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

Asked on by kikiri

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

21x^2+22x= 8.

Actually we do not factor an equation. We solve an equation for the value of x which will satisfy the equation. We factor an expression if it is possible to write it as the product of two expressions.

Therefore 21x^2+22x-8 = 0.

21x^2 +28x -6x-8 = 0

7x(3x+4) -2(3x+4) = 0

(3x+4)(7x-2) = 0

Equate the factors to zero.

3x+4 = 0 or 7x-2 = 0

3x+4 = 0 gives 3x = -4, or x = -4/3

7x-2 = 0 gives 7x= 2, or x= 2/7.

Therefore x = -4/3, or x = 2/7 are the solution.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll move all terms to one side:

21x^2 + 22x - 8 = 0

We'll use the quadratic formula to factorize the equation:

21x^2+22x-8=0

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

We'll identify the coefficients a,b,c:

a = 21

b = 22

c = -8

b^2 - 4ac = 484 + 672

sqrt (b^2 - 4ac) = sqrt 1156

sqrt (b^2 - 4ac) = 34

x1 = (-22+34)/2*21

x1 = 12/42

x1 = 6/21

x2 =  (-22-34)/2*21

x2 = -56/42

x2 = -28/21

We can now factorize the quadratic:

21x^2+22x-8 = 21(x - x1)(x - x2)

We'll substitute x1 and x2

21x^2+22x-8 = 21(x - 6/21)(x + 28/21)

or

21x^2+22x-8 = (21x - 6)(21x + 28)

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