Solve 12x^2+28x-5=

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

Factor

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

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.

$$a+b=28 ab=12\left(-5\right)=-60$$

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 -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.

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

Square 28.

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

Multiply -4 times 12.

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

Multiply -48 times -5.

$$x=\frac{-28±\sqrt{784+240}}{2\times 12}$$

Add 784 to 240.

$$x=\frac{-28±\sqrt{1024}}{2\times 12}$$

Take the square root of 1024.

$$x=\frac{-28±32}{2\times 12}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply 6 times 2.

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

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

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

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

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

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

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