Solve z^2-19z-150

$$z ^ { 2 } - 19 z - 150$$

Factor

$$\left(z-25\right)\left(z+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-150. To find a and b, set up a system to be solved.

$$a+b=-19 ab=1\left(-150\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -19.

$$a=-25 b=6$$

Rewrite z^{2}-19z-150 as \left(z^{2}-25z\right)+\left(6z-150\right).

$$\left(z^{2}-25z\right)+\left(6z-150\right)$$

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

$$z\left(z-25\right)+6\left(z-25\right)$$

Factor out common term z-25 by using distributive property.

$$\left(z-25\right)\left(z+6\right)$$

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.

$$z^{2}-19z-150=0$$

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.

$$z=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\left(-150\right)}}{2}$$

Square -19.

$$z=\frac{-\left(-19\right)±\sqrt{361-4\left(-150\right)}}{2}$$

Multiply -4 times -150.

$$z=\frac{-\left(-19\right)±\sqrt{361+600}}{2}$$

Add 361 to 600.

$$z=\frac{-\left(-19\right)±\sqrt{961}}{2}$$

Take the square root of 961.

$$z=\frac{-\left(-19\right)±31}{2}$$

The opposite of -19 is 19.

$$z=\frac{19±31}{2}$$

Now solve the equation z=\frac{19±31}{2} when ± is plus. Add 19 to 31.

$$z=\frac{50}{2}$$

Divide 50 by 2.

$$z=25$$

Now solve the equation z=\frac{19±31}{2} when ± is minus. Subtract 31 from 19.

$$z=-\frac{12}{2}$$

Divide -12 by 2.

$$z=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 25 for x_{1} and -6 for x_{2}.

$$z^{2}-19z-150=\left(z-25\right)\left(z-\left(-6\right)\right)$$

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

$$z^{2}-19z-150=\left(z-25\right)\left(z+6\right)$$

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.

$$x ^ 2 -19x -150 = 0$$

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

$$r + s = 19 rs = -150$$

Two numbers r and s sum up to 19 exactly when the average of the two numbers is \frac{1}{2}*19 = \frac{19}{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>

$$r = \frac{19}{2} - u s = \frac{19}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -150

$$(\frac{19}{2} - u) (\frac{19}{2} + u) = -150$$

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

$$\frac{361}{4} - u^2 = -150$$

Simplify the expression by subtracting \frac{361}{4} on both sides

$$-u^2 = -150-\frac{361}{4} = -\frac{961}{4}$$

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

$$u^2 = \frac{961}{4} u = \pm\sqrt{\frac{961}{4}} = \pm \frac{31}{2}$$

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

$$r =\frac{19}{2} - \frac{31}{2} = -6 s = \frac{19}{2} + \frac{31}{2} = 25$$

Evaluate

$$\left(z-25\right)\left(z+6\right)$$