Solve 15x^2-34x-16

$$15 x ^ { 2 } - 34 x - 16$$

Factor

$$\left(3x-8\right)\left(5x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-34 ab=15\left(-16\right)=-240$$

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 -240.

$$1,-240 2,-120 3,-80 4,-60 5,-48 6,-40 8,-30 10,-24 12,-20 15,-16$$

Calculate the sum for each pair.

$$1-240=-239 2-120=-118 3-80=-77 4-60=-56 5-48=-43 6-40=-34 8-30=-22 10-24=-14 12-20=-8 15-16=-1$$

The solution is the pair that gives sum -34.

$$a=-40 b=6$$

Rewrite 15x^{2}-34x-16 as \left(15x^{2}-40x\right)+\left(6x-16\right).

$$\left(15x^{2}-40x\right)+\left(6x-16\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(3x-8\right)+2\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(5x+2\right)$$

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.

$$15x^{2}-34x-16=0$$

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.

$$x=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 15\left(-16\right)}}{2\times 15}$$

Square -34.

$$x=\frac{-\left(-34\right)±\sqrt{1156-4\times 15\left(-16\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-34\right)±\sqrt{1156-60\left(-16\right)}}{2\times 15}$$

Multiply -60 times -16.

$$x=\frac{-\left(-34\right)±\sqrt{1156+960}}{2\times 15}$$

Add 1156 to 960.

$$x=\frac{-\left(-34\right)±\sqrt{2116}}{2\times 15}$$

Take the square root of 2116.

$$x=\frac{-\left(-34\right)±46}{2\times 15}$$

The opposite of -34 is 34.

$$x=\frac{34±46}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{34±46}{30}$$

Now solve the equation x=\frac{34±46}{30} when ± is plus. Add 34 to 46.

$$x=\frac{80}{30}$$

Reduce the fraction \frac{80}{30} to lowest terms by extracting and canceling out 10.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{34±46}{30} when ± is minus. Subtract 46 from 34.

$$x=-\frac{12}{30}$$

Reduce the fraction \frac{-12}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and -\frac{2}{5} for x_{2}.

$$15x^{2}-34x-16=15\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{2}{5}\right)\right)$$

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

$$15x^{2}-34x-16=15\left(x-\frac{8}{3}\right)\left(x+\frac{2}{5}\right)$$

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-34x-16=15\times \frac{3x-8}{3}\left(x+\frac{2}{5}\right)$$

Add \frac{2}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-34x-16=15\times \frac{3x-8}{3}\times \frac{5x+2}{5}$$

Multiply \frac{3x-8}{3} times \frac{5x+2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-34x-16=15\times \frac{\left(3x-8\right)\left(5x+2\right)}{3\times 5}$$

Multiply 3 times 5.

$$15x^{2}-34x-16=15\times \frac{\left(3x-8\right)\left(5x+2\right)}{15}$$

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

$$15x^{2}-34x-16=\left(3x-8\right)\left(5x+2\right)$$

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

$$x ^ 2 -\frac{34}{15}x -\frac{16}{15} = 0$$

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

$$r + s = \frac{34}{15} rs = -\frac{16}{15}$$

Two numbers r and s sum up to \frac{34}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{34}{15} = \frac{17}{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>

$$r = \frac{17}{15} - u s = \frac{17}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{15}

$$(\frac{17}{15} - u) (\frac{17}{15} + u) = -\frac{16}{15}$$

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

$$\frac{289}{225} - u^2 = -\frac{16}{15}$$

Simplify the expression by subtracting \frac{289}{225} on both sides

$$-u^2 = -\frac{16}{15}-\frac{289}{225} = -\frac{529}{225}$$

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

$$u^2 = \frac{529}{225} u = \pm\sqrt{\frac{529}{225}} = \pm \frac{23}{15}$$

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

$$r =\frac{17}{15} - \frac{23}{15} = -0.400 s = \frac{17}{15} + \frac{23}{15} = 2.667$$

Evaluate

$$\left(3x-8\right)\left(5x+2\right)$$