Solve z^2+14z+24

$$z ^ { 2 } + 14 z + 24$$

Factor

$$\left(z+2\right)\left(z+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz+24. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 24=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 14.

$$a=2 b=12$$

Rewrite z^{2}+14z+24 as \left(z^{2}+2z\right)+\left(12z+24\right).

$$\left(z^{2}+2z\right)+\left(12z+24\right)$$

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

$$z\left(z+2\right)+12\left(z+2\right)$$

Factor out common term z+2 by using distributive property.

$$\left(z+2\right)\left(z+12\right)$$

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.

$$z^{2}+14z+24=0$$

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.

$$z=\frac{-14±\sqrt{14^{2}-4\times 24}}{2}$$

Square 14.

$$z=\frac{-14±\sqrt{196-4\times 24}}{2}$$

Multiply -4 times 24.

$$z=\frac{-14±\sqrt{196-96}}{2}$$

Add 196 to -96.

$$z=\frac{-14±\sqrt{100}}{2}$$

Take the square root of 100.

$$z=\frac{-14±10}{2}$$

Now solve the equation z=\frac{-14±10}{2} when ± is plus. Add -14 to 10.

$$z=-\frac{4}{2}$$

Divide -4 by 2.

$$z=-2$$

Now solve the equation z=\frac{-14±10}{2} when ± is minus. Subtract 10 from -14.

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

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 -2 for x_{1} and -12 for x_{2}.

$$z^{2}+14z+24=\left(z-\left(-2\right)\right)\left(z-\left(-12\right)\right)$$

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

$$z^{2}+14z+24=\left(z+2\right)\left(z+12\right)$$

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.

$$x ^ 2 +14x +24 = 0$$

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

$$r + s = -14 rs = 24$$

Two numbers r and s sum up to -14 exactly when the average of the two numbers is \frac{1}{2}*-14 = -7. 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>

$$r = -7 - u s = -7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 24

$$(-7 - u) (-7 + u) = 24$$

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

$$49 - u^2 = 24$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 24-49 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =-7 - 5 = -12 s = -7 + 5 = -2$$

Evaluate

$$\left(z+2\right)\left(z+12\right)$$