Solve -3p^2+15p-18

$$- 3 p ^ { 2 } + 15 p - 18$$

Factor

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

Solution Steps

Factor out 3.

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

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

$$a+b=5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

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

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

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

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

Factor out common term p-3 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$-3p^{2}+15p-18=0$$

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.

$$p=\frac{-15±\sqrt{15^{2}-4\left(-3\right)\left(-18\right)}}{2\left(-3\right)}$$

Square 15.

$$p=\frac{-15±\sqrt{225-4\left(-3\right)\left(-18\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$p=\frac{-15±\sqrt{225+12\left(-18\right)}}{2\left(-3\right)}$$

Multiply 12 times -18.

$$p=\frac{-15±\sqrt{225-216}}{2\left(-3\right)}$$

Add 225 to -216.

$$p=\frac{-15±\sqrt{9}}{2\left(-3\right)}$$

Take the square root of 9.

$$p=\frac{-15±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$p=\frac{-15±3}{-6}$$

Now solve the equation p=\frac{-15±3}{-6} when ± is plus. Add -15 to 3.

$$p=-\frac{12}{-6}$$

Divide -12 by -6.

$$p=2$$

Now solve the equation p=\frac{-15±3}{-6} when ± is minus. Subtract 3 from -15.

$$p=-\frac{18}{-6}$$

Divide -18 by -6.

$$p=3$$

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

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

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.

$$x ^ 2 -5x +6 = 0$$

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

$$r + s = 5 rs = 6$$

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

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 6

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = 6$$

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

$$\frac{25}{4} - u^2 = 6$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = 6-\frac{25}{4} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{5}{2} - \frac{1}{2} = 2 s = \frac{5}{2} + \frac{1}{2} = 3$$

Evaluate

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