Solve 12u^2+9u-3

$$12 u ^ { 2 } + 9 u - 3$$

Factor

$$3\left(4u-1\right)\left(u+1\right)$$

Solution Steps

Factor out 3.

$$3\left(4u^{2}+3u-1\right)$$

Consider 4u^{2}+3u-1. Factor the expression by grouping. First, the expression needs to be rewritten as 4u^{2}+au+bu-1. To find a and b, set up a system to be solved.

$$a+b=3 ab=4\left(-1\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 4u^{2}+3u-1 as \left(4u^{2}-u\right)+\left(4u-1\right).

$$\left(4u^{2}-u\right)+\left(4u-1\right)$$

Factor out u in 4u^{2}-u.

$$u\left(4u-1\right)+4u-1$$

Factor out common term 4u-1 by using distributive property.

$$\left(4u-1\right)\left(u+1\right)$$

Rewrite the complete factored expression.

$$3\left(4u-1\right)\left(u+1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$12u^{2}+9u-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$u=\frac{-9±\sqrt{9^{2}-4\times 12\left(-3\right)}}{2\times 12}$$

Square 9.

$$u=\frac{-9±\sqrt{81-4\times 12\left(-3\right)}}{2\times 12}$$

Multiply -4 times 12.

$$u=\frac{-9±\sqrt{81-48\left(-3\right)}}{2\times 12}$$

Multiply -48 times -3.

$$u=\frac{-9±\sqrt{81+144}}{2\times 12}$$

Add 81 to 144.

$$u=\frac{-9±\sqrt{225}}{2\times 12}$$

Take the square root of 225.

$$u=\frac{-9±15}{2\times 12}$$

Multiply 2 times 12.

$$u=\frac{-9±15}{24}$$

Now solve the equation u=\frac{-9±15}{24} when ± is plus. Add -9 to 15.

$$u=\frac{6}{24}$$

Reduce the fraction \frac{6}{24} to lowest terms by extracting and canceling out 6.

$$u=\frac{1}{4}$$

Now solve the equation u=\frac{-9±15}{24} when ± is minus. Subtract 15 from -9.

$$u=-\frac{24}{24}$$

Divide -24 by 24.

$$u=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and -1 for x_{2}.

$$12u^{2}+9u-3=12\left(u-\frac{1}{4}\right)\left(u-\left(-1\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12u^{2}+9u-3=12\left(u-\frac{1}{4}\right)\left(u+1\right)$$

Subtract \frac{1}{4} from u by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12u^{2}+9u-3=12\times \frac{4u-1}{4}\left(u+1\right)$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12u^{2}+9u-3=3\left(4u-1\right)\left(u+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 12

$$x ^ 2 +\frac{3}{4}x -\frac{1}{4} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{3}{4} rs = -\frac{1}{4}$$

Two numbers r and s sum up to -\frac{3}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{4} = -\frac{3}{8}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{3}{8} - u s = -\frac{3}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}

$$(-\frac{3}{8} - u) (-\frac{3}{8} + u) = -\frac{1}{4}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{64} - u^2 = -\frac{1}{4}$$

Simplify the expression by subtracting \frac{9}{64} on both sides

$$-u^2 = -\frac{1}{4}-\frac{9}{64} = -\frac{25}{64}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{64} u = \pm\sqrt{\frac{25}{64}} = \pm \frac{5}{8}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{3}{8} - \frac{5}{8} = -1 s = -\frac{3}{8} + \frac{5}{8} = 0.250$$

Evaluate

$$3\left(4u-1\right)\left(u+1\right)$$