# FACTORING QUADRATICS - FACTORING QUADRATIC EXPRESSIONS, EXAMPLES, EXERCISE

Please study

Method of Factoring Quadratics

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 :

Factorize 12 - 4x - 5x2

Solution to Example 1 of Factoring Quadratics :

Let P = 12 - 4x - 5x2 = -5x2 - 4x + 12

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

Step 1: Coefficient of x2 x constant term
= -5 x 12 = -60

Step 2: We have to express -60 as two factors whose sum
= coefficient of x = -4 ;
-60 = -10 x 6; (-10 + 6 = 4)

Step 3: P = -5x2 - 4x + 12 = -5x2 - 10x + 6x + 12

Step 4: P = -5x(x + 2) + 6(x + 2)

Step 5: P = (x + 2)(-5x + 6)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 12 - 4x - 5x2,we get the Factors as (x + 2)(-5x + 6) Ans.

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### Example 2 of Factoring Quadratics

Factorize 3x2 - 17x - 20

Solution to Example 2 of Factoring Quadratics :

Let P = 3x2 - 17x - 20

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

Step 1: Coefficient of x2 x constant term
= 3 x -20 = -60

Step 2: We have to express -60 as two factors whose sum
= coefficient of x = -17 ;
-60 = -20 x 3; (-20 + 3 = -17)

Step 3: P = 3x2 - 17x - 20 = 3x2 - 20x + 3x - 20

Step 4: P = x(3x - 20) + 1(3x - 20)

Step 5: P = (3x - 20)(x + 1)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 3x2 - 17x - 20, we get the Factors as (3x - 20)(x + 1) Ans.

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### Example 3 of Factoring Quadratics

Factorize 4x4 - 5x2 + 1

Solution to Example 3 of Factoring Quadratics :

Let P = 4x4 - 5x2 + 1
Put x2 = t; then P = 4t2 - 5t + 1
In Factoring of Trinomials (Quadratics) , follow the five steps listed above.

Step 1: Coefficient of t2 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 = 4t2 - 5t + 1 = 4t2 - 4t - t + 1

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

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

But t = x2; ∴ P = (t - 1)(4t - 1) = (x2 - 1)(4x2 - 1)
P = (x2 - 12){(2x)2 - 12}

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 4x4 - 5x2 + 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 :

1. Factorize 6 - x - 2x2
2. Factorize 7x2 - 8x - 12
3. Factorize 4x4 - 25x2 + 36

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### Answers to Exercise on Factoring Quadratics

Answers to Problems on Factoring Quadratics :

1. (-2x + 3)(x + 2)
2. (7x + 6)(x - 2)
3. (2x + 3)(2x - 3)(x + 2)(x - 2)

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