Solve 6b^2-27b-15

$$6 b ^ { 2 } - 27 b - 15$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(2b^{2}-9b-5\right)$$

Consider 2b^{2}-9b-5. Factor the expression by grouping. First, the expression needs to be rewritten as 2b^{2}+pb+qb-5. To find p and q, set up a system to be solved.

$$p+q=-9 pq=2\left(-5\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$p=-10 q=1$$

Rewrite 2b^{2}-9b-5 as \left(2b^{2}-10b\right)+\left(b-5\right).

$$\left(2b^{2}-10b\right)+\left(b-5\right)$$

Factor out 2b in 2b^{2}-10b.

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

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

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

Rewrite the complete factored expression.

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

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.

$$6b^{2}-27b-15=0$$

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.

$$b=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 6\left(-15\right)}}{2\times 6}$$

Square -27.

$$b=\frac{-\left(-27\right)±\sqrt{729-4\times 6\left(-15\right)}}{2\times 6}$$

Multiply -4 times 6.

$$b=\frac{-\left(-27\right)±\sqrt{729-24\left(-15\right)}}{2\times 6}$$

Multiply -24 times -15.

$$b=\frac{-\left(-27\right)±\sqrt{729+360}}{2\times 6}$$

Add 729 to 360.

$$b=\frac{-\left(-27\right)±\sqrt{1089}}{2\times 6}$$

Take the square root of 1089.

$$b=\frac{-\left(-27\right)±33}{2\times 6}$$

The opposite of -27 is 27.

$$b=\frac{27±33}{2\times 6}$$

Multiply 2 times 6.

$$b=\frac{27±33}{12}$$

Now solve the equation b=\frac{27±33}{12} when ± is plus. Add 27 to 33.

$$b=\frac{60}{12}$$

Divide 60 by 12.

$$b=5$$

Now solve the equation b=\frac{27±33}{12} when ± is minus. Subtract 33 from 27.

$$b=-\frac{6}{12}$$

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

$$b=-\frac{1}{2}$$

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

$$6b^{2}-27b-15=6\left(b-5\right)\left(b-\left(-\frac{1}{2}\right)\right)$$

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

$$6b^{2}-27b-15=6\left(b-5\right)\left(b+\frac{1}{2}\right)$$

Add \frac{1}{2} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6b^{2}-27b-15=6\left(b-5\right)\times \frac{2b+1}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6b^{2}-27b-15=3\left(b-5\right)\left(2b+1\right)$$

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 6

$$x ^ 2 -\frac{9}{2}x -\frac{5}{2} = 0$$

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

$$r + s = \frac{9}{2} rs = -\frac{5}{2}$$

Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{4}. 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>

$$r = \frac{9}{4} - u s = \frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(\frac{9}{4} - u) (\frac{9}{4} + u) = -\frac{5}{2}$$

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

$$\frac{81}{16} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = -\frac{5}{2}-\frac{81}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =\frac{9}{4} - \frac{11}{4} = -0.500 s = \frac{9}{4} + \frac{11}{4} = 5$$

Evaluate

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