Factorization by Cross-method

Quadratic Polynomials in One Variable x2 + qx + r Consider the following quadratic polynomial : x2 – 10x + 9 Can you factorize it? We cannot directly factorize it by taking out the common factors, grouping terms or using identities…

Let me introduce another method of factorization Let me introduce another method of factorization. Consider how we expand the expression (x + 2)(x + 3) by long multiplication first … …by using the same method, expand the expression (x + a)(x + b).

We note that: Coefficient of x: 2 + 3 = 5 Coefficient of x: a + b Constant term: 2 × 3 = 6 Constant term: ab

In fact, when the expression (x + a)(x + b) is expanded, the coefficients of x2, x and the constant term can be found in the following way: a x b ) + × x 2 x b a ) ( + ab + i.e. When a polynomial can be expressed as x2 + (a + b)x + ab, it can be factorized into (x + a)(x + b). This method is called the cross-method. It is very useful in factorization of quadratic polynomials.

Now, let’s try to factorize x2 – 10x + 9 by cross-method. Step 1: Suppose x2 – 10x + 9 can be factorized as (x + a)(x + b), where a + b = –10 and ab = +9. Step 2: Since the constant term 9 equals ab, a and b are factors of 9. All the possible pairs of factors of 9 are: 3 9 1 - + 3) 3)( ( 9) 1)( - + x Step 3: Among the four pairs of a and b above, only one pair (a = –1, b = –9) satisfies the condition a + b = –10.

Follow-up question Factorize x2 + 2x – 15. Solution List all the possible pairs of factors of –15. Check each pair of factors by the cross-method as follows:

Quadratic Polynomials in One Variable px2 + qx + r We can also apply the cross-method to polynomials in the form px2 + qx + r, where In this case, we should consider both the factors of p and r during the process of factorization.

Let’s try to factorize 6x2 – 7x + 1. Since 6x2 can be written as either (x)(6x) or (2x)(3x) and the constant term +1 can be written as either (+1) × (+1) or (–1) × (–1), we have: The factors of 6x2 The factors of +1 The corresponding possible results are (x + 1)(6x + 1), (x – 1)(6x – 1), (2x + 1)(3x + 1) and (2x – 1)(3x – 1).

By the cross-method: Not all quadratic polynomials can be factorized. For example, x2 – x + 1 cannot be factorized.

Follow-up question Factorize 2x2 + x – 1. Solution 2x2 can be written as (x)(2x), –1 can be written as (+1) × (–1) or (–1) × (+1). The trials can be done as follows:

Quadratic Polynomials in Two Variables A quadratic polynomial in two variables in the form px2 + qxy + ry2, where p, q and r are non-zero constants, can be factorized in a similar way. Take x2 – 3xy – 4y2 as an example. –4y2 can be written as (+y)(–4y), (+2y)(–2y) or (+4y)(–y). The trials can be done as follows:

Follow-up question Factorize 3x2 + 5xy + 2y2. Solution The possible pairs of factors are shown below: The factors of 3x2 The factors of 2y2 Since the coefficient of xy is positive, we do not consider the pairs and . -y -2y -2y -y