Math Lesson 6.4.3 - Factorising Quadratics of the Form x² + bx + c (a = 1)

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Welcome to our Math lesson on Factorising Quadratics of the Form x² + bx + c (a = 1), this is the third lesson of our suite of math lessons covering the topic of Factorising, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Factorising Quadratics of the Form x² + bx + c (a = 1)

This is the simple form of quadratics, where the coefficient a is 1 (a = 1). For example,

belongs to this type of quadratics where a = 1, b = -5 and c = 4.

If such quadratics are factorised, the expression will be written in the form

where m and n are two numbers such that when the factorised quadratics expands, we can obtain the original form of this expression. Thus, when expanding the last expression, we obtain

In this way, if the two conditions

and

meet simultaneously in a given quadratic expression, then it is possible to factorise it.

Let's see whether the quadratic we took as an example at the beginning of this paragraph can be factorised or not. For this, we must try to find two numbers m and n that fit the above description. Since b = -5 and c = 4, we have

or

The two numbers that meet these conditions are 1 and 4 (or 4 and 1). Therefore, the original quadratics

is expressed in the form

or

Example 3

Factorise the following quadratics

1. x² + 7x + 6
2. x² - 4x + 4

Solution 3

1. We have a = 1, b = 7 and c = 6. In addition, we must have
   -(m + n) = b and mn = c
   Hence, substituting the known values, yields
   -(m + n) = 7 and m ∙ n = 6
   or
   m + n = -7 and m ∙ n = 6
   The only two numbers that meet both these requirements are m = -1 and n = -6 (or m = -6 and n = -1). Therefore, the factorised form of the expression
   x² + 7x + 6
   is
   (x - m)(x - n) = [x - (-1)][x - (-6)]
   = (x + 1)(x + 6)
2. We have a = 1, b = -4 and c = 4. In addition, we must have
   -(m + n) = b and mn = c
   Hence, substituting the known values, yields
   -(m + n) = -4 and m ∙ n = 4
   or
   m + n = 4 and m ∙ n = 4
   There is only one possibility from two numbers to meet both these conditions: they must be equal. Thus, we have m = 2 and n = 2. Therefore, the factorised version of the original expression is
   (x - m)(x - n) = (x - 2)(x - 2)
   This is the long form of the second special algebraic identity (x - 2)². Indeed, expanding this expression according to the rule
   (a - b)² = a² + 2ab + b²
   we obtain
   (x - 2)² = x² - 2 ∙ x ∙ 2 + 2²
   = x² - 4x + 4
   which is identical to the original expression.

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7. Continuing learning expressions - read our next math tutorial: Algebraic Fractions

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