Solve (4n^2+4n-3)

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

Factor

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

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.

$$a+b=4 ab=4\left(-3\right)=-12$$

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.

$$a=-2 b=6$$

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

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

Factor out 2n in the first and 3 in the second group.

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

Factor out common term 2n-1 by using distributive property.

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

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.

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

Square 4.

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

Multiply -4 times 4.

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

Multiply -16 times -3.

$$n=\frac{-4±\sqrt{16+48}}{2\times 4}$$

Add 16 to 48.

$$n=\frac{-4±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$n=\frac{-4±8}{2\times 4}$$

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}.

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

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

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

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

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

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

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

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.

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

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.

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

Evaluate

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