Solve 5+8x-4x^2

$$5 + 8 x - 4 x ^ { 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}+8x+5$$

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.

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

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.

$$-4x^{2}+8x+5=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-8±\sqrt{64+16\times 5}}{2\left(-4\right)}$$

Multiply 16 times 5.

$$x=\frac{-8±\sqrt{64+80}}{2\left(-4\right)}$$

Add 64 to 80.

$$x=\frac{-8±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$x=\frac{-8±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-8±12}{-8}$$

Now solve the equation x=\frac{-8±12}{-8} when ± is plus. Add -8 to 12.

$$x=\frac{4}{-8}$$

Reduce the fraction \frac{4}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-8±12}{-8} when ± is minus. Subtract 12 from -8.

$$x=-\frac{20}{-8}$$

Reduce the fraction \frac{-20}{-8} to lowest terms by extracting and canceling out 4.

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

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}.

$$-4x^{2}+8x+5=-4\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\frac{5}{2}\right)$$

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

$$-4x^{2}+8x+5=-4\left(x+\frac{1}{2}\right)\left(x-\frac{5}{2}\right)$$

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

$$-4x^{2}+8x+5=-4\times \frac{-2x-1}{-2}\left(x-\frac{5}{2}\right)$$

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

$$-4x^{2}+8x+5=-4\times \frac{-2x-1}{-2}\times \frac{-2x+5}{-2}$$

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.

$$-4x^{2}+8x+5=-4\times \frac{\left(-2x-1\right)\left(-2x+5\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-4x^{2}+8x+5=-4\times \frac{\left(-2x-1\right)\left(-2x+5\right)}{4}$$

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

$$-4x^{2}+8x+5=-\left(-2x-1\right)\left(-2x+5\right)$$

Evaluate

$$5+8x-4x^{2}$$