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 8x^{6} - 65x^{3} + 8

Solution to Example 1 of Factoring Quadratic Equations :

Let P = 8x^{6} - 65x^{3} + 8 Put x^{3} = t; Then P = 8t^{2} - 65t + 8

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

Step 1: Coefficient of t^{2} 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 = 8t^{2} - 65t + 8 = 8t^{2} - 64t - t + 8

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

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

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

In each of the two brackets, there is difference of cubes of two terms for which we have a formula a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) [See Formula 5]

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

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 8x^{6} - 65x^{3} + 8 and then applying Algebra formulas, we get the Factors as (x - 2)(x^{2} + 2x + 4)(2x - 1)(4x^{2} + 2x + 1) Ans.

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

Exercise on Factoring Quadratic Equations

Solve the following problems on Factoring Quadratic Equations

Factorize 8x^{6} - 9x^{3} + 1

Factorize (x^{2} + 3x)^{2} - 14(x^{2} + 3x) + 40

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