Solve 8x^2-6x-9

$$8 x ^ { 2 } - 6 x - 9$$

Factor

$$\left(2x-3\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=8\left(-9\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -6.

$$a=-12 b=6$$

Rewrite 8x^{2}-6x-9 as \left(8x^{2}-12x\right)+\left(6x-9\right).

$$\left(8x^{2}-12x\right)+\left(6x-9\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(2x-3\right)+3\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(4x+3\right)$$

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.

$$8x^{2}-6x-9=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\left(-9\right)}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\left(-9\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\left(-9\right)}}{2\times 8}$$

Multiply -32 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+288}}{2\times 8}$$

Add 36 to 288.

$$x=\frac{-\left(-6\right)±\sqrt{324}}{2\times 8}$$

Take the square root of 324.

$$x=\frac{-\left(-6\right)±18}{2\times 8}$$

The opposite of -6 is 6.

$$x=\frac{6±18}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±18}{16}$$

Now solve the equation x=\frac{6±18}{16} when ± is plus. Add 6 to 18.

$$x=\frac{24}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{6±18}{16} when ± is minus. Subtract 18 from 6.

$$x=-\frac{12}{16}$$

Reduce the fraction \frac{-12}{16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -\frac{3}{4} for x_{2}.

$$8x^{2}-6x-9=8\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$8x^{2}-6x-9=8\left(x-\frac{3}{2}\right)\left(x+\frac{3}{4}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-6x-9=8\times \frac{2x-3}{2}\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-6x-9=8\times \frac{2x-3}{2}\times \frac{4x+3}{4}$$

Multiply \frac{2x-3}{2} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-6x-9=8\times \frac{\left(2x-3\right)\left(4x+3\right)}{2\times 4}$$

Multiply 2 times 4.

$$8x^{2}-6x-9=8\times \frac{\left(2x-3\right)\left(4x+3\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8x^{2}-6x-9=\left(2x-3\right)\left(4x+3\right)$$

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 8

$$x ^ 2 -\frac{3}{4}x -\frac{9}{8} = 0$$

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

$$r + s = \frac{3}{4} rs = -\frac{9}{8}$$

Two numbers r and s sum up to \frac{3}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{4} = \frac{3}{8}. 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>

$$r = \frac{3}{8} - u s = \frac{3}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{8}

$$(\frac{3}{8} - u) (\frac{3}{8} + u) = -\frac{9}{8}$$

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

$$\frac{9}{64} - u^2 = -\frac{9}{8}$$

Simplify the expression by subtracting \frac{9}{64} on both sides

$$-u^2 = -\frac{9}{8}-\frac{9}{64} = -\frac{81}{64}$$

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

$$u^2 = \frac{81}{64} u = \pm\sqrt{\frac{81}{64}} = \pm \frac{9}{8}$$

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

$$r =\frac{3}{8} - \frac{9}{8} = -0.750 s = \frac{3}{8} + \frac{9}{8} = 1.500$$

Evaluate

$$\left(2x-3\right)\left(4x+3\right)$$