Solve 12x^2+3x-15

$$12 x ^ { 2 } + 3 x - 15$$

Factor

$$3\left(x-1\right)\left(4x+5\right)$$

Solution Steps

Factor out 3.

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

Consider 4x^{2}+x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=1 ab=4\left(-5\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 4x^{2}+x-5 as \left(4x^{2}-4x\right)+\left(5x-5\right).

$$\left(4x^{2}-4x\right)+\left(5x-5\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(4x+5\right)$$

Rewrite the complete factored expression.

$$3\left(x-1\right)\left(4x+5\right)$$

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.

$$12x^{2}+3x-15=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 12\left(-15\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-3±\sqrt{9-48\left(-15\right)}}{2\times 12}$$

Multiply -48 times -15.

$$x=\frac{-3±\sqrt{9+720}}{2\times 12}$$

Add 9 to 720.

$$x=\frac{-3±\sqrt{729}}{2\times 12}$$

Take the square root of 729.

$$x=\frac{-3±27}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-3±27}{24}$$

Now solve the equation x=\frac{-3±27}{24} when ± is plus. Add -3 to 27.

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

Divide 24 by 24.

$$x=1$$

Now solve the equation x=\frac{-3±27}{24} when ± is minus. Subtract 27 from -3.

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

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

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

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

$$12x^{2}+3x-15=12\left(x-1\right)\left(x-\left(-\frac{5}{4}\right)\right)$$

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

$$12x^{2}+3x-15=12\left(x-1\right)\left(x+\frac{5}{4}\right)$$

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

$$12x^{2}+3x-15=12\left(x-1\right)\times \frac{4x+5}{4}$$

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

$$12x^{2}+3x-15=3\left(x-1\right)\left(4x+5\right)$$

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

$$x ^ 2 +\frac{1}{4}x -\frac{5}{4} = 0$$

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

$$r + s = -\frac{1}{4} rs = -\frac{5}{4}$$

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

$$r = -\frac{1}{8} - u s = -\frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{4}

$$(-\frac{1}{8} - u) (-\frac{1}{8} + u) = -\frac{5}{4}$$

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

$$\frac{1}{64} - u^2 = -\frac{5}{4}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -\frac{5}{4}-\frac{1}{64} = -\frac{81}{64}$$

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

$$u^2 = \frac{81}{64} u = \pm\sqrt{\frac{81}{64}} = \pm \frac{9}{8}$$

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

$$r =-\frac{1}{8} - \frac{9}{8} = -1.250 s = -\frac{1}{8} + \frac{9}{8} = 1$$

Evaluate

$$3\left(x-1\right)\left(4x+5\right)$$