Solve 4b^2-9b-9

$$4 b ^ { 2 } - 9 b - 9$$

Factor

$$\left(b-3\right)\left(4b+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4b^{2}+pb+qb-9. To find p and q, set up a system to be solved.

$$p+q=-9 pq=4\left(-9\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$p=-12 q=3$$

Rewrite 4b^{2}-9b-9 as \left(4b^{2}-12b\right)+\left(3b-9\right).

$$\left(4b^{2}-12b\right)+\left(3b-9\right)$$

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

$$4b\left(b-3\right)+3\left(b-3\right)$$

Factor out common term b-3 by using distributive property.

$$\left(b-3\right)\left(4b+3\right)$$

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.

$$4b^{2}-9b-9=0$$

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.

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

Square -9.

$$b=\frac{-\left(-9\right)±\sqrt{81-4\times 4\left(-9\right)}}{2\times 4}$$

Multiply -4 times 4.

$$b=\frac{-\left(-9\right)±\sqrt{81-16\left(-9\right)}}{2\times 4}$$

Multiply -16 times -9.

$$b=\frac{-\left(-9\right)±\sqrt{81+144}}{2\times 4}$$

Add 81 to 144.

$$b=\frac{-\left(-9\right)±\sqrt{225}}{2\times 4}$$

Take the square root of 225.

$$b=\frac{-\left(-9\right)±15}{2\times 4}$$

The opposite of -9 is 9.

$$b=\frac{9±15}{2\times 4}$$

Multiply 2 times 4.

$$b=\frac{9±15}{8}$$

Now solve the equation b=\frac{9±15}{8} when ± is plus. Add 9 to 15.

$$b=\frac{24}{8}$$

Divide 24 by 8.

$$b=3$$

Now solve the equation b=\frac{9±15}{8} when ± is minus. Subtract 15 from 9.

$$b=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

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

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

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

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

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

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

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

Cancel out 4, the greatest common factor in 4 and 4.

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

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 4

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

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

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

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

$$r = \frac{9}{8} - u s = \frac{9}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{4}

$$(\frac{9}{8} - u) (\frac{9}{8} + u) = -\frac{9}{4}$$

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

$$\frac{81}{64} - u^2 = -\frac{9}{4}$$

Simplify the expression by subtracting \frac{81}{64} on both sides

$$-u^2 = -\frac{9}{4}-\frac{81}{64} = -\frac{225}{64}$$

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

$$u^2 = \frac{225}{64} u = \pm\sqrt{\frac{225}{64}} = \pm \frac{15}{8}$$

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

$$r =\frac{9}{8} - \frac{15}{8} = -0.750 s = \frac{9}{8} + \frac{15}{8} = 3$$

Evaluate

$$\left(b-3\right)\left(4b+3\right)$$