Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 12 - 4x - 5x^{2},we get the Factors as (x + 2)(-5x + 6) Ans.

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Let P = 4x^{4} - 5x^{2} + 1 Put x^{2} = t; then P = 4t^{2} - 5t + 1 In Factoring of Trinomials (Quadratics) , follow the five steps listed above.

Step 1: Coefficient of t^{2} x constant term = 4 x 1 = 4

Step 2: We have to express 4 as two factors whose sum = coefficient of t = -5 ; 4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]

Step 3: P = 4t^{2} - 5t + 1 = 4t^{2} - 4t - t + 1

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

Step 5: P = (t - 1)(4t - 1)

But t = x^{2}; ∴ P = (t - 1)(4t - 1) = (x^{2} - 1)(4x^{2} - 1) P = (x^{2} - 1^{2}){(2x)^{2} - 1^{2}}

In each of the two brackets, there is difference of squares of two terms which is equal to the product of the sum and difference of the terms.[See Formula 3]

∴ P = (x + 1)(x - 1)(2x + 1)(2x - 1)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 4x^{4} - 5x^{2} + 1 and then applying Algebra formulas, we get the Factors as (x + 1)(x - 1)(2x + 1)(2x - 1) Ans.

Exercise on Factoring Quadratics

Problems on Factoring Quadratics :

Factorize 6 - x - 2x^{2}

Factorize 7x^{2} - 8x - 12

Factorize 4x^{4} - 25x^{2} + 36

Progressive Learning of Math : Factoring Quadratics

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