Solve -12x^2-28x+5

$$- 12 x ^ { 2 } - 28 x + 5$$

Factor

$$\left(1-6x\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-28 ab=-12\times 5=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -28.

$$a=2 b=-30$$

Rewrite -12x^{2}-28x+5 as \left(-12x^{2}+2x\right)+\left(-30x+5\right).

$$\left(-12x^{2}+2x\right)+\left(-30x+5\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(-6x+1\right)+5\left(-6x+1\right)$$

Factor out common term -6x+1 by using distributive property.

$$\left(-6x+1\right)\left(2x+5\right)$$

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.

$$-12x^{2}-28x+5=0$$

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.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\left(-12\right)\times 5}}{2\left(-12\right)}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\left(-12\right)\times 5}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-28\right)±\sqrt{784+48\times 5}}{2\left(-12\right)}$$

Multiply 48 times 5.

$$x=\frac{-\left(-28\right)±\sqrt{784+240}}{2\left(-12\right)}$$

Add 784 to 240.

$$x=\frac{-\left(-28\right)±\sqrt{1024}}{2\left(-12\right)}$$

Take the square root of 1024.

$$x=\frac{-\left(-28\right)±32}{2\left(-12\right)}$$

The opposite of -28 is 28.

$$x=\frac{28±32}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{28±32}{-24}$$

Now solve the equation x=\frac{28±32}{-24} when ± is plus. Add 28 to 32.

$$x=\frac{60}{-24}$$

Reduce the fraction \frac{60}{-24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{28±32}{-24} when ± is minus. Subtract 32 from 28.

$$x=-\frac{4}{-24}$$

Reduce the fraction \frac{-4}{-24} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{2} for x_{1} and \frac{1}{6} for x_{2}.

$$-12x^{2}-28x+5=-12\left(x-\left(-\frac{5}{2}\right)\right)\left(x-\frac{1}{6}\right)$$

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

$$-12x^{2}-28x+5=-12\left(x+\frac{5}{2}\right)\left(x-\frac{1}{6}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}-28x+5=-12\times \frac{-2x-5}{-2}\left(x-\frac{1}{6}\right)$$

Subtract \frac{1}{6} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}-28x+5=-12\times \frac{-2x-5}{-2}\times \frac{-6x+1}{-6}$$

Multiply \frac{-2x-5}{-2} times \frac{-6x+1}{-6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}-28x+5=-12\times \frac{\left(-2x-5\right)\left(-6x+1\right)}{-2\left(-6\right)}$$

Multiply -2 times -6.

$$-12x^{2}-28x+5=-12\times \frac{\left(-2x-5\right)\left(-6x+1\right)}{12}$$

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

$$-12x^{2}-28x+5=-\left(-2x-5\right)\left(-6x+1\right)$$

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.

$$x ^ 2 +\frac{7}{3}x -\frac{5}{12} = 0$$

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

$$r + s = -\frac{7}{3} rs = -\frac{5}{12}$$

Two numbers r and s sum up to -\frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{3} = -\frac{7}{6}. 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>

$$r = -\frac{7}{6} - u s = -\frac{7}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{12}

$$(-\frac{7}{6} - u) (-\frac{7}{6} + u) = -\frac{5}{12}$$

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

$$\frac{49}{36} - u^2 = -\frac{5}{12}$$

Simplify the expression by subtracting \frac{49}{36} on both sides

$$-u^2 = -\frac{5}{12}-\frac{49}{36} = -\frac{16}{9}$$

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.

$$r =-\frac{7}{6} - \frac{4}{3} = -2.500 s = -\frac{7}{6} + \frac{4}{3} = 0.167$$

Evaluate

$$\left(1-6x\right)\left(2x+5\right)$$