Solve 16x^2-26x+9

$$16 x ^ { 2 } - 26 x + 9$$

Factor

$$\left(2x-1\right)\left(8x-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-26 ab=16\times 9=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -26.

$$a=-18 b=-8$$

Rewrite 16x^{2}-26x+9 as \left(16x^{2}-18x\right)+\left(-8x+9\right).

$$\left(16x^{2}-18x\right)+\left(-8x+9\right)$$

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

$$2x\left(8x-9\right)-\left(8x-9\right)$$

Factor out common term 8x-9 by using distributive property.

$$\left(8x-9\right)\left(2x-1\right)$$

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.

$$16x^{2}-26x+9=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 16\times 9}}{2\times 16}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 16\times 9}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-26\right)±\sqrt{676-64\times 9}}{2\times 16}$$

Multiply -64 times 9.

$$x=\frac{-\left(-26\right)±\sqrt{676-576}}{2\times 16}$$

Add 676 to -576.

$$x=\frac{-\left(-26\right)±\sqrt{100}}{2\times 16}$$

Take the square root of 100.

$$x=\frac{-\left(-26\right)±10}{2\times 16}$$

The opposite of -26 is 26.

$$x=\frac{26±10}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{26±10}{32}$$

Now solve the equation x=\frac{26±10}{32} when ± is plus. Add 26 to 10.

$$x=\frac{36}{32}$$

Reduce the fraction \frac{36}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{9}{8}$$

Now solve the equation x=\frac{26±10}{32} when ± is minus. Subtract 10 from 26.

$$x=\frac{16}{32}$$

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

$$x=\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{8} for x_{1} and \frac{1}{2} for x_{2}.

$$16x^{2}-26x+9=16\left(x-\frac{9}{8}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{9}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-26x+9=16\times \frac{8x-9}{8}\left(x-\frac{1}{2}\right)$$

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

$$16x^{2}-26x+9=16\times \frac{8x-9}{8}\times \frac{2x-1}{2}$$

Multiply \frac{8x-9}{8} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$16x^{2}-26x+9=16\times \frac{\left(8x-9\right)\left(2x-1\right)}{8\times 2}$$

Multiply 8 times 2.

$$16x^{2}-26x+9=16\times \frac{\left(8x-9\right)\left(2x-1\right)}{16}$$

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

$$16x^{2}-26x+9=\left(8x-9\right)\left(2x-1\right)$$

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

$$x ^ 2 -\frac{13}{8}x +\frac{9}{16} = 0$$

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

$$r + s = \frac{13}{8} rs = \frac{9}{16}$$

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{9}{16}

$$(\frac{13}{16} - u) (\frac{13}{16} + u) = \frac{9}{16}$$

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

$$\frac{169}{256} - u^2 = \frac{9}{16}$$

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

$$-u^2 = \frac{9}{16}-\frac{169}{256} = -\frac{25}{256}$$

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

$$u^2 = \frac{25}{256} u = \pm\sqrt{\frac{25}{256}} = \pm \frac{5}{16}$$

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

$$r =\frac{13}{16} - \frac{5}{16} = 0.500 s = \frac{13}{16} + \frac{5}{16} = 1.125$$

Evaluate

$$\left(2x-1\right)\left(8x-9\right)$$