Solve 15z^2+2z-8

$$15 z ^ { 2 } + 2 z - 8$$

Factor

$$\left(3z-2\right)\left(5z+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15z^{2}+az+bz-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=15\left(-8\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 2.

$$a=-10 b=12$$

Rewrite 15z^{2}+2z-8 as \left(15z^{2}-10z\right)+\left(12z-8\right).

$$\left(15z^{2}-10z\right)+\left(12z-8\right)$$

Factor out 5z in the first and 4 in the second group.

$$5z\left(3z-2\right)+4\left(3z-2\right)$$

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

$$\left(3z-2\right)\left(5z+4\right)$$

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.

$$15z^{2}+2z-8=0$$

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.

$$z=\frac{-2±\sqrt{2^{2}-4\times 15\left(-8\right)}}{2\times 15}$$

Square 2.

$$z=\frac{-2±\sqrt{4-4\times 15\left(-8\right)}}{2\times 15}$$

Multiply -4 times 15.

$$z=\frac{-2±\sqrt{4-60\left(-8\right)}}{2\times 15}$$

Multiply -60 times -8.

$$z=\frac{-2±\sqrt{4+480}}{2\times 15}$$

Add 4 to 480.

$$z=\frac{-2±\sqrt{484}}{2\times 15}$$

Take the square root of 484.

$$z=\frac{-2±22}{2\times 15}$$

Multiply 2 times 15.

$$z=\frac{-2±22}{30}$$

Now solve the equation z=\frac{-2±22}{30} when ± is plus. Add -2 to 22.

$$z=\frac{20}{30}$$

Reduce the fraction \frac{20}{30} to lowest terms by extracting and canceling out 10.

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

Now solve the equation z=\frac{-2±22}{30} when ± is minus. Subtract 22 from -2.

$$z=-\frac{24}{30}$$

Reduce the fraction \frac{-24}{30} to lowest terms by extracting and canceling out 6.

$$z=-\frac{4}{5}$$

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{4}{5} for x_{2}.

$$15z^{2}+2z-8=15\left(z-\frac{2}{3}\right)\left(z-\left(-\frac{4}{5}\right)\right)$$

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

$$15z^{2}+2z-8=15\left(z-\frac{2}{3}\right)\left(z+\frac{4}{5}\right)$$

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

$$15z^{2}+2z-8=15\times \frac{3z-2}{3}\left(z+\frac{4}{5}\right)$$

Add \frac{4}{5} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15z^{2}+2z-8=15\times \frac{3z-2}{3}\times \frac{5z+4}{5}$$

Multiply \frac{3z-2}{3} times \frac{5z+4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15z^{2}+2z-8=15\times \frac{\left(3z-2\right)\left(5z+4\right)}{3\times 5}$$

Multiply 3 times 5.

$$15z^{2}+2z-8=15\times \frac{\left(3z-2\right)\left(5z+4\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15z^{2}+2z-8=\left(3z-2\right)\left(5z+4\right)$$

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.This is achieved by dividing both sides of the equation by 15

$$x ^ 2 +\frac{2}{15}x -\frac{8}{15} = 0$$

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

$$r + s = -\frac{2}{15} rs = -\frac{8}{15}$$

Two numbers r and s sum up to -\frac{2}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{15} = -\frac{1}{15}. 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>

$$r = -\frac{1}{15} - u s = -\frac{1}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{15}

$$(-\frac{1}{15} - u) (-\frac{1}{15} + u) = -\frac{8}{15}$$

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

$$\frac{1}{225} - u^2 = -\frac{8}{15}$$

Simplify the expression by subtracting \frac{1}{225} on both sides

$$-u^2 = -\frac{8}{15}-\frac{1}{225} = -\frac{121}{225}$$

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

$$u^2 = \frac{121}{225} u = \pm\sqrt{\frac{121}{225}} = \pm \frac{11}{15}$$

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

$$r =-\frac{1}{15} - \frac{11}{15} = -0.800 s = -\frac{1}{15} + \frac{11}{15} = 0.667$$

Evaluate

$$\left(3z-2\right)\left(5z+4\right)$$