Solve 12p^2+8p-15

$$12 p ^ { 2 } + 8 p - 15$$

Factor

$$\left(6p-5\right)\left(2p+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12p^{2}+ap+bp-15. To find a and b, set up a system to be solved.

$$a+b=8 ab=12\left(-15\right)=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 8.

$$a=-10 b=18$$

Rewrite 12p^{2}+8p-15 as \left(12p^{2}-10p\right)+\left(18p-15\right).

$$\left(12p^{2}-10p\right)+\left(18p-15\right)$$

Factor out 2p in the first and 3 in the second group.

$$2p\left(6p-5\right)+3\left(6p-5\right)$$

Factor out common term 6p-5 by using distributive property.

$$\left(6p-5\right)\left(2p+3\right)$$

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.

$$12p^{2}+8p-15=0$$

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.

$$p=\frac{-8±\sqrt{8^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

Square 8.

$$p=\frac{-8±\sqrt{64-4\times 12\left(-15\right)}}{2\times 12}$$

Multiply -4 times 12.

$$p=\frac{-8±\sqrt{64-48\left(-15\right)}}{2\times 12}$$

Multiply -48 times -15.

$$p=\frac{-8±\sqrt{64+720}}{2\times 12}$$

Add 64 to 720.

$$p=\frac{-8±\sqrt{784}}{2\times 12}$$

Take the square root of 784.

$$p=\frac{-8±28}{2\times 12}$$

Multiply 2 times 12.

$$p=\frac{-8±28}{24}$$

Now solve the equation p=\frac{-8±28}{24} when ± is plus. Add -8 to 28.

$$p=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$p=\frac{5}{6}$$

Now solve the equation p=\frac{-8±28}{24} when ± is minus. Subtract 28 from -8.

$$p=-\frac{36}{24}$$

Reduce the fraction \frac{-36}{24} to lowest terms by extracting and canceling out 12.

$$p=-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and -\frac{3}{2} for x_{2}.

$$12p^{2}+8p-15=12\left(p-\frac{5}{6}\right)\left(p-\left(-\frac{3}{2}\right)\right)$$

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

$$12p^{2}+8p-15=12\left(p-\frac{5}{6}\right)\left(p+\frac{3}{2}\right)$$

Subtract \frac{5}{6} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12p^{2}+8p-15=12\times \frac{6p-5}{6}\left(p+\frac{3}{2}\right)$$

Add \frac{3}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12p^{2}+8p-15=12\times \frac{6p-5}{6}\times \frac{2p+3}{2}$$

Multiply \frac{6p-5}{6} times \frac{2p+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12p^{2}+8p-15=12\times \frac{\left(6p-5\right)\left(2p+3\right)}{6\times 2}$$

Multiply 6 times 2.

$$12p^{2}+8p-15=12\times \frac{\left(6p-5\right)\left(2p+3\right)}{12}$$

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

$$12p^{2}+8p-15=\left(6p-5\right)\left(2p+3\right)$$

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

$$x ^ 2 +\frac{2}{3}x -\frac{5}{4} = 0$$

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

$$r + s = -\frac{2}{3} rs = -\frac{5}{4}$$

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

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

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

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -\frac{5}{4}$$

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

$$\frac{1}{9} - u^2 = -\frac{5}{4}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{5}{4}-\frac{1}{9} = -\frac{49}{36}$$

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

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

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

$$r =-\frac{1}{3} - \frac{7}{6} = -1.500 s = -\frac{1}{3} + \frac{7}{6} = 0.833$$

Evaluate

$$\left(6p-5\right)\left(2p+3\right)$$