Solve 3p^2-3p-90

$$3 p ^ { 2 } - 3 p - 90$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(p^{2}-p-30\right)$$

Consider p^{2}-p-30. Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp-30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite p^{2}-p-30 as \left(p^{2}-6p\right)+\left(5p-30\right).

$$\left(p^{2}-6p\right)+\left(5p-30\right)$$

Factor out p in the first and 5 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$3p^{2}-3p-90=0$$

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.

$$p=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-90\right)}}{2\times 3}$$

Square -3.

$$p=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-90\right)}}{2\times 3}$$

Multiply -4 times 3.

$$p=\frac{-\left(-3\right)±\sqrt{9-12\left(-90\right)}}{2\times 3}$$

Multiply -12 times -90.

$$p=\frac{-\left(-3\right)±\sqrt{9+1080}}{2\times 3}$$

Add 9 to 1080.

$$p=\frac{-\left(-3\right)±\sqrt{1089}}{2\times 3}$$

Take the square root of 1089.

$$p=\frac{-\left(-3\right)±33}{2\times 3}$$

The opposite of -3 is 3.

$$p=\frac{3±33}{2\times 3}$$

Multiply 2 times 3.

$$p=\frac{3±33}{6}$$

Now solve the equation p=\frac{3±33}{6} when ± is plus. Add 3 to 33.

$$p=\frac{36}{6}$$

Divide 36 by 6.

$$p=6$$

Now solve the equation p=\frac{3±33}{6} when ± is minus. Subtract 33 from 3.

$$p=-\frac{30}{6}$$

Divide -30 by 6.

$$p=-5$$

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

$$3p^{2}-3p-90=3\left(p-6\right)\left(p-\left(-5\right)\right)$$

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

$$3p^{2}-3p-90=3\left(p-6\right)\left(p+5\right)$$

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 3

$$x ^ 2 -1x -30 = 0$$

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

$$r + s = 1 rs = -30$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -30

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -30$$

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

$$\frac{1}{4} - u^2 = -30$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -30-\frac{1}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =\frac{1}{2} - \frac{11}{2} = -5 s = \frac{1}{2} + \frac{11}{2} = 6$$

Evaluate

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