Solve 15b^2-100b-160

$$15 b ^ { 2 } - 100 b - 160$$

Factor

$$5\left(b-8\right)\left(3b+4\right)$$

Solution Steps

Factor out 5.

$$5\left(3b^{2}-20b-32\right)$$

Consider 3b^{2}-20b-32. Factor the expression by grouping. First, the expression needs to be rewritten as 3b^{2}+pb+qb-32. To find p and q, set up a system to be solved.

$$p+q=-20 pq=3\left(-32\right)=-96$$

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 -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -20.

$$p=-24 q=4$$

Rewrite 3b^{2}-20b-32 as \left(3b^{2}-24b\right)+\left(4b-32\right).

$$\left(3b^{2}-24b\right)+\left(4b-32\right)$$

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

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

Factor out common term b-8 by using distributive property.

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

Rewrite the complete factored expression.

$$5\left(b-8\right)\left(3b+4\right)$$

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.

$$15b^{2}-100b-160=0$$

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.

$$b=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 15\left(-160\right)}}{2\times 15}$$

Square -100.

$$b=\frac{-\left(-100\right)±\sqrt{10000-4\times 15\left(-160\right)}}{2\times 15}$$

Multiply -4 times 15.

$$b=\frac{-\left(-100\right)±\sqrt{10000-60\left(-160\right)}}{2\times 15}$$

Multiply -60 times -160.

$$b=\frac{-\left(-100\right)±\sqrt{10000+9600}}{2\times 15}$$

Add 10000 to 9600.

$$b=\frac{-\left(-100\right)±\sqrt{19600}}{2\times 15}$$

Take the square root of 19600.

$$b=\frac{-\left(-100\right)±140}{2\times 15}$$

The opposite of -100 is 100.

$$b=\frac{100±140}{2\times 15}$$

Multiply 2 times 15.

$$b=\frac{100±140}{30}$$

Now solve the equation b=\frac{100±140}{30} when ± is plus. Add 100 to 140.

$$b=\frac{240}{30}$$

Divide 240 by 30.

$$b=8$$

Now solve the equation b=\frac{100±140}{30} when ± is minus. Subtract 140 from 100.

$$b=-\frac{40}{30}$$

Reduce the fraction \frac{-40}{30} to lowest terms by extracting and canceling out 10.

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

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

$$15b^{2}-100b-160=15\left(b-8\right)\left(b-\left(-\frac{4}{3}\right)\right)$$

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

$$15b^{2}-100b-160=15\left(b-8\right)\left(b+\frac{4}{3}\right)$$

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

$$15b^{2}-100b-160=15\left(b-8\right)\times \frac{3b+4}{3}$$

Cancel out 3, the greatest common factor in 15 and 3.

$$15b^{2}-100b-160=5\left(b-8\right)\left(3b+4\right)$$

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 15

$$x ^ 2 -\frac{20}{3}x -\frac{32}{3} = 0$$

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

$$r + s = \frac{20}{3} rs = -\frac{32}{3}$$

Two numbers r and s sum up to \frac{20}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{20}{3} = \frac{10}{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>

$$r = \frac{10}{3} - u s = \frac{10}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{32}{3}

$$(\frac{10}{3} - u) (\frac{10}{3} + u) = -\frac{32}{3}$$

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

$$\frac{100}{9} - u^2 = -\frac{32}{3}$$

Simplify the expression by subtracting \frac{100}{9} on both sides

$$-u^2 = -\frac{32}{3}-\frac{100}{9} = -\frac{196}{9}$$

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

$$u^2 = \frac{196}{9} u = \pm\sqrt{\frac{196}{9}} = \pm \frac{14}{3}$$

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

$$r =\frac{10}{3} - \frac{14}{3} = -1.333 s = \frac{10}{3} + \frac{14}{3} = 8$$

Evaluate

$$5\left(b-8\right)\left(3b+4\right)$$