Solve 9b^2+12b-32

$$9 b ^ { 2 } + 12 b - 32$$

Factor

$$\left(3b-4\right)\left(3b+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9b^{2}+pb+qb-32. To find p and q, set up a system to be solved.

$$p+q=12 pq=9\left(-32\right)=-288$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 12.

$$p=-12 q=24$$

Rewrite 9b^{2}+12b-32 as \left(9b^{2}-12b\right)+\left(24b-32\right).

$$\left(9b^{2}-12b\right)+\left(24b-32\right)$$

Factor out 3b in the first and 8 in the second group.

$$3b\left(3b-4\right)+8\left(3b-4\right)$$

Factor out common term 3b-4 by using distributive property.

$$\left(3b-4\right)\left(3b+8\right)$$

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.

$$9b^{2}+12b-32=0$$

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.

$$b=\frac{-12±\sqrt{12^{2}-4\times 9\left(-32\right)}}{2\times 9}$$

Square 12.

$$b=\frac{-12±\sqrt{144-4\times 9\left(-32\right)}}{2\times 9}$$

Multiply -4 times 9.

$$b=\frac{-12±\sqrt{144-36\left(-32\right)}}{2\times 9}$$

Multiply -36 times -32.

$$b=\frac{-12±\sqrt{144+1152}}{2\times 9}$$

Add 144 to 1152.

$$b=\frac{-12±\sqrt{1296}}{2\times 9}$$

Take the square root of 1296.

$$b=\frac{-12±36}{2\times 9}$$

Multiply 2 times 9.

$$b=\frac{-12±36}{18}$$

Now solve the equation b=\frac{-12±36}{18} when ± is plus. Add -12 to 36.

$$b=\frac{24}{18}$$

Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation b=\frac{-12±36}{18} when ± is minus. Subtract 36 from -12.

$$b=-\frac{48}{18}$$

Reduce the fraction \frac{-48}{18} to lowest terms by extracting and canceling out 6.

$$b=-\frac{8}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{8}{3} for x_{2}.

$$9b^{2}+12b-32=9\left(b-\frac{4}{3}\right)\left(b-\left(-\frac{8}{3}\right)\right)$$

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

$$9b^{2}+12b-32=9\left(b-\frac{4}{3}\right)\left(b+\frac{8}{3}\right)$$

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

$$9b^{2}+12b-32=9\times \frac{3b-4}{3}\left(b+\frac{8}{3}\right)$$

Add \frac{8}{3} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9b^{2}+12b-32=9\times \frac{3b-4}{3}\times \frac{3b+8}{3}$$

Multiply \frac{3b-4}{3} times \frac{3b+8}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9b^{2}+12b-32=9\times \frac{\left(3b-4\right)\left(3b+8\right)}{3\times 3}$$

Multiply 3 times 3.

$$9b^{2}+12b-32=9\times \frac{\left(3b-4\right)\left(3b+8\right)}{9}$$

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

$$9b^{2}+12b-32=\left(3b-4\right)\left(3b+8\right)$$

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

$$x ^ 2 +\frac{4}{3}x -\frac{32}{9} = 0$$

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

$$r + s = -\frac{4}{3} rs = -\frac{32}{9}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{32}{9}

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -\frac{32}{9}$$

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

$$\frac{4}{9} - u^2 = -\frac{32}{9}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = -\frac{32}{9}-\frac{4}{9} = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =-\frac{2}{3} - 2 = -2.667 s = -\frac{2}{3} + 2 = 1.333$$

Evaluate

$$\left(3b-4\right)\left(3b+8\right)$$