Solve -4n^2+4n+3

$$- 4 n ^ { 2 } + 4 n + 3$$

Factor

$$-\left(2n-3\right)\left(2n+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -4n^{2}+an+bn+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-4\times 3=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=6 b=-2$$

Rewrite -4n^{2}+4n+3 as \left(-4n^{2}+6n\right)+\left(-2n+3\right).

$$\left(-4n^{2}+6n\right)+\left(-2n+3\right)$$

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

$$-2n\left(2n-3\right)-\left(2n-3\right)$$

Factor out common term 2n-3 by using distributive property.

$$\left(2n-3\right)\left(-2n-1\right)$$

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.

$$-4n^{2}+4n+3=0$$

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.

$$n=\frac{-4±\sqrt{4^{2}-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Square 4.

$$n=\frac{-4±\sqrt{16-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$n=\frac{-4±\sqrt{16+16\times 3}}{2\left(-4\right)}$$

Multiply 16 times 3.

$$n=\frac{-4±\sqrt{16+48}}{2\left(-4\right)}$$

Add 16 to 48.

$$n=\frac{-4±\sqrt{64}}{2\left(-4\right)}$$

Take the square root of 64.

$$n=\frac{-4±8}{2\left(-4\right)}$$

Multiply 2 times -4.

$$n=\frac{-4±8}{-8}$$

Now solve the equation n=\frac{-4±8}{-8} when ± is plus. Add -4 to 8.

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

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

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

Now solve the equation n=\frac{-4±8}{-8} when ± is minus. Subtract 8 from -4.

$$n=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$n=\frac{3}{2}$$

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{3}{2} for x_{2}.

$$-4n^{2}+4n+3=-4\left(n-\left(-\frac{1}{2}\right)\right)\left(n-\frac{3}{2}\right)$$

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

$$-4n^{2}+4n+3=-4\left(n+\frac{1}{2}\right)\left(n-\frac{3}{2}\right)$$

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

$$-4n^{2}+4n+3=-4\times \frac{-2n-1}{-2}\left(n-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4n^{2}+4n+3=-4\times \frac{-2n-1}{-2}\times \frac{-2n+3}{-2}$$

Multiply \frac{-2n-1}{-2} times \frac{-2n+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-4n^{2}+4n+3=-4\times \frac{\left(-2n-1\right)\left(-2n+3\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-4n^{2}+4n+3=-4\times \frac{\left(-2n-1\right)\left(-2n+3\right)}{4}$$

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

$$-4n^{2}+4n+3=-\left(-2n-1\right)\left(-2n+3\right)$$

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.

$$x ^ 2 -1x -\frac{3}{4} = 0$$

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

$$r + s = 1 rs = -\frac{3}{4}$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{3}{4}$$

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

$$\frac{1}{4} - u^2 = -\frac{3}{4}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{3}{4}-\frac{1}{4} = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =\frac{1}{2} - 1 = -0.500 s = \frac{1}{2} + 1 = 1.500$$

Evaluate

$$3+4n-4n^{2}$$