Math Lesson 9.5.3 - Solving Quadratic Equations by Factorization Part Two

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Welcome to our Math lesson on Solving Quadratic Equations by Factorization Part Two, this is the third lesson of our suite of math lessons covering the topic of Quadratic Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Solving Quadratic Equations by Factorization II

The second method for solving a quadratic equation by factorization consists of factorizing the left part of the equation so that the part containing the variable takes the form of one of the first two special algebraic identities. The rest must also fit this form but in the first power. In other words, we must try to express the equation in the form

(kx ± t)² ± r(kx ± t) = 0

where k = √a while t and r are numbers.

After doing this, we may make a new factorization in the form

(kx ± t)[(kx ± t) ± r] = 0

For example, the quadratic equation

x² + 3x + 2 = 0

can be written as

x² + 2x + 1 + x + 1 = 0

This is because we can write the first three terms as (x + 1)². In this way, we obtain

(x + 1)² + (x + 1) = 0

Therefore, now we have a common factor to factorize. It is (x + 1). When writing this common factor separately, we have another (x + 1) left from the first term and 1 from the second term. Hence, factorizing this common factor yields:

(x + 1)[(x + 1) + 1] = 0

We can make further operations in the square brackets to simplify the expression in it. Thus, we have

(x + 1)[x + 1 + 1] = 0
(x + 1)(x + 2) = 0

In this way, we obtained a factorization of the original equation that helps calculate the roots. As in the previous examples, when a quadratic equation is factorized, one of the brackets must be zero to have a correct solution. Hence, in this example too, we have two distinct roots: one is obtained for x + 1 = 0, i.e. x¹ = - 1 and the other for x + 2 = 0, i.e. x² = -2.

Example 3

Solve the following quadratic equations by factorization.

x² - 5x + 4 = 0
4x² - 10x + 6 = 0

Solution 3

We can try to factorize this equation by expressing it first in two ways:
x² - 4x + 4 - x = 0
or
x² - 2x + 1 - 3x + 3 = 0
The first disintegration is futile; it doesn't provide us with a factorization. As for the second disintegration, we can write
(x² - 2x + 1) - (3x - 3) = 0
(It is known that a negative sign before a pair of brackets makes all terms in the brackets change sign.)
The first brackets represent the expanded form of (x - 1)². As for the second pair of brackets, we can factorize 3 from them. In this way, we obtain
(x - 1)² - 3(x - 1) = 0
Obviously, there is a new expression to factorize: (x - 1). Hence, we obtain
(x - 1)[(x - 1) - 3] = 0
(x - 1)(x - 1 - 3) = 0
(x - 1)(x - 4) = 0
Therefore, the two roots of the equation are
x - 1 = 0
x₁ = 1
x - 4 = 0
x₂ = 4
It is clear that to factorize this equation, we must have as coefficient before x² a number that has an integer square root. Therefore, we have to express 4x² as 2x · 2x. We have the following options available to express this quadratic equation:
4x² - 8x + 4 - 2x + 2 = 0
and
4x² - 4x + 1 - 6x + 5 = 0
Only the first disintegration is useful in the sense that it can allow us to factorize further. We have
4x² - 8x + 4 - 2x + 2 = 0
(4x² - 8x + 4) - (2x - 2) = 0
Factorizing 4 in the first pairs of brackets and 2 in the second, yields
4(x² - 2x + 1) - 2(x - 1) = 0
The first pairs of brackets contain the expanded form of (x - 1)². Hence, we obtain
4(x - 1)² - 2(x - 1) = 0
Factorizing (x - 1) yields
(x - 1)[4(x - 1) - 2] = 0
(x - 1)[4x - 4 - 2] = 0
(x - 1)(4x - 6) = 0
Therefore, the two roots of this quadratic equation are
x - 1 = 0
x₁ = 1
and
4x - 6 = 0
4x = 6
x₂ = 6/4
= 3/2
We could have used another approach to factorize this quadratic equation. Thus, we could have written
4x² - 10x + 6 = 0
4x² - 4x + 6x + 6 = 0
4x(x - 1) - 6(x - 1) = 0
(x - 1)(4x - 6) = 0

More Quadratic Equations Lessons and Learning Resources

Equations Learning Material
Tutorial ID	Math Tutorial Title	Revision Notes	Revision Questions
9.5	Quadratic Equations
Lesson ID	Math Lesson Title	Lesson	Video Lesson
9.5.1	Solving Quadratic Equations by Factorizing
9.5.2	Special Cases of Quadratic Equations
9.5.3	Solving Quadratic Equations by Factorization Part Two
9.5.4	Solving a Quadratic Equation by Completing the Square

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