Solve z^2-7z-44

$$z ^ { 2 } - 7 z - 44$$

Factor

$$\left(z-11\right)\left(z+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-44. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\left(-44\right)=-44$$

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 -44.

$$1,-44 2,-22 4,-11$$

Calculate the sum for each pair.

$$1-44=-43 2-22=-20 4-11=-7$$

The solution is the pair that gives sum -7.

$$a=-11 b=4$$

Rewrite z^{2}-7z-44 as \left(z^{2}-11z\right)+\left(4z-44\right).

$$\left(z^{2}-11z\right)+\left(4z-44\right)$$

Factor out z in the first and 4 in the second group.

$$z\left(z-11\right)+4\left(z-11\right)$$

Factor out common term z-11 by using distributive property.

$$\left(z-11\right)\left(z+4\right)$$

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.

$$z^{2}-7z-44=0$$

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.

$$z=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-44\right)}}{2}$$

Square -7.

$$z=\frac{-\left(-7\right)±\sqrt{49-4\left(-44\right)}}{2}$$

Multiply -4 times -44.

$$z=\frac{-\left(-7\right)±\sqrt{49+176}}{2}$$

Add 49 to 176.

$$z=\frac{-\left(-7\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$z=\frac{-\left(-7\right)±15}{2}$$

The opposite of -7 is 7.

$$z=\frac{7±15}{2}$$

Now solve the equation z=\frac{7±15}{2} when ± is plus. Add 7 to 15.

$$z=\frac{22}{2}$$

Divide 22 by 2.

$$z=11$$

Now solve the equation z=\frac{7±15}{2} when ± is minus. Subtract 15 from 7.

$$z=-\frac{8}{2}$$

Divide -8 by 2.

$$z=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and -4 for x_{2}.

$$z^{2}-7z-44=\left(z-11\right)\left(z-\left(-4\right)\right)$$

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

$$z^{2}-7z-44=\left(z-11\right)\left(z+4\right)$$

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.

$$x ^ 2 -7x -44 = 0$$

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

$$r + s = 7 rs = -44$$

Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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>

$$r = \frac{7}{2} - u s = \frac{7}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -44

$$(\frac{7}{2} - u) (\frac{7}{2} + u) = -44$$

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

$$\frac{49}{4} - u^2 = -44$$

Simplify the expression by subtracting \frac{49}{4} on both sides

$$-u^2 = -44-\frac{49}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =\frac{7}{2} - \frac{15}{2} = -4 s = \frac{7}{2} + \frac{15}{2} = 11$$

Evaluate

$$\left(z-11\right)\left(z+4\right)$$