Solve 16x^2+8x-3

$$16 { x }^{ 2 } +8x-3$$

Factor

$$\left(4x-1\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=8 ab=16\left(-3\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite 16x^{2}+8x-3 as \left(16x^{2}-4x\right)+\left(12x-3\right).

$$\left(16x^{2}-4x\right)+\left(12x-3\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(4x-1\right)+3\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(4x+3\right)$$

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.

$$16x^{2}+8x-3=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16\left(-3\right)}}{2\times 16}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16\left(-3\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64\left(-3\right)}}{2\times 16}$$

Multiply -64 times -3.

$$x=\frac{-8±\sqrt{64+192}}{2\times 16}$$

Add 64 to 192.

$$x=\frac{-8±\sqrt{256}}{2\times 16}$$

Take the square root of 256.

$$x=\frac{-8±16}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-8±16}{32}$$

Now solve the equation x=\frac{-8±16}{32} when ± is plus. Add -8 to 16.

$$x=\frac{8}{32}$$

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-8±16}{32} when ± is minus. Subtract 16 from -8.

$$x=-\frac{24}{32}$$

Reduce the fraction \frac{-24}{32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and -\frac{3}{4} for x_{2}.

$$16x^{2}+8x-3=16\left(x-\frac{1}{4}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$16x^{2}+8x-3=16\left(x-\frac{1}{4}\right)\left(x+\frac{3}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+8x-3=16\times \frac{4x-1}{4}\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+8x-3=16\times \frac{4x-1}{4}\times \frac{4x+3}{4}$$

Multiply \frac{4x-1}{4} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}+8x-3=16\times \frac{\left(4x-1\right)\left(4x+3\right)}{4\times 4}$$

Multiply 4 times 4.

$$16x^{2}+8x-3=16\times \frac{\left(4x-1\right)\left(4x+3\right)}{16}$$

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

$$16x^{2}+8x-3=\left(4x-1\right)\left(4x+3\right)$$

Evaluate

$$\left(4x-1\right)\left(4x+3\right)$$