Solve 15w^2-2w-1=

$$15 w ^ { 2 } - 2 w - 1 =$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15w^{2}+aw+bw-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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-5 b=3$$

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

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

Factor out 5w in 15w^{2}-5w.

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

Factor out common term 3w-1 by using distributive property.

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

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.

$$15w^{2}-2w-1=0$$

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.

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

Square -2.

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

Multiply -4 times 15.

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

Multiply -60 times -1.

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

Add 4 to 60.

$$w=\frac{-\left(-2\right)±\sqrt{64}}{2\times 15}$$

Take the square root of 64.

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

The opposite of -2 is 2.

$$w=\frac{2±8}{2\times 15}$$

Multiply 2 times 15.

$$w=\frac{2±8}{30}$$

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

$$w=\frac{10}{30}$$

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

$$w=\frac{1}{3}$$

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

$$w=-\frac{6}{30}$$

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

$$w=-\frac{1}{5}$$

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

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

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

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

Subtract \frac{1}{3} from w by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Add \frac{1}{5} to w by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Multiply 3 times 5.

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

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

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

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.

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

Evaluate

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