FACTORING QUADRATIC EQUATIONS - SOLVED EXAMPLES, EXERCISE PROBLEMS

Please study

Method of Factoring Quadratic Equations

if you have not already done so.

There, we explained a Five Step Method,
with each step explained in detail.

That knowledge is a prerequisite here.

Example 1 of Factoring Quadratic Equations

Factorize 8x⁶ - 65x³ + 8

Solution to Example 1 of Factoring Quadratic Equations :

Let P = 8x⁶ - 65x³ + 8
Put x³ = t; Then P = 8t² - 65t + 8

In Factoring of Trinomials (Quadratics) , follow the five steps listed above.

Step 1: Coefficient of t² x constant term
= 8 x 8 = 64

Step 2: We have to express 64 as two factors whose sum
= coefficient of t = -65 ;
64 = 64 x 1 = -64 x -1; {(-64) + (-1) = -65}

Step 3: P = 8t² - 65t + 8 = 8t² - 64t - t + 8

Step 4: P = 8t(t - 8) - 1(t - 8)

Step 5: P = (t - 8)(8t - 1)

But t = x³; ∴ P = (t - 8)(8t - 1) = (x³ - 8)(8x³ - 1)
⇒ P = (x³ - 2³){(2x)³ - 1³}

In each of the two brackets, there is difference
of cubes of two terms for which we have a formula
a³ - b³ = (a - b)(a² + ab + b²) [See Formula 5]

Applying this Formula here, we get
P = (x - 2){x² + x(2) + 2²}(2x - 1){(2x)² + (2x)(1) + 1²}
= (x - 2)(x² + 2x + 4)(2x - 1)(4x² + 2x + 1)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 8x⁶ - 65x³ + 8 and then applying Algebra formulas, we get the Factors as (x - 2)(x² + 2x + 4)(2x - 1)(4x² + 2x + 1) Ans.

Example 2 of Factoring Quadratic Equations

Factorize (x² + x)² -18(x² + x) + 72

Solution to Example 2 of Factoring Quadratic Equations :

Let P = (x² + x)² -18(x² + x) + 72
Put (x² + x) = t; Then P = t² -18t + 72

In Factoring of Trinomials (Quadratics), follow the five steps listed above.

Step 1: Coefficient of t² x constant term
= 1 x 72 = 72

Step 2: We have to express 72 as two factors whose sum
= coefficient of t = -18 ;
72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]

Step 3: P = t² -18t + 72 = t² - 12t - 6t + 72

Step 4: P = t(t - 12) - 6(t - 12)

Step 5: P = (t - 12)(t - 6)

But t = (x² + x); ∴ P = (t - 12)(t - 6) = (x² + x - 12)(x² + x - 6)

In each of these two brackets, there is a Quadratic Polynomial
which can be factorised using the five steps above.

x² + x - 12 = x² + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x + 4)(x - 3)

x² + x - 6 = x² + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2)

See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps. You might have mastered the 5 stepsof factorisation by this time, to write directly like this.

Thus P = (x² + x)² -18(x² + x) + 72
= (x² + x - 12)(x² + x - 6)
= (x + 4)(x - 3)(x + 3)(x - 2)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to (x² + x)² -18(x² + x) + 72 twice,
we get the Factors as (x + 4)(x - 3)(x + 3)(x - 2) Ans.

Exercise on Factoring Quadratic Equations

1. Factorize 8x⁶ - 9x³ + 1
2. Factorize (x² + 3x)² - 14(x² + 3x) + 40

Answers to Exercise on Factoring Quadratic Equations

1. (2x - 1)(4x² + 2x + 1)(x - 1)(x² + x + 1)
2. (x + 5)(x - 2)(x + 4)(x - 1)