Solve -18p^2-15p+12

$$- 18 p ^ { 2 } - 15 p + 12$$

Factor

$$3\left(-3p-4\right)\left(2p-1\right)$$

Solution Steps

Factor out 3.

$$3\left(-6p^{2}-5p+4\right)$$

Consider -6p^{2}-5p+4. Factor the expression by grouping. First, the expression needs to be rewritten as -6p^{2}+ap+bp+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6\times 4=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=3 b=-8$$

Rewrite -6p^{2}-5p+4 as \left(-6p^{2}+3p\right)+\left(-8p+4\right).

$$\left(-6p^{2}+3p\right)+\left(-8p+4\right)$$

Factor out -3p in the first and -4 in the second group.

$$-3p\left(2p-1\right)-4\left(2p-1\right)$$

Factor out common term 2p-1 by using distributive property.

$$\left(2p-1\right)\left(-3p-4\right)$$

Rewrite the complete factored expression.

$$3\left(2p-1\right)\left(-3p-4\right)$$

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.

$$-18p^{2}-15p+12=0$$

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.

$$p=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-18\right)\times 12}}{2\left(-18\right)}$$

Square -15.

$$p=\frac{-\left(-15\right)±\sqrt{225-4\left(-18\right)\times 12}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$p=\frac{-\left(-15\right)±\sqrt{225+72\times 12}}{2\left(-18\right)}$$

Multiply 72 times 12.

$$p=\frac{-\left(-15\right)±\sqrt{225+864}}{2\left(-18\right)}$$

Add 225 to 864.

$$p=\frac{-\left(-15\right)±\sqrt{1089}}{2\left(-18\right)}$$

Take the square root of 1089.

$$p=\frac{-\left(-15\right)±33}{2\left(-18\right)}$$

The opposite of -15 is 15.

$$p=\frac{15±33}{2\left(-18\right)}$$

Multiply 2 times -18.

$$p=\frac{15±33}{-36}$$

Now solve the equation p=\frac{15±33}{-36} when ± is plus. Add 15 to 33.

$$p=\frac{48}{-36}$$

Reduce the fraction \frac{48}{-36} to lowest terms by extracting and canceling out 12.

$$p=-\frac{4}{3}$$

Now solve the equation p=\frac{15±33}{-36} when ± is minus. Subtract 33 from 15.

$$p=-\frac{18}{-36}$$

Reduce the fraction \frac{-18}{-36} to lowest terms by extracting and canceling out 18.

$$p=\frac{1}{2}$$

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

$$-18p^{2}-15p+12=-18\left(p-\left(-\frac{4}{3}\right)\right)\left(p-\frac{1}{2}\right)$$

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

$$-18p^{2}-15p+12=-18\left(p+\frac{4}{3}\right)\left(p-\frac{1}{2}\right)$$

Add \frac{4}{3} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-18p^{2}-15p+12=-18\times \frac{-3p-4}{-3}\left(p-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-18p^{2}-15p+12=-18\times \frac{-3p-4}{-3}\times \frac{-2p+1}{-2}$$

Multiply \frac{-3p-4}{-3} times \frac{-2p+1}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-18p^{2}-15p+12=-18\times \frac{\left(-3p-4\right)\left(-2p+1\right)}{-3\left(-2\right)}$$

Multiply -3 times -2.

$$-18p^{2}-15p+12=-18\times \frac{\left(-3p-4\right)\left(-2p+1\right)}{6}$$

Cancel out 6, the greatest common factor in -18 and 6.

$$-18p^{2}-15p+12=-3\left(-3p-4\right)\left(-2p+1\right)$$

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.

$$x ^ 2 +\frac{5}{6}x -\frac{2}{3} = 0$$

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

$$r + s = -\frac{5}{6} rs = -\frac{2}{3}$$

Two numbers r and s sum up to -\frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{6} = -\frac{5}{12}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}

$$(-\frac{5}{12} - u) (-\frac{5}{12} + u) = -\frac{2}{3}$$

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

$$\frac{25}{144} - u^2 = -\frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{144} on both sides

$$-u^2 = -\frac{2}{3}-\frac{25}{144} = -\frac{121}{144}$$

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

$$u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}$$

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

$$r =-\frac{5}{12} - \frac{11}{12} = -1.333 s = -\frac{5}{12} + \frac{11}{12} = 0.500$$

Evaluate

$$12-15p-18p^{2}$$