Consider the following quadratic equation:

12x^2=−2x
Step 1 of 2 : Using the standard form ax2+bx+c=0
of the given quadratic equation, factor the left hand side of the equation into two linear factors

1 answer

To factorize the left hand side of the equation, we need to write the quadratic expression in the form (px + q)(rx + s) where p, q, r, and s are constants.

Given equation:

12x^2 = -2x

To factorize, let's first simplify the equation by dividing both sides by the greatest common factor, which is 2:

6x^2 = -x

Now we can see that the equation already has only one term on both sides. To factorize, we need to write it as a product of two linear factors.

Let's try the two linear factors to be (ax + b)(cx + d):

(6x^2) = (-2x)
(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd

We can see that for the coefficient of x^2, ac must be equal to 6.

Let's try a = 2 and c = 3:

(2x + b)(3x + d) = 6x^2 + (6d + 3b)x + bd

By comparing the coefficients, we can obtain:

6d + 3b = -1 (Coefficient of the x term)
bd = 0 (Constant term)

To satisfy the first equation, one possibility is to let b = 0 and d = -1/6. However, this would make bd = 0, not -1.

Let's try another possibility:

Let b = -1 and d = 1/6:

(2x - 1)(3x + 1/6) = 6x^2 + (3/6 - 2/6)x - 1/6

Simplifying the equation:

(2x - 1)(3x + 1/6) = 6x^2 + (1/6)x - 1/6

Now we can rewrite the quadratic equation in the factored form:

(2x - 1)(3x + 1/6) = 0

So the factored form of the left hand side of the equation is (2x - 1)(3x + 1/6).