Solve z^2+8z-48

$$z ^ { 2 } + 8 z - 48$$

Factor

$$\left(z-4\right)\left(z+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-48. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-48\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite z^{2}+8z-48 as \left(z^{2}-4z\right)+\left(12z-48\right).

$$\left(z^{2}-4z\right)+\left(12z-48\right)$$

Factor out z in the first and 12 in the second group.

$$z\left(z-4\right)+12\left(z-4\right)$$

Factor out common term z-4 by using distributive property.

$$\left(z-4\right)\left(z+12\right)$$

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.

$$z^{2}+8z-48=0$$

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.

$$z=\frac{-8±\sqrt{8^{2}-4\left(-48\right)}}{2}$$

Square 8.

$$z=\frac{-8±\sqrt{64-4\left(-48\right)}}{2}$$

Multiply -4 times -48.

$$z=\frac{-8±\sqrt{64+192}}{2}$$

Add 64 to 192.

$$z=\frac{-8±\sqrt{256}}{2}$$

Take the square root of 256.

$$z=\frac{-8±16}{2}$$

Now solve the equation z=\frac{-8±16}{2} when ± is plus. Add -8 to 16.

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

Divide 8 by 2.

$$z=4$$

Now solve the equation z=\frac{-8±16}{2} when ± is minus. Subtract 16 from -8.

$$z=-\frac{24}{2}$$

Divide -24 by 2.

$$z=-12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -12 for x_{2}.

$$z^{2}+8z-48=\left(z-4\right)\left(z-\left(-12\right)\right)$$

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

$$z^{2}+8z-48=\left(z-4\right)\left(z+12\right)$$

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.

$$x ^ 2 +8x -48 = 0$$

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

$$r + s = -8 rs = -48$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -48

$$(-4 - u) (-4 + u) = -48$$

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

$$16 - u^2 = -48$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -48-16 = -64$$

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

$$u^2 = 64 u = \pm\sqrt{64} = \pm 8$$

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

$$r =-4 - 8 = -12 s = -4 + 8 = 4$$

Evaluate

$$\left(z-4\right)\left(z+12\right)$$