Solve 12x^2+5x-25

$$12 x ^ { 2 } + 5 x - 25$$

Factor

$$\left(4x-5\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=5 ab=12\left(-25\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 5.

$$a=-15 b=20$$

Rewrite 12x^{2}+5x-25 as \left(12x^{2}-15x\right)+\left(20x-25\right).

$$\left(12x^{2}-15x\right)+\left(20x-25\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(4x-5\right)+5\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(3x+5\right)$$

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.

$$12x^{2}+5x-25=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 12\left(-25\right)}}{2\times 12}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 12\left(-25\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-5±\sqrt{25-48\left(-25\right)}}{2\times 12}$$

Multiply -48 times -25.

$$x=\frac{-5±\sqrt{25+1200}}{2\times 12}$$

Add 25 to 1200.

$$x=\frac{-5±\sqrt{1225}}{2\times 12}$$

Take the square root of 1225.

$$x=\frac{-5±35}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-5±35}{24}$$

Now solve the equation x=\frac{-5±35}{24} when ± is plus. Add -5 to 35.

$$x=\frac{30}{24}$$

Reduce the fraction \frac{30}{24} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{-5±35}{24} when ± is minus. Subtract 35 from -5.

$$x=-\frac{40}{24}$$

Reduce the fraction \frac{-40}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and -\frac{5}{3} for x_{2}.

$$12x^{2}+5x-25=12\left(x-\frac{5}{4}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$12x^{2}+5x-25=12\left(x-\frac{5}{4}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+5x-25=12\times \frac{4x-5}{4}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+5x-25=12\times \frac{4x-5}{4}\times \frac{3x+5}{3}$$

Multiply \frac{4x-5}{4} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+5x-25=12\times \frac{\left(4x-5\right)\left(3x+5\right)}{4\times 3}$$

Multiply 4 times 3.

$$12x^{2}+5x-25=12\times \frac{\left(4x-5\right)\left(3x+5\right)}{12}$$

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

$$12x^{2}+5x-25=\left(4x-5\right)\left(3x+5\right)$$

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

$$x ^ 2 +\frac{5}{12}x -\frac{25}{12} = 0$$

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

$$r + s = -\frac{5}{12} rs = -\frac{25}{12}$$

Two numbers r and s sum up to -\frac{5}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{12} = -\frac{5}{24}. 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>

$$r = -\frac{5}{24} - u s = -\frac{5}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{25}{12}

$$(-\frac{5}{24} - u) (-\frac{5}{24} + u) = -\frac{25}{12}$$

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

$$\frac{25}{576} - u^2 = -\frac{25}{12}$$

Simplify the expression by subtracting \frac{25}{576} on both sides

$$-u^2 = -\frac{25}{12}-\frac{25}{576} = -\frac{1225}{576}$$

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

$$u^2 = \frac{1225}{576} u = \pm\sqrt{\frac{1225}{576}} = \pm \frac{35}{24}$$

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

$$r =-\frac{5}{24} - \frac{35}{24} = -1.667 s = -\frac{5}{24} + \frac{35}{24} = 1.250$$

Evaluate

$$\left(4x-5\right)\left(3x+5\right)$$