Factorization of Quadratic Equations

Solving Quadratic Equations by Factoring:

The standard form of a quadratic expression is \(ax^2~+~bx~+~c \), where a, b and c are constants with a ≠ 0. This can be easily expressed in the form of its factors by the method of factorization. In this method, any expression can be expressed as the product of its factors, which when multiplied will give back the original expression.

Consider a Quadratic expression \( x^2~+~5x~+~6\) . A visualization of the same expression can bring out its factors in an easier way. First, observe that the expression is actually the sum of the three terms of the given expression, i.e., \(x^2\), 5x and 6.

\(x^2\) can be represented on a graph as a square of side x units, whose area is given by \(x^2\), as shown in the given figure.

Area of a square of side x

Figure 1 Area of a square of side x

Similarly, x can be visualized as a rectangle with length x units and breadth as 1 unit. So, 5x would be the area of 5 such rectangles stacked together. This is represented as shown:

(x2 + 5x)

Figure 2

This figure above represents a “shape” with total area \( (x^2~+~5x) \) . The area remains the same if we change the positions of the x by 1 rectangle as shown:

fig3(x2 + 5x)

Figure 3

This figure still represents \((x^2~+~5x)\).

Now, the constant term ‘6’ can be represented as a rectangle with sides 6 units and 1 unit or as a rectangle with one side as 3 units and other as 2 units. While factorization it must be noted that the factors of the constant term must add equally to the coefficient of x.

fig 4.1 Factorization of constant term 6

Figure 4.1 Factorization of constant term 6

fig 4.2 Factorization of constant term 6

Figure 4.2 Factorization of constant term 6

We can fit the figure shown by Fig 4.2 neatly in the above figure, which would then look like:

fig 5 (x+2)(x+3)

Figure 5

The area of this rectangle can be given by (x+2)(x+3), which is our original expression \(x^2~+~5x~+~6\). Thus, visualization makes it more convenient to understand factorization of quadratic expression.

Any polynomial expression which contains equality sign becomes an equation. So, when this expression \((x^2~+~5x~+~6)\) is equated to zero, this becomes a quadratic equation. The roots or solution of this equation can be easily pictured here using figure 5. If any sides of the rectangle formed, becomes 0, then the area would become 0 and that particular value where the area would be zero will represent roots.


⇒ (x + 2)(x + 3) = 0

⇒ x = -2 or x = -3

If the value of x becomes equal to -2 or -3 in the above equation, then the equation becomes 0, which are required roots of given quadratic equation.

Graphs of Quadratic Equations:

The graphs of quadratic equations are represented using parabolas. For any equation y = \(ax^2\), three cases are possible;

Case i) a > 0

This would represent an upward facing parabola and depending upon value of a, the concavity of the graph or the rate of increase or decrease of function is affected. Simply, if a decreases, graph opens up and vice –versa.

Consider y = \(100x^2\)

Quadratic Polynomial

Figure 6 Graph of Quadratic Polynomial

Now, if a is increased or decreased, say a = 10 and a = 1000 then graphically it would be represented as:

Transformation of Quadratic Functions

Figure 7 Transformation of Quadratic Functions

Similarly if a is negative then, a downward facing parabola is obtained and concavity depends upon value of a.

fig 8 Transformation of Quadratic Functions

Figure 8 Transformation of Quadratic Functions

To shift the graph horizontally changes are done in x and for vertical shifting, changes are done in y.
The equation, y = \(100(x~-~2)^2\) represents a change in x, it can be represented graphically as:

y = 100(x − 2)2

Figure 9

Similarly, y + 2 = \(x^2\) can be represented graphically as:

 y + 2 = x2

Figure 10

Thus by using graphs and by visualization, quadratic equations can be understood in a better way.

Practise This Question

For the damped oscillator shown, the mass m of the block is 200 g, k = 90 N m1 and the damping constant b is 40 g s1. What is the period of oscillation, time taken for its amplitude of vibrations to drop to half of its initial value.