Solve 15x^2+2x-1

$$15 x ^ { 2 } + 2 x - 1$$

Factor

$$\left(5x-1\right)\left(3x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=15\left(-1\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 15x^{2}+2x-1 as \left(15x^{2}-3x\right)+\left(5x-1\right).

$$\left(15x^{2}-3x\right)+\left(5x-1\right)$$

Factor out 3x in 15x^{2}-3x.

$$3x\left(5x-1\right)+5x-1$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(3x+1\right)$$

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.

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

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 15\left(-1\right)}}{2\times 15}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 15\left(-1\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-2±\sqrt{4-60\left(-1\right)}}{2\times 15}$$

Multiply -60 times -1.

$$x=\frac{-2±\sqrt{4+60}}{2\times 15}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2\times 15}$$

Take the square root of 64.

$$x=\frac{-2±8}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-2±8}{30}$$

Now solve the equation x=\frac{-2±8}{30} when ± is plus. Add -2 to 8.

$$x=\frac{6}{30}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-2±8}{30} when ± is minus. Subtract 8 from -2.

$$x=-\frac{10}{30}$$

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

$$x=-\frac{1}{3}$$

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

$$15x^{2}+2x-1=15\left(x-\frac{1}{5}\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$15x^{2}+2x-1=15\left(x-\frac{1}{5}\right)\left(x+\frac{1}{3}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+2x-1=15\times \frac{5x-1}{5}\left(x+\frac{1}{3}\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+2x-1=15\times \frac{5x-1}{5}\times \frac{3x+1}{3}$$

Multiply \frac{5x-1}{5} times \frac{3x+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}+2x-1=15\times \frac{\left(5x-1\right)\left(3x+1\right)}{5\times 3}$$

Multiply 5 times 3.

$$15x^{2}+2x-1=15\times \frac{\left(5x-1\right)\left(3x+1\right)}{15}$$

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

$$15x^{2}+2x-1=\left(5x-1\right)\left(3x+1\right)$$

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

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

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

$$r + s = -\frac{2}{15} rs = -\frac{1}{15}$$

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{1}{15}

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

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

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

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

$$-u^2 = -\frac{1}{15}-\frac{1}{225} = -\frac{16}{225}$$

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

$$u^2 = \frac{16}{225} u = \pm\sqrt{\frac{16}{225}} = \pm \frac{4}{15}$$

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

$$r =-\frac{1}{15} - \frac{4}{15} = -0.333 s = -\frac{1}{15} + \frac{4}{15} = 0.200$$

Evaluate

$$\left(5x-1\right)\left(3x+1\right)$$