FACTORING QUADRATIC EQUATIONS - SOLVED EXAMPLES, EXERCISE PROBLEMS

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.

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Example 2 of Factoring Quadratic Equations

Solve the following problem 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

Solve the following problems on Factoring Quadratic Equations

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

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Answers to Exercise on Factoring Quadratic Equations

Answers to Problems on Factoring Quadratic Equations :

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

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