Solve -3q^2+3q+90

$$- 3 q ^ { 2 } + 3 q + 90$$

Factor

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

Solution Steps

Factor out 3.

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

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

$$a+b=1 ab=-30=-30$$

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

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

Factor out -q in the first and -5 in the second group.

$$-q\left(q-6\right)-5\left(q-6\right)$$

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

$$\left(q-6\right)\left(-q-5\right)$$

Rewrite the complete factored expression.

$$3\left(q-6\right)\left(-q-5\right)$$

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.

$$-3q^{2}+3q+90=0$$

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.

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

Square 3.

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

Multiply -4 times -3.

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

Multiply 12 times 90.

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

Add 9 to 1080.

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

Take the square root of 1089.

$$q=\frac{-3±33}{2\left(-3\right)}$$

Multiply 2 times -3.

$$q=\frac{-3±33}{-6}$$

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

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

Divide 30 by -6.

$$q=-5$$

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

$$q=-\frac{36}{-6}$$

Divide -36 by -6.

$$q=6$$

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

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

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

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

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.

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

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

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