Solve 16x^2-26x+3

$$16 x ^ { 2 } - 26 x + 3$$

Factor

$$\left(2x-3\right)\left(8x-1\right)$$

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.

$$a+b=-26 ab=16\times 3=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both 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.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -26.

$$a=-24 b=-2$$

Rewrite 16x^{2}-26x+3 as \left(16x^{2}-24x\right)+\left(-2x+3\right).

$$\left(16x^{2}-24x\right)+\left(-2x+3\right)$$

Factor out 8x in the first and -1 in the second group.

$$8x\left(2x-3\right)-\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(8x-1\right)$$

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.

$$16x^{2}-26x+3=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 16\times 3}}{2\times 16}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 16\times 3}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-26\right)±\sqrt{676-64\times 3}}{2\times 16}$$

Multiply -64 times 3.

$$x=\frac{-\left(-26\right)±\sqrt{676-192}}{2\times 16}$$

Add 676 to -192.

$$x=\frac{-\left(-26\right)±\sqrt{484}}{2\times 16}$$

Take the square root of 484.

$$x=\frac{-\left(-26\right)±22}{2\times 16}$$

The opposite of -26 is 26.

$$x=\frac{26±22}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{26±22}{32}$$

Now solve the equation x=\frac{26±22}{32} when ± is plus. Add 26 to 22.

$$x=\frac{48}{32}$$

Reduce the fraction \frac{48}{32} to lowest terms by extracting and canceling out 16.

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

Now solve the equation x=\frac{26±22}{32} when ± is minus. Subtract 22 from 26.

$$x=\frac{4}{32}$$

Reduce the fraction \frac{4}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{1}{8} for x_{2}.

$$16x^{2}-26x+3=16\left(x-\frac{3}{2}\right)\left(x-\frac{1}{8}\right)$$

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

$$16x^{2}-26x+3=16\times \frac{2x-3}{2}\left(x-\frac{1}{8}\right)$$

Subtract \frac{1}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-26x+3=16\times \frac{2x-3}{2}\times \frac{8x-1}{8}$$

Multiply \frac{2x-3}{2} times \frac{8x-1}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-26x+3=16\times \frac{\left(2x-3\right)\left(8x-1\right)}{2\times 8}$$

Multiply 2 times 8.

$$16x^{2}-26x+3=16\times \frac{\left(2x-3\right)\left(8x-1\right)}{16}$$

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

$$16x^{2}-26x+3=\left(2x-3\right)\left(8x-1\right)$$

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

$$x ^ 2 -\frac{13}{8}x +\frac{3}{16} = 0$$

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

$$r + s = \frac{13}{8} rs = \frac{3}{16}$$

Two numbers r and s sum up to \frac{13}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{8} = \frac{13}{16}. 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>

$$r = \frac{13}{16} - u s = \frac{13}{16} + u$$

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

$$(\frac{13}{16} - u) (\frac{13}{16} + u) = \frac{3}{16}$$

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

$$\frac{169}{256} - u^2 = \frac{3}{16}$$

Simplify the expression by subtracting \frac{169}{256} on both sides

$$-u^2 = \frac{3}{16}-\frac{169}{256} = -\frac{121}{256}$$

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

$$u^2 = \frac{121}{256} u = \pm\sqrt{\frac{121}{256}} = \pm \frac{11}{16}$$

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

$$r =\frac{13}{16} - \frac{11}{16} = 0.125 s = \frac{13}{16} + \frac{11}{16} = 1.500$$

Evaluate

$$\left(2x-3\right)\left(8x-1\right)$$