Math Lesson 6.4.4 - Factorising Quadratics of the Form ax² + bx + c (a ≠ 1)

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Welcome to our Math lesson on Factorising Quadratics of the Form ax² + bx + c (a ≠ 1), this is the fourth 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 ax² + bx + c (a ≠ 1)

This factorisation involves an extra step compared to the previous one. You have to find another combination of numbers that gives the coefficient a in addition to the two combinations used in the previous paragraph. In this way, we have to find other two numbers p and q which must meet the condition.

Giving that the general form of a quadratic expression is

then, if we are sure that one of the numbers between p and q is 1 (or -1), the expression must be written in the form

or

In this way, the following two conditions must meet simultaneously

and

or

Hence, we now have three conditions: one for each coefficient (constant). They are:

For example, if we are asked to factorise the quadratic expression

first, we have to point out the coefficients and constant (here we have a = 3, b = 7 and c = 2), then we try to find the other numbers (p, q, m and n).

From the third condition we have

Hence, we have four possible combinations for m and n: m = 2, n = 1; m = 1, n = 2; m = -1, n = -2; m = -2, n = -1.

Likewise, for p and q we have four possible combinations as well, given the first condition. They are: p = 1, q = 3; p = 3, q = 1; p = -1, q = -3; p = -3, q = -1. However, this is not important, as we always have them multiplied together and the result is always the same.

Now, we have to look at the second condition. We have

Thus, from all combinations for m and n, only m = -1 and n = -2 are suitable, as

Therefore, the original quadratic expression is factorised as

Example 4

Factorise the following quadratics.

1. 2x² - 5x + 2
2. 4x² + 9x + 2

Solution 4

1. We have a = 2, b = -5, c = 2. In addition, we have the following combinations
   p ∙ q = a = 2
   m ∙ n = c = 2
   Likewise,
   p ∙ q ∙ n + m = -b
   2 ∙ n + m = -b = -(-5)
   2 ∙ n + m = 5
   Therefore, we have n = 2 and m = 1. Hence, the factorised form of the quadratic expression 2x² - 5x + 2 is
   (pqx - m)(x - n)
   or
   (2x - 1)(x - 2)
   When expanded, this expression gives
   (2x - 1)(x - 2)
   = 2x ∙ x + 2x ∙ (-2) - 1 ∙ x-1 ∙ (-2)
   = 2x² - 4x - x + 2
   = 2x² - 5x + 2
   which is identical to the original expression.
2. We have a = 4, b = 9, c = 2. In addition, we have the following combinations
   p ∙ q = a = 4
   m ∙ n = c = 2
   Likewise,
   p ∙ q ∙ n + m = -b
   4 ∙ n + m = -b = -9
   4 ∙ n + m = -9
   Thus, n = -2 and m = -1.
   Moreover, since p and q are usually different, we have p = 4 and q = 1. Hence, the factorised form of the quadratic expression 4x² + 9x + 2 is
   (pqx - m)(x - n)
   or
   [4x - (-1)][(x - (-2)]
   = (4x + 1)(x + 2)
   Let's prove this by expanding the last expression to confirm the result obtained. Thus,
   (4x + 1)(x + 2)
   = 4x ∙ x + 4x ∙ 2 + 1 ∙ x + 1 ∙ 2
   = 4x² + 8x + x + 2
   = 4x² + 9x + 2
   which is identical to the original expression.
   So far, one of numbers p and q was always 1. But, what happens if both p and q are different from 1? Let's consider the following quadratic expression.
   6x² - 11x - 10
   where a = 6, b = -11 and c = -10.
   Starting from the constant c, yields
   m ∙ n = c = -10
   There are eight possible combinations that give this result: m = -1, n = 10; m = 1; n = -10; m = -2, n = 5; m = 2, n = -5; m = -5, n = 2; m = 5, n = -2; m = -10, n = 1 and m = 10, n = -1. We must find the suitable combination that gives
   p ∙ q ∙ n + m = -b
   Thus,
   6 ∙ n + m = -(-10)
   or
   6 ∙ n + m = 10
   None of the combinations can be used here. Therefore, we must look for another way to express the given quadratics. We must try to write it in the form
   (px - m)(qx - n)
   where both p and q are different from 1. Now, let's discover the relationship between p, q, m and n. Expanding the expression above yields
   (px-m)(qx-n)
   = px ∙ qx + px ∙ (-n) - m ∙ qx - m ∙ (-n)
   = pqx² - npx - mqx + mn
   = pqx² + (-np - mq)x + mn
   Therefore, we obtain the new expression for the coefficient b:
   -np - mq = b
   Substituting the value of b in our example yields
   -np - mq = -11
   or
   np + mq = 11
   From all combinations that give p · q = 6, (given that m · n = -10) we choose n = 5, m = -2, p = 3 and q = 2, as
   5 ∙ 3 + (-2) ∙ 2 = 15 - 4 = 11
   Therefore, the original expression is factorised as
   (px - m)(qx - n)
   = [3x - (-2)] ∙ [2x - 5]
   = (3x + 2)(2x - 5)
   Proof: Expanding the last expression, gives
   (3x + 2)(2x - 5)
   = 3x ∙ 2x + 3x ∙ (-5) + 2 ∙ 2x + 2 ∙ (-5)
   = 6x² - 15x + 4x - 10
   = 6x² - 11x - 10
   which is identical to the original expression.

It is clear that this method is more inclusive. We only use the first method when we are sure that either p or q are equal to 1; otherwise, we use the second method for finding all coefficients required.

Example 5

Factorise the expression

Solution 5

We have a = 8, b = -10 and c = 3. Most probably, p and n are different from 1 as there are four other possible combinations (4 and 2; 2 and 4 and their negative counterparts). Hence, we are not going to take the risk of using the first method, as we may find ourselves blocked on the way. Hence, we aim to write the expression in the form

We have

so, m may be one of the following numbers: -3, -1, 1 and 3, as well as n (m and n must have the same sign to give 3 when multiplied).

In addition, we have p · q = 8, and

The only combination possible is the one that produces the arithmetic expression

because the value of this expression is 10. Thus, based on this description, we have n = 3, m = 1, p = 2 and q = 4. Therefore, the expression becomes

Proof:

which is identical to the original expression.

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