Solve -16x^2+54x+40

$$- 16 x ^ { 2 } + 54 x + 40$$

Factor

$$2\left(4-x\right)\left(8x+5\right)$$

Solution Steps

Factor out 2.

$$2\left(-8x^{2}+27x+20\right)$$

Consider -8x^{2}+27x+20. Factor the expression by grouping. First, the expression needs to be rewritten as -8x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=27 ab=-8\times 20=-160$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -160.

$$-1,160 -2,80 -4,40 -5,32 -8,20 -10,16$$

Calculate the sum for each pair.

$$-1+160=159 -2+80=78 -4+40=36 -5+32=27 -8+20=12 -10+16=6$$

The solution is the pair that gives sum 27.

$$a=32 b=-5$$

Rewrite -8x^{2}+27x+20 as \left(-8x^{2}+32x\right)+\left(-5x+20\right).

$$\left(-8x^{2}+32x\right)+\left(-5x+20\right)$$

Factor out 8x in the first and 5 in the second group.

$$8x\left(-x+4\right)+5\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(8x+5\right)$$

Rewrite the complete factored expression.

$$2\left(-x+4\right)\left(8x+5\right)$$

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.

$$-16x^{2}+54x+40=0$$

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.

$$x=\frac{-54±\sqrt{54^{2}-4\left(-16\right)\times 40}}{2\left(-16\right)}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\left(-16\right)\times 40}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-54±\sqrt{2916+64\times 40}}{2\left(-16\right)}$$

Multiply 64 times 40.

$$x=\frac{-54±\sqrt{2916+2560}}{2\left(-16\right)}$$

Add 2916 to 2560.

$$x=\frac{-54±\sqrt{5476}}{2\left(-16\right)}$$

Take the square root of 5476.

$$x=\frac{-54±74}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-54±74}{-32}$$

Now solve the equation x=\frac{-54±74}{-32} when ± is plus. Add -54 to 74.

$$x=\frac{20}{-32}$$

Reduce the fraction \frac{20}{-32} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{8}$$

Now solve the equation x=\frac{-54±74}{-32} when ± is minus. Subtract 74 from -54.

$$x=-\frac{128}{-32}$$

Divide -128 by -32.

$$x=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{8} for x_{1} and 4 for x_{2}.

$$-16x^{2}+54x+40=-16\left(x-\left(-\frac{5}{8}\right)\right)\left(x-4\right)$$

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

$$-16x^{2}+54x+40=-16\left(x+\frac{5}{8}\right)\left(x-4\right)$$

Add \frac{5}{8} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}+54x+40=-16\times \frac{-8x-5}{-8}\left(x-4\right)$$

Cancel out 8, the greatest common factor in -16 and 8.

$$-16x^{2}+54x+40=2\left(-8x-5\right)\left(x-4\right)$$

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.

$$x ^ 2 -\frac{27}{8}x -\frac{5}{2} = 0$$

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

$$r + s = \frac{27}{8} rs = -\frac{5}{2}$$

Two numbers r and s sum up to \frac{27}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{27}{8} = \frac{27}{16}. 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>

$$r = \frac{27}{16} - u s = \frac{27}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(\frac{27}{16} - u) (\frac{27}{16} + u) = -\frac{5}{2}$$

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

$$\frac{729}{256} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{729}{256} on both sides

$$-u^2 = -\frac{5}{2}-\frac{729}{256} = -\frac{1369}{256}$$

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

$$u^2 = \frac{1369}{256} u = \pm\sqrt{\frac{1369}{256}} = \pm \frac{37}{16}$$

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

$$r =\frac{27}{16} - \frac{37}{16} = -0.625 s = \frac{27}{16} + \frac{37}{16} = 4$$

Evaluate

$$40+54x-16x^{2}$$