Solve -3u^2-4u+4

$$- 3 u ^ { 2 } - 4 u + 4$$

Factor

$$-\left(3u-2\right)\left(u+2\right)$$

Solution Steps

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

$$a+b=-4 ab=-3\times 4=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=2 b=-6$$

Rewrite -3u^{2}-4u+4 as \left(-3u^{2}+2u\right)+\left(-6u+4\right).

$$\left(-3u^{2}+2u\right)+\left(-6u+4\right)$$

Factor out -u in the first and -2 in the second group.

$$-u\left(3u-2\right)-2\left(3u-2\right)$$

Factor out common term 3u-2 by using distributive property.

$$\left(3u-2\right)\left(-u-2\right)$$

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.

$$-3u^{2}-4u+4=0$$

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.

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

Square -4.

$$u=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$u=\frac{-\left(-4\right)±\sqrt{16+12\times 4}}{2\left(-3\right)}$$

Multiply 12 times 4.

$$u=\frac{-\left(-4\right)±\sqrt{16+48}}{2\left(-3\right)}$$

Add 16 to 48.

$$u=\frac{-\left(-4\right)±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$u=\frac{-\left(-4\right)±8}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$u=\frac{4±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$u=\frac{4±8}{-6}$$

Now solve the equation u=\frac{4±8}{-6} when ± is plus. Add 4 to 8.

$$u=\frac{12}{-6}$$

Divide 12 by -6.

$$u=-2$$

Now solve the equation u=\frac{4±8}{-6} when ± is minus. Subtract 8 from 4.

$$u=-\frac{4}{-6}$$

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out 2.

$$u=\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and \frac{2}{3} for x_{2}.

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

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

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

Subtract \frac{2}{3} from u by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 3, the greatest common factor in -3 and 3.

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

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.

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

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

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

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

$$r = -\frac{2}{3} - u s = -\frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{3}

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -\frac{4}{3}$$

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

$$\frac{4}{9} - u^2 = -\frac{4}{3}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = -\frac{4}{3}-\frac{4}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-\frac{2}{3} - \frac{4}{3} = -2 s = -\frac{2}{3} + \frac{4}{3} = 0.667$$

Evaluate

$$-\left(3u-2\right)\left(u+2\right)$$