# Please factor this trinomial: 12x²-18x-21

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### 2 Answers

We have to factor 12x^2 - 18x - 21

12x^2 - 18x - 21

factor 3

=>3*(4x^2 - 6x - 7)

The roots of 4x^2 - 6x - 7 = 0 are

x1 = 6/8 + sqrt (36 + 112) / 8

=> 6/8 + sqrt 148 / 8

x2 = 6/8 - sqrt 148 / 8

This gives 4x^2 - 6x - 7 = (x - 3/4 - sqrt 37/4)(x - 3/4 + sqrt 37/4)

The factors of the given expression are:

**3*(x - 3/4 - (sqrt 37)/4)(x + 3/4 - (sqrt 37)/4)**

To factor the trinomial, we'll consider the equation:

12x^2 - 18x - 21 = 0

We'll divide by 3 the trinomial:

4x^2 - 6x - 7 = 0

To write the quadratic as a product of linear factors, we'll have to determine the roots of the quadratic.

We'll apply the quadratic formula:

x1= [-b+sqrt(b^2 - 4ac)]/2a

a,b,c, are the coefficients of the quadratic:

x1 = [6+sqrt(36 + 112)]/8

x1 = (6+sqrt148)/8

x1 = (6+2sqrt37)/8

x1 = (3+sqrt37)/4

x2 = (3-sqrt37)/4

We'll write the quadratic as a product of linear factors:

**12x^2 - 18x - 21 = 3([x-(3+sqrt37)/4]*[x-(3-sqrt37)/4]**