Solve 15x^2+16x-15=

$$15 x ^ { 2 } + 16 x - 15 =$$

Factor

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

Solution Steps

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

$$a+b=16 ab=15\left(-15\right)=-225$$

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 -225.

$$-1,225 -3,75 -5,45 -9,25 -15,15$$

Calculate the sum for each pair.

$$-1+225=224 -3+75=72 -5+45=40 -9+25=16 -15+15=0$$

The solution is the pair that gives sum 16.

$$a=-9 b=25$$

Rewrite 15x^{2}+16x-15 as \left(15x^{2}-9x\right)+\left(25x-15\right).

$$\left(15x^{2}-9x\right)+\left(25x-15\right)$$

Factor out 3x in the first and 5 in the second group.

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

Factor out common term 5x-3 by using distributive property.

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

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.

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

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.

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

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 15\left(-15\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-16±\sqrt{256-60\left(-15\right)}}{2\times 15}$$

Multiply -60 times -15.

$$x=\frac{-16±\sqrt{256+900}}{2\times 15}$$

Add 256 to 900.

$$x=\frac{-16±\sqrt{1156}}{2\times 15}$$

Take the square root of 1156.

$$x=\frac{-16±34}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-16±34}{30}$$

Now solve the equation x=\frac{-16±34}{30} when ± is plus. Add -16 to 34.

$$x=\frac{18}{30}$$

Reduce the fraction \frac{18}{30} to lowest terms by extracting and canceling out 6.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-16±34}{30} when ± is minus. Subtract 34 from -16.

$$x=-\frac{50}{30}$$

Reduce the fraction \frac{-50}{30} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{3}$$

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

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

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

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

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Multiply 5 times 3.

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

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

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

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

$$x ^ 2 +\frac{16}{15}x -1 = 0$$

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

$$r + s = -\frac{16}{15} rs = -1$$

Two numbers r and s sum up to -\frac{16}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{15} = -\frac{8}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -1

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

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

$$\frac{64}{225} - u^2 = -1$$

Simplify the expression by subtracting \frac{64}{225} on both sides

$$-u^2 = -1-\frac{64}{225} = -\frac{289}{225}$$

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

$$u^2 = \frac{289}{225} u = \pm\sqrt{\frac{289}{225}} = \pm \frac{17}{15}$$

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

$$r =-\frac{8}{15} - \frac{17}{15} = -1.667 s = -\frac{8}{15} + \frac{17}{15} = 0.600$$

Evaluate

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