Solve 9z^2+18z-16

$$9 z ^ { 2 } + 18 z - 16$$

Factor

$$\left(3z-2\right)\left(3z+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9z^{2}+az+bz-16. To find a and b, set up a system to be solved.

$$a+b=18 ab=9\left(-16\right)=-144$$

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 -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 18.

$$a=-6 b=24$$

Rewrite 9z^{2}+18z-16 as \left(9z^{2}-6z\right)+\left(24z-16\right).

$$\left(9z^{2}-6z\right)+\left(24z-16\right)$$

Factor out 3z in the first and 8 in the second group.

$$3z\left(3z-2\right)+8\left(3z-2\right)$$

Factor out common term 3z-2 by using distributive property.

$$\left(3z-2\right)\left(3z+8\right)$$

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.

$$9z^{2}+18z-16=0$$

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.

$$z=\frac{-18±\sqrt{18^{2}-4\times 9\left(-16\right)}}{2\times 9}$$

Square 18.

$$z=\frac{-18±\sqrt{324-4\times 9\left(-16\right)}}{2\times 9}$$

Multiply -4 times 9.

$$z=\frac{-18±\sqrt{324-36\left(-16\right)}}{2\times 9}$$

Multiply -36 times -16.

$$z=\frac{-18±\sqrt{324+576}}{2\times 9}$$

Add 324 to 576.

$$z=\frac{-18±\sqrt{900}}{2\times 9}$$

Take the square root of 900.

$$z=\frac{-18±30}{2\times 9}$$

Multiply 2 times 9.

$$z=\frac{-18±30}{18}$$

Now solve the equation z=\frac{-18±30}{18} when ± is plus. Add -18 to 30.

$$z=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$z=\frac{2}{3}$$

Now solve the equation z=\frac{-18±30}{18} when ± is minus. Subtract 30 from -18.

$$z=-\frac{48}{18}$$

Reduce the fraction \frac{-48}{18} to lowest terms by extracting and canceling out 6.

$$z=-\frac{8}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and -\frac{8}{3} for x_{2}.

$$9z^{2}+18z-16=9\left(z-\frac{2}{3}\right)\left(z-\left(-\frac{8}{3}\right)\right)$$

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

$$9z^{2}+18z-16=9\left(z-\frac{2}{3}\right)\left(z+\frac{8}{3}\right)$$

Subtract \frac{2}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9z^{2}+18z-16=9\times \frac{3z-2}{3}\left(z+\frac{8}{3}\right)$$

Add \frac{8}{3} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9z^{2}+18z-16=9\times \frac{3z-2}{3}\times \frac{3z+8}{3}$$

Multiply \frac{3z-2}{3} times \frac{3z+8}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9z^{2}+18z-16=9\times \frac{\left(3z-2\right)\left(3z+8\right)}{3\times 3}$$

Multiply 3 times 3.

$$9z^{2}+18z-16=9\times \frac{\left(3z-2\right)\left(3z+8\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9z^{2}+18z-16=\left(3z-2\right)\left(3z+8\right)$$

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 9

$$x ^ 2 +2x -\frac{16}{9} = 0$$

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

$$r + s = -2 rs = -\frac{16}{9}$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{9}

$$(-1 - u) (-1 + u) = -\frac{16}{9}$$

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

$$1 - u^2 = -\frac{16}{9}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{16}{9}-1 = -\frac{25}{9}$$

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

$$u^2 = \frac{25}{9} u = \pm\sqrt{\frac{25}{9}} = \pm \frac{5}{3}$$

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

$$r =-1 - \frac{5}{3} = -2.667 s = -1 + \frac{5}{3} = 0.667$$

Evaluate

$$\left(3z-2\right)\left(3z+8\right)$$