Solve 16x^2-8x-3=

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

Factor

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-12 b=4$$

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

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

Factor out 4x in 16x^{2}-12x.

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

Factor out common term 4x-3 by using distributive property.

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

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.

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

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.

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

Square -8.

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

Multiply -4 times 16.

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

Multiply -64 times -3.

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

Add 64 to 192.

$$x=\frac{-\left(-8\right)±\sqrt{256}}{2\times 16}$$

Take the square root of 256.

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

The opposite of -8 is 8.

$$x=\frac{8±16}{2\times 16}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply 4 times 4.

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

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

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

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.This is achieved by dividing both sides of the equation by 16

$$x ^ 2 -\frac{1}{2}x -\frac{3}{16} = 0$$

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

$$r + s = \frac{1}{2} rs = -\frac{3}{16}$$

Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{16}

$$(\frac{1}{4} - u) (\frac{1}{4} + u) = -\frac{3}{16}$$

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

$$\frac{1}{16} - u^2 = -\frac{3}{16}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{3}{16}-\frac{1}{16} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{1}{4} - \frac{1}{2} = -0.250 s = \frac{1}{4} + \frac{1}{2} = 0.750$$

Evaluate

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