Solve 2z^2+2z-60

$$2 z ^ { 2 } + 2 z - 60$$

Factor

$$2\left(z-5\right)\left(z+6\right)$$

Solution Steps

Factor out 2.

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

Consider z^{2}+z-30. Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-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 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.

$$a=-5 b=6$$

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

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

Factor out z in the first and 6 in the second group.

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

Factor out common term z-5 by using distributive property.

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

Rewrite the complete factored expression.

$$2\left(z-5\right)\left(z+6\right)$$

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.

$$2z^{2}+2z-60=0$$

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.

$$z=\frac{-2±\sqrt{2^{2}-4\times 2\left(-60\right)}}{2\times 2}$$

Square 2.

$$z=\frac{-2±\sqrt{4-4\times 2\left(-60\right)}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-2±\sqrt{4-8\left(-60\right)}}{2\times 2}$$

Multiply -8 times -60.

$$z=\frac{-2±\sqrt{4+480}}{2\times 2}$$

Add 4 to 480.

$$z=\frac{-2±\sqrt{484}}{2\times 2}$$

Take the square root of 484.

$$z=\frac{-2±22}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{-2±22}{4}$$

Now solve the equation z=\frac{-2±22}{4} when ± is plus. Add -2 to 22.

$$z=\frac{20}{4}$$

Divide 20 by 4.

$$z=5$$

Now solve the equation z=\frac{-2±22}{4} when ± is minus. Subtract 22 from -2.

$$z=-\frac{24}{4}$$

Divide -24 by 4.

$$z=-6$$

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}.

$$2z^{2}+2z-60=2\left(z-5\right)\left(z-\left(-6\right)\right)$$

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

$$2z^{2}+2z-60=2\left(z-5\right)\left(z+6\right)$$

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.This is achieved by dividing both sides of the equation by 2

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

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.

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

Evaluate

$$2\left(z-5\right)\left(z+6\right)$$