Solve 22b^2-124b-48

$$22 b ^ { 2 } - 124 b - 48$$

Factor

$$2\left(b-6\right)\left(11b+4\right)$$

Solution Steps

Factor out 2.

$$2\left(11b^{2}-62b-24\right)$$

Consider 11b^{2}-62b-24. Factor the expression by grouping. First, the expression needs to be rewritten as 11b^{2}+pb+qb-24. To find p and q, set up a system to be solved.

$$p+q=-62 pq=11\left(-24\right)=-264$$

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 -264.

$$1,-264 2,-132 3,-88 4,-66 6,-44 8,-33 11,-24 12,-22$$

Calculate the sum for each pair.

$$1-264=-263 2-132=-130 3-88=-85 4-66=-62 6-44=-38 8-33=-25 11-24=-13 12-22=-10$$

The solution is the pair that gives sum -62.

$$p=-66 q=4$$

Rewrite 11b^{2}-62b-24 as \left(11b^{2}-66b\right)+\left(4b-24\right).

$$\left(11b^{2}-66b\right)+\left(4b-24\right)$$

Factor out 11b in the first and 4 in the second group.

$$11b\left(b-6\right)+4\left(b-6\right)$$

Factor out common term b-6 by using distributive property.

$$\left(b-6\right)\left(11b+4\right)$$

Rewrite the complete factored expression.

$$2\left(b-6\right)\left(11b+4\right)$$

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.

$$22b^{2}-124b-48=0$$

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.

$$b=\frac{-\left(-124\right)±\sqrt{\left(-124\right)^{2}-4\times 22\left(-48\right)}}{2\times 22}$$

Square -124.

$$b=\frac{-\left(-124\right)±\sqrt{15376-4\times 22\left(-48\right)}}{2\times 22}$$

Multiply -4 times 22.

$$b=\frac{-\left(-124\right)±\sqrt{15376-88\left(-48\right)}}{2\times 22}$$

Multiply -88 times -48.

$$b=\frac{-\left(-124\right)±\sqrt{15376+4224}}{2\times 22}$$

Add 15376 to 4224.

$$b=\frac{-\left(-124\right)±\sqrt{19600}}{2\times 22}$$

Take the square root of 19600.

$$b=\frac{-\left(-124\right)±140}{2\times 22}$$

The opposite of -124 is 124.

$$b=\frac{124±140}{2\times 22}$$

Multiply 2 times 22.

$$b=\frac{124±140}{44}$$

Now solve the equation b=\frac{124±140}{44} when ± is plus. Add 124 to 140.

$$b=\frac{264}{44}$$

Divide 264 by 44.

$$b=6$$

Now solve the equation b=\frac{124±140}{44} when ± is minus. Subtract 140 from 124.

$$b=-\frac{16}{44}$$

Reduce the fraction \frac{-16}{44} to lowest terms by extracting and canceling out 4.

$$b=-\frac{4}{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{4}{11} for x_{2}.

$$22b^{2}-124b-48=22\left(b-6\right)\left(b-\left(-\frac{4}{11}\right)\right)$$

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

$$22b^{2}-124b-48=22\left(b-6\right)\left(b+\frac{4}{11}\right)$$

Add \frac{4}{11} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$22b^{2}-124b-48=22\left(b-6\right)\times \frac{11b+4}{11}$$

Cancel out 11, the greatest common factor in 22 and 11.

$$22b^{2}-124b-48=2\left(b-6\right)\left(11b+4\right)$$

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 22

$$x ^ 2 -\frac{62}{11}x -\frac{24}{11} = 0$$

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

$$r + s = \frac{62}{11} rs = -\frac{24}{11}$$

Two numbers r and s sum up to \frac{62}{11} exactly when the average of the two numbers is \frac{1}{2}*\frac{62}{11} = \frac{31}{11}. 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>

$$r = \frac{31}{11} - u s = \frac{31}{11} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{24}{11}

$$(\frac{31}{11} - u) (\frac{31}{11} + u) = -\frac{24}{11}$$

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

$$\frac{961}{121} - u^2 = -\frac{24}{11}$$

Simplify the expression by subtracting \frac{961}{121} on both sides

$$-u^2 = -\frac{24}{11}-\frac{961}{121} = -\frac{1225}{121}$$

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

$$u^2 = \frac{1225}{121} u = \pm\sqrt{\frac{1225}{121}} = \pm \frac{35}{11}$$

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

$$r =\frac{31}{11} - \frac{35}{11} = -0.364 s = \frac{31}{11} + \frac{35}{11} = 6$$

Evaluate

$$2\left(b-6\right)\left(11b+4\right)$$