Solve 15b^2-b-2=

$$15 b ^ { 2 } - b - 2 =$$

Factor

$$\left(5b-2\right)\left(3b+1\right)$$

Solution Steps

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

$$p+q=-1 pq=15\left(-2\right)=-30$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$p=-6 q=5$$

Rewrite 15b^{2}-b-2 as \left(15b^{2}-6b\right)+\left(5b-2\right).

$$\left(15b^{2}-6b\right)+\left(5b-2\right)$$

Factor out 3b in 15b^{2}-6b.

$$3b\left(5b-2\right)+5b-2$$

Factor out common term 5b-2 by using distributive property.

$$\left(5b-2\right)\left(3b+1\right)$$

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.

$$15b^{2}-b-2=0$$

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.

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

Multiply -4 times 15.

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

Multiply -60 times -2.

$$b=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 15}$$

Add 1 to 120.

$$b=\frac{-\left(-1\right)±\sqrt{121}}{2\times 15}$$

Take the square root of 121.

$$b=\frac{-\left(-1\right)±11}{2\times 15}$$

The opposite of -1 is 1.

$$b=\frac{1±11}{2\times 15}$$

Multiply 2 times 15.

$$b=\frac{1±11}{30}$$

Now solve the equation b=\frac{1±11}{30} when ± is plus. Add 1 to 11.

$$b=\frac{12}{30}$$

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

$$b=\frac{2}{5}$$

Now solve the equation b=\frac{1±11}{30} when ± is minus. Subtract 11 from 1.

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

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

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

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

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

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

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

Subtract \frac{2}{5} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Multiply \frac{5b-2}{5} times \frac{3b+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 5 times 3.

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

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

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

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

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

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

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

Two numbers r and s sum up to \frac{1}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{15} = \frac{1}{30}. 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>

$$r = \frac{1}{30} - u s = \frac{1}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{15}

$$(\frac{1}{30} - u) (\frac{1}{30} + u) = -\frac{2}{15}$$

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

$$\frac{1}{900} - u^2 = -\frac{2}{15}$$

Simplify the expression by subtracting \frac{1}{900} on both sides

$$-u^2 = -\frac{2}{15}-\frac{1}{900} = -\frac{121}{900}$$

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

$$u^2 = \frac{121}{900} u = \pm\sqrt{\frac{121}{900}} = \pm \frac{11}{30}$$

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

$$r =\frac{1}{30} - \frac{11}{30} = -0.333 s = \frac{1}{30} + \frac{11}{30} = 0.400$$

Evaluate

$$\left(5b-2\right)\left(3b+1\right)$$