Solve B(x)=9x^2-6x-15

$$B ( x ) = 9 x ^ { 2 } - 6 x - 15$$

Factor

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

Solution Steps

Factor out 3.

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

Consider 3x^{2}-2x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

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

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

Factor out x in 3x^{2}-5x.

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

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

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

Rewrite the complete factored expression.

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

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.

$$9x^{2}-6x-15=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-15\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-15\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-15\right)}}{2\times 9}$$

Multiply -36 times -15.

$$x=\frac{-\left(-6\right)±\sqrt{36+540}}{2\times 9}$$

Add 36 to 540.

$$x=\frac{-\left(-6\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{-\left(-6\right)±24}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±24}{18}$$

Now solve the equation x=\frac{6±24}{18} when ± is plus. Add 6 to 24.

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

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

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

Now solve the equation x=\frac{6±24}{18} when ± is minus. Subtract 24 from 6.

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

Divide -18 by 18.

$$x=-1$$

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

$$9x^{2}-6x-15=9\left(x-\frac{5}{3}\right)\left(x-\left(-1\right)\right)$$

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

$$9x^{2}-6x-15=9\left(x-\frac{5}{3}\right)\left(x+1\right)$$

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

$$9x^{2}-6x-15=9\times \frac{3x-5}{3}\left(x+1\right)$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9x^{2}-6x-15=3\left(3x-5\right)\left(x+1\right)$$

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 9

$$x ^ 2 -\frac{2}{3}x -\frac{5}{3} = 0$$

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

$$r + s = \frac{2}{3} rs = -\frac{5}{3}$$

Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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>

$$r = \frac{1}{3} - u s = \frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = -\frac{5}{3}$$

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

$$\frac{1}{9} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{5}{3}-\frac{1}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =\frac{1}{3} - \frac{4}{3} = -1.000 s = \frac{1}{3} + \frac{4}{3} = 1.667$$

Evaluate

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