Solve f(x)=-16x^2+32x+20

$$f ( x ) = - 16 x ^ { 2 } + 32 x + 20$$

Factor

$$4\left(-2x-1\right)\left(2x-5\right)$$

Solution Steps

Factor out 4.

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

Consider -4x^{2}+8x+5. Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=8 ab=-4\times 5=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 8.

$$a=10 b=-2$$

Rewrite -4x^{2}+8x+5 as \left(-4x^{2}+10x\right)+\left(-2x+5\right).

$$\left(-4x^{2}+10x\right)+\left(-2x+5\right)$$

Factor out -2x in the first and -1 in the second group.

$$-2x\left(2x-5\right)-\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(-2x-1\right)$$

Rewrite the complete factored expression.

$$4\left(2x-5\right)\left(-2x-1\right)$$

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.

$$-16x^{2}+32x+20=0$$

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.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-16\right)\times 20}}{2\left(-16\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-16\right)\times 20}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-32±\sqrt{1024+64\times 20}}{2\left(-16\right)}$$

Multiply 64 times 20.

$$x=\frac{-32±\sqrt{1024+1280}}{2\left(-16\right)}$$

Add 1024 to 1280.

$$x=\frac{-32±\sqrt{2304}}{2\left(-16\right)}$$

Take the square root of 2304.

$$x=\frac{-32±48}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-32±48}{-32}$$

Now solve the equation x=\frac{-32±48}{-32} when ± is plus. Add -32 to 48.

$$x=\frac{16}{-32}$$

Reduce the fraction \frac{16}{-32} to lowest terms by extracting and canceling out 16.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-32±48}{-32} when ± is minus. Subtract 48 from -32.

$$x=-\frac{80}{-32}$$

Reduce the fraction \frac{-80}{-32} to lowest terms by extracting and canceling out 16.

$$x=\frac{5}{2}$$

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

$$-16x^{2}+32x+20=-16\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\frac{5}{2}\right)$$

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

$$-16x^{2}+32x+20=-16\left(x+\frac{1}{2}\right)\left(x-\frac{5}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}+32x+20=-16\times \frac{-2x-1}{-2}\left(x-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}+32x+20=-16\times \frac{-2x-1}{-2}\times \frac{-2x+5}{-2}$$

Multiply \frac{-2x-1}{-2} times \frac{-2x+5}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}+32x+20=-16\times \frac{\left(-2x-1\right)\left(-2x+5\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-16x^{2}+32x+20=-16\times \frac{\left(-2x-1\right)\left(-2x+5\right)}{4}$$

Cancel out 4, the greatest common factor in -16 and 4.

$$-16x^{2}+32x+20=-4\left(-2x-1\right)\left(-2x+5\right)$$

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.

$$x ^ 2 -2x -\frac{5}{4} = 0$$

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

$$r + s = 2 rs = -\frac{5}{4}$$

Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{4}

$$(1 - u) (1 + u) = -\frac{5}{4}$$

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

$$1 - u^2 = -\frac{5}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{5}{4}-1 = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =1 - \frac{3}{2} = -0.500 s = 1 + \frac{3}{2} = 2.500$$

Evaluate

$$20+32x-16x^{2}$$