There, we gave introduction to Quadratic polynomial, Quadratic Equation, Methods to solve the Quadratic Equations and about Method of Solving by Factoring.

That knowledge is a prerequisite here.

Here, we apply the Method to solve problems. Solved Examples and Exercise problems are given.

Example 1 : Solving Quadratic Equations

Solve 8x^{6} - 65x^{3} + 8 = 0 with the knowledge of Solving Quadratic Equations.

Solution of Example 1 of Solving Quadratic Equations :

Factoring of LHS of the given Equation is given below.

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

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

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, Factoring of Quadratic Polynomial, 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) -*-*-*-*-*-*-*

8x^{6} - 65x^{3} + 8 = 0 ⇒ (x - 2)(x^{2} + 2x + 4)(2x - 1)(4x^{2} + 2x + 1) = 0 ⇒ (x - 2) = 0 or (2x - 1) = 0 or (x^{2} + 2x + 4) = 0 or (4x^{2} + 2x + 1) = 0 (x - 2) = 0 ⇒ x = 2 (2x - 1) = 0 ⇒ 2x = 1 ⇒ x = 1⁄2 (x^{2} + 2x + 4) = 0 has no real roots as (x^{2} + 2x + 4) has no real factors. (4x^{2} + 2x + 1) = 0 has no real roots as (4x^{2} + 2x + 1) has no real factors.

Thus, Knowledge of Solving Quadratic Equations helped in presenting the solution as x = 2, 1⁄2 are the two roots of the given 6th Degree equation. The other four roots are not real. 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, Knowledge of Solving Quadratic Equations helped in presenting the solution as x = -4, 3, -3, 2 are the four roots of the given 4th Degree equation. Ans.
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With the knowledge of Solving Quadratic Equations,

Solve 8x^{6} - 9x^{3} + 1 = 0.

Solve (x^{2} + 3x)^{2} - 14(x^{2} + 3x) + 40 = 0.

For Answers See at the bottom of the Page.

NOTE: You may solve these problems of Exercise using Quadratic Formula, after learning it.

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