# Factor the equation: 21x^2 + 22x = 8

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### 2 Answers

A simpler method involving the quadratic equation may be used.

`21x^2+22x=8`

`21x^2+22x-8=0`

a = 21 ; b = 22 ; c = -8

Quadratice Equation: `x=(-b+-sqrt(b^2-4ac))/(2a)`

Substituting in the a, b, and c values:

`x=(-22+-sqrt(22^2-4(21)(-8)))/(2*21)`

`x=(-22+-sqrt(1156))/42`

`x=(-22+34)/42` or `x=(-22-34)/42`

Upon simplification, `x=2/7 , x= -4/3`

These are the two values that make `21x^2+22x-8=0`

In order to find the factors, we try to get all the values to one side so that the expression can equal zero.

`x=2/7 => 7x=2 => 7x-2=0`

`x=-4/3 => 3x=-4 => 3x+4=0`

**Therefore, factoring the original equation results in `(7x-2)(3x+4)=0` **

You should complete the square `21x^2 + 22x` using the following formula, such that:

`a^2 + 2ab + b^2 = (a + b)^2`

Considering `a^2 = 21x^2` and `2ab = 22x` yields:

`a = x*sqrt 21`

You need to complete the square such that:

`21x^2 + 22x + 121/21 = 8 + 121/21`

`(sqrt 21*x + 11/sqrt 21)^2 = (168 + 121)/21`

`(sqrt 21*x + 11/sqrt 21)^2 - 289/21 = 0`

Converting the difference of squares into a product yields:

`(sqrt 21*x + 11/sqrt 21 - 17/sqrt21)(sqrt 21*x + 11/sqrt 21 + 17/sqrt21) = 0`

`(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0`

**Hence, evaluating the factored form of the given equation yields `(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0` .**