Factorise non-monic quadratics using the cross and fraction methods - Teacher explains

In this video, I explain how to factorise non-monic quadratic trinomials using the cross method and the fraction method.... factorize for our US friends.

Timestamps:
0:00 Intro
0:26 The cross method example 1
3:27 The cross method example 2
5:11 The fraction method example 1
7:48 The fraction method example 2

[bairie]

Thank you for your help sir. Clear explanation and easy to understand.

[NadiehFan]

OK. You asked for a comment, so here it is. The bottom line is that I actually like neither the cross nor the fraction method, although I like the fraction method better because there is too much guessing involved in the cross method and too much potential for confusion there.

The fraction method works because if we multiply a quadratic ax² + bx + c (where a, b, c are assumed to be integers, a ≠ 0) by a we get

a²x² + abx + ac = (ax)² + b(ax) + ac

which is a monic polynomial in ax. Evidently, if f and g are two integers such that f + g = b and fg = ac then we have

(ax)² + b(ax) + ac = (ax + f)(ax + g)

Now, c = fg/a is an integer and if a divides f or g we can take out the factor a from (ax + f) or from (ax + g) and we get (x + f/a)(ax + g) or (ax + f)(x + g/a) as factorisation over the integers of the original polynomial ax² + bx + c. But if a divides neither f nor g, then any prime factor of a must be a prime factor of either f or g since fg/a = c is an integer. This means that a must then be a product a₁a₂ of two integers a₁ and a₂ where a₁ divides f and a₂ divides g. In this case we can take out a factor a₁ from (ax + f) and a factor a₂ from (ax + g) to get (a₂x + f/a₁)(a₁x + g/a₂) as factorisation over the integers of the original polynomial ax² + bx + c.

However, in my country a method was taught in schools more than a century ago for factoring non-monic quadratics with integer coefficients that is known in English speaking countries as factoring by grouping and that is my favourite method. Let's take your first quadratic

5x² − 13x + 6

as an example. To factor a quadratic ax² + bx + c with integer coefficients over the integers we need to find two integers with product ac and sum b. Incidentally, it is not hard to prove that such integers exist and, consequently, that the quadratic can be factored over the integers, if and only if b² − 4ac is the square of an integer, so it easy to check ahead of time whether a factorisation over the integers is actually possible.

For the present example we need to find two integers with product 30 and sum −13 and it is not hard to see, even without pen or paper, that these numbers are −10 and −3. Now, since we have −13 = (−10) + (−3) we can write the linear term −13x as the sum of −10x and −3x which gives us

5x² − 10x − 3x + 6

Now we can factor out 5x in the first two terms and a constant −3 in the last two terms, which gives us

5x(x − 2) − 3(x − 2)

and lo and behold, we now have two terms which again have a common factor (x − 2) which we can take out, and if we do that we get

(5x − 3)(x − 2)

and we are done. In my opinion this easily beats both the cross and the fraction method.

It is not hard to see why this always works. If ax² + bx + c is a quadratic with integer coefficients, a ≠ 0, and f and g are integers such that f + g = b and fg = ac, then we have

ax² + fx + gx + fg/a

If a divides f, then we can factor out ax in the first two terms and g in the last two terms and we get

ax(x + f/a) + g(x + f/a) = (ax + g)(x + f/a)

If a divides g, then we can factor out x in the first two terms and g/a in the last two terms and we get

x(ax + f) + (g/a)(ax + f) = (x + g/a)(ax + f)

And finally, if a divides neither f nor g then a must be a product a₁a₂ of two integers such that a₁ divides f and a₂ divides g since a divides fg. Then we can factor out a₁x in the first two terms and g/a₂ in the last two terms and we get

a₁x(a₂x + f/a₁) + (g/a₂)(a₂x + f/a₁) = (a₁x + g/a₂)(a₂x + f/a₁)

and we have proved that factoring by grouping of a quadratic ax² + bx + c with integer coefficients, a ≠ 0, is always possible if two integers exist with product ac and sum b.

It is easy to prove that the converse is also true, that is, if a quadratic ax² + bx + c with integer coefficients, a ≠ 0, can be factored over the integers, then there must exist two integers with product ac and sum b. To see this, suppose that

ax² + bx + c = (px + q)(rx + s)

where p, q, r, s are integers. Expanding (px + q)(rx + s) we get

ax² + bx + c = prx² + (ps + qr)x + qs

which implies that we have b = ps + qr and ac = pr·qs = ps·qr so indeed two integers ps and qr exist with product ac and sum b.

So, we have proved that a quadratic ax² + bx + c with integer coefficients, a ≠ 0, can be factored over the integers if and only if two integers exist with sum b and product ac.

If a quadratic ax² + bx + c with integer coefficients, a ≠ 0, is factorable over the integers and if, therefore, two integers f and g exist such that f + g = b and fg = ac, then b² − 4ac = (f + g)² − 4fg = (f − g)² is the square of an integer. Conversely, if ax² + bx + c is a quadratic with integer coefficients, a ≠ 0, and b² − 4ac is the square of an integer, then b and √(b² − 4ac) are either both odd integers or both even integers so f = (b + √(b² − 4ac))/2 and g = (b − √(b² − 4ac))/2 are two integers with sum b and product ac, implying that ax² + bx + c is factorable over the integers.

So, we have also proved that a quadratic ax² + bx + c with integer coefficients, a ≠ 0, can be factored over the integers if and only if the discriminant b² − 4ac of this quadratic is the square of an integer.

If you really want to, you could even completely eliminate the guesswork in finding the appropriate integers with a given product and sum, as follows. In the example above we want to find two integers with sum −13 which implies that the average (arithmetic mean) of these integers is −¹³⁄₂. The two numbers are obviously equidistant from their mean. Let's call this distance u, then the numbers we want to find are

−¹³⁄₂ − u and −¹³⁄₂ + u

and since the product of these numbers is required to be 30 we have

(−¹³⁄₂ − u)(−¹³⁄₂ + u) = 30

Applying the difference of two squares identity this gives

¹⁶⁹⁄₄ − u² = ¹²⁰⁄₄
u² = ⁴⁹⁄₄
u = ⁷⁄₂

so the numbers are

−¹³⁄₂ − ⁷⁄₂ = −10 and −¹³⁄₂ + ⁷⁄₂ = −3

Of course, u² = ⁴⁹⁄₄ implies that u is either ⁷⁄₂ or −⁷⁄₂ but inverting the sign of u only swaps the values of −¹³⁄₂ − u and −¹³⁄₂ + u so we only need to consider one value of u.

Alternatively, we can simply calculate the appropriate integers with a given product and sum directly. In the example above we have a = 5, b = −13, c = 6 so b² − 4ac = (−13)² − 4·5·6 = 169 − 120 = 49 = 7² and so the two integers with product ac = 30 and sum b = −13 are (−13 − 7)/2 = −20/2 = −10 and (−13 + 7)/2 = −6/2 = −3.

[monishrules]

They teach the second method in india kinda