2, and that x2 = x3 = 2 is the other root of Q (x), which is repeated.

Third exercise

Factor R (x) = x² - x - 6.

Solution

When a remarkable product cannot be detected, or the necessary experience to manipulate the expression is not available, we proceed with the use of the resolvent. The values ​​are as follows a = 1, b = -1, and c = -6.

Substituting them in the formula results in x = (-1 ± √ ((- 1) ² - 4 * 1 * (- 6))) / 2 * 1 = (-1 ± √25) / 2 = (-1 ± 5)/two.

From here there are two solutions that are the following:

x1 = (-1 + 5) / 2 = 2

x2 = (-1-5) / 2 = -3.

Therefore, the polynomial R (x) can be factored as R (x) = (x-2) (x - (- 3)) = (x-2) (x + 3).

Fourth exercise

Factor H (x) = x³ - x² - 2x.

Solution

In this exercise we can start by taking the common factor x and we obtain that H (x) = x (x²-x-2).

Therefore, it only remains to factor the quadratic polynomial. Using the resolvent again, we have that the roots are:

x = (-1 ± √ ((-1) ²-4 * 1 * (- 2))) / 2 * 1 = (-1 ± √9) / 2 = (-1 ± 3) / 2.

Therefore the roots of the quadratic polynomial are x1 = 1 and x2 = -2.

In conclusion, the factorization of the polynomial H (x) is given by H (x) = x (x-1) (x + 2).

References

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