Solving Quadratic Equation – Factorisation method

What is Quadratic Equation by Factorization? 

To factorize an expression means to write the expression as the product of its factors which usually involves the introduction of brackets. So we can say that factorization is the opposite of expansion and vice-versa. There are various ways of factorization. These methods depend on the nature of the expression to be factorized. These methods are:

Factorizing by taking out common factors

Factorizing by grouping of terms of expression

Factorizing quadratic expressions

Factorizing perfect squares

Factorizing difference of two squares

Factorizing by taking out common factors

Factorize the following

12x – 3y, 2x + bx – 3x

12x – 3y

The H C F of 12x and 3y is 3

Taking out the H C F: 12x + 3y = 3(4x – y)

2x + bx – 3x

H C F of 2x, bx and 3x is x

2x + bx  – 3x = x(2 + b – 3) = x(b – 1)

Factorizing by grouping of Terms of Expression

This is done by grouping the terms of the expressions in pairs and then factoring out the common quantities.

Factorize the following expressions

2dx + 2dy + cx + cy, ab + b² – ay – yb

Solution:

The first and second terms, 2dx and 2dy are grouped (since 2d is common), while the third and fourth terms, cx and cy are grouped (since c is common).

i.e. 2dx + 2dy + cx + cy = (2dx +2dy) + (cx + cy)

Hence the method is to look for terms that have common factor in them and group them together

= 2d(x + y)+(x + y)

(x + y) is common so we bring it out: (x + y)(2d + c)

= Hence 2dx + 2dy + cx + cy = (x + y)(2d + c)

ab + b² – ay – yb

Grouping = (ab + b²) – (ay – yb)

Factorizing = b(a + b) – y(a + b)

= (a + b)(b – y)

ab + b² – ay – yb = (a + b)(b – y)

Factorization of Quadratic Equation

Expression of the form ax² + bx + c is called general quadratic expression in x where a, b and c are real constants, a ¹ 0. In this expression ax² + bx + c, ax² is the first term with a, the coefficient of x². Bx is the second or middle term with b the coefficient of x and c the third or constant term.

In a nutshell, a quadratic equation must have a squared term as its highest power.

Now consider the product of two linear expressions (x + a) and (x + b) where a and b are constants.

i.e. (x + a)(x + b) = x(x +b) + (x + B)

= x² + bx + ax + ab

= x² + (a + b)x + ab

Observe that the coefficient of x in the result is the sum of a and b, i.e. (a + b) and the constant term is the product of a and b, ab.

The factors of x² + 3x – 4 are:

(x+4) and (x-1)

Let us multiply them to see:

(x + 4)(x – 1) = x(x – 1) + 4(x – 1)

= x² – x + 4x – 4

= x² + 3x – 4

Example: 6x² + 5x – 6

List the positive factor Example: 6x² + 5x – 6

Step 1: a × c is 6× (-6) = -36, and b is 5

One of the numbers has to be negative to make -36, so by playing with a few different numbers I find that -4 and 9 work nicely:

-4×9 = -36 and -4+9 = 5

Step 2: Rewrite 5x with -4x and 9x:

6x² – 4x + 9x – 6

Step 3: Factor first two and last two:

2x(3x – 2) + 3(3x -2)

Step 4: Common Factor is (3x – 2):

(2x+3)(3x – 2)

Check: (2x+3)(3x – 2) = 6x² – 4x + 9x – 6 = 6x² + 5x – 6

Factorization of perfect squares

In some cases recognizing some common patterns in the trinomial will help you to factor it faster. For example, we could check whether the trinomial is a perfect square.

A perfect square trinomial is of the form:

(ax)² + 2abx + b²

 Take note that

1. The first term and the last term are perfect squares
   2. The coefficient of the middle term is twice the square root of the last term multiplied by the square root of the coefficient of the first term.

When we factor a perfect square trinomial, we will get

(ax)² + 2abx + b² = (ax + b)²

The perfect square trinomial can also be in the form:

(ax)² – 2abx + b²

In which case it will factor as follows:

h(ax)² – 2abx + b² = (ax – b)²

Example

1. x²+ 2x + 1 = 0

(x + 1)² = 0

1. x² + 6x + 9 = 0

X² + 2(3)x + 3² = 0

(x + 3)² = 0

Factor the following trinomials:

1. a) x²+ 8x+ 16
   b) 4x²– 20x + 25

Solution:

1. a) x²+ 8x+ 16
   = x² + 2(x)(4) + 4²
   = (x + 4)²
2. b) 4x²– 20x+ 25
   = (2x)²– 2(2x)(5) + 5²
   = (2x– 5)²

Factorization of Difference of Two Squares

Remember from your translation skills that “difference” means “subtraction”. So a difference of squares is something that looks like x² – 4. That’s because 4 = 2², so you really have x² – 2², a difference of squares. To factor this, do your parentheses, same as usual:

x² – 4 = (x      )(x     )

You need factors of –4 that add up to zero, so use –2 and +2:

x² – 4 = (x – 2)(x + 2)

Note that we had x² – 2², and ended up with (x – 2)(x + 2). Differences of squares (something squared minus something else squared) always work this way:

For a² – b², do the parentheses:

(         )(        )

…put the first squared thing in front:

(a      )(a       )

…put the second squared thing in back:

(a    b)(a     b)

…and alternate the signs in the middles:

(a – b)(a + b)

Factor x² – 16

This is x² – 4², so I get:

x² – 16 = x² – 4² = (x – 4)(x + 4)

Factor 4x² – 25

This is (2x)² – 5², so I get:

4x² – 25 = (2x)² – 5² = (2x – 5)(2x + 5)

Factor 9x⁶ – y⁸

This is (3x³)² – (y⁴)², so I get:

9x⁶ – y⁸ = (3x³)² – (y⁴)² = (3×3 – y4)(3×3 + y4)

Factor 4x – 1

This is (x²)² – 1², so I get:

x⁴ – 1 = (x²)² – 1² = (x² – 1)(x² + 1)

Note that I’m not done yet, because x² – 1 is itself a difference of squares, so I need to apply the formula again to get the fully-factored form. Since x² – 1 = (x – 1)(x + 1), then:

x⁴ – 1 = (x²)² – 1² = (x2 – 1)(x² + 1)

= ((x)2 – (1)2)(x² + 1)

= (x – 1)(x + 1)(x2 + 1)