Solve -20b^2+136b+192

$$- 20 b ^ { 2 } + 136 b + 192$$

Factor

$$4\left(8-b\right)\left(5b+6\right)$$

Solution Steps

Factor out 4.

$$4\left(-5b^{2}+34b+48\right)$$

Consider -5b^{2}+34b+48. Factor the expression by grouping. First, the expression needs to be rewritten as -5b^{2}+pb+qb+48. To find p and q, set up a system to be solved.

$$p+q=34 pq=-5\times 48=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 34.

$$p=40 q=-6$$

Rewrite -5b^{2}+34b+48 as \left(-5b^{2}+40b\right)+\left(-6b+48\right).

$$\left(-5b^{2}+40b\right)+\left(-6b+48\right)$$

Factor out 5b in the first and 6 in the second group.

$$5b\left(-b+8\right)+6\left(-b+8\right)$$

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

$$\left(-b+8\right)\left(5b+6\right)$$

Rewrite the complete factored expression.

$$4\left(-b+8\right)\left(5b+6\right)$$

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.

$$-20b^{2}+136b+192=0$$

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.

$$b=\frac{-136±\sqrt{136^{2}-4\left(-20\right)\times 192}}{2\left(-20\right)}$$

Square 136.

$$b=\frac{-136±\sqrt{18496-4\left(-20\right)\times 192}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$b=\frac{-136±\sqrt{18496+80\times 192}}{2\left(-20\right)}$$

Multiply 80 times 192.

$$b=\frac{-136±\sqrt{18496+15360}}{2\left(-20\right)}$$

Add 18496 to 15360.

$$b=\frac{-136±\sqrt{33856}}{2\left(-20\right)}$$

Take the square root of 33856.

$$b=\frac{-136±184}{2\left(-20\right)}$$

Multiply 2 times -20.

$$b=\frac{-136±184}{-40}$$

Now solve the equation b=\frac{-136±184}{-40} when ± is plus. Add -136 to 184.

$$b=\frac{48}{-40}$$

Reduce the fraction \frac{48}{-40} to lowest terms by extracting and canceling out 8.

$$b=-\frac{6}{5}$$

Now solve the equation b=\frac{-136±184}{-40} when ± is minus. Subtract 184 from -136.

$$b=-\frac{320}{-40}$$

Divide -320 by -40.

$$b=8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{6}{5} for x_{1} and 8 for x_{2}.

$$-20b^{2}+136b+192=-20\left(b-\left(-\frac{6}{5}\right)\right)\left(b-8\right)$$

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

$$-20b^{2}+136b+192=-20\left(b+\frac{6}{5}\right)\left(b-8\right)$$

Add \frac{6}{5} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-20b^{2}+136b+192=-20\times \frac{-5b-6}{-5}\left(b-8\right)$$

Cancel out 5, the greatest common factor in -20 and 5.

$$-20b^{2}+136b+192=4\left(-5b-6\right)\left(b-8\right)$$

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.

$$x ^ 2 -\frac{34}{5}x -\frac{48}{5} = 0$$

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

$$r + s = \frac{34}{5} rs = -\frac{48}{5}$$

Two numbers r and s sum up to \frac{34}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{34}{5} = \frac{17}{5}. 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>

$$r = \frac{17}{5} - u s = \frac{17}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{48}{5}

$$(\frac{17}{5} - u) (\frac{17}{5} + u) = -\frac{48}{5}$$

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

$$\frac{289}{25} - u^2 = -\frac{48}{5}$$

Simplify the expression by subtracting \frac{289}{25} on both sides

$$-u^2 = -\frac{48}{5}-\frac{289}{25} = -\frac{529}{25}$$

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

$$u^2 = \frac{529}{25} u = \pm\sqrt{\frac{529}{25}} = \pm \frac{23}{5}$$

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

$$r =\frac{17}{5} - \frac{23}{5} = -1.200 s = \frac{17}{5} + \frac{23}{5} = 8$$

Evaluate

$$192+136b-20b^{2}$$