Solve 26y^2-35y+9

$$26 y ^ { 2 } - 35 y + 9$$

Factor

$$\left(y-1\right)\left(26y-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 26y^{2}+ay+by+9. To find a and b, set up a system to be solved.

$$a+b=-35 ab=26\times 9=234$$

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 234.

$$-1,-234 -2,-117 -3,-78 -6,-39 -9,-26 -13,-18$$

Calculate the sum for each pair.

$$-1-234=-235 -2-117=-119 -3-78=-81 -6-39=-45 -9-26=-35 -13-18=-31$$

The solution is the pair that gives sum -35.

$$a=-26 b=-9$$

Rewrite 26y^{2}-35y+9 as \left(26y^{2}-26y\right)+\left(-9y+9\right).

$$\left(26y^{2}-26y\right)+\left(-9y+9\right)$$

Factor out 26y in the first and -9 in the second group.

$$26y\left(y-1\right)-9\left(y-1\right)$$

Factor out common term y-1 by using distributive property.

$$\left(y-1\right)\left(26y-9\right)$$

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.

$$26y^{2}-35y+9=0$$

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.

$$y=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 26\times 9}}{2\times 26}$$

Square -35.

$$y=\frac{-\left(-35\right)±\sqrt{1225-4\times 26\times 9}}{2\times 26}$$

Multiply -4 times 26.

$$y=\frac{-\left(-35\right)±\sqrt{1225-104\times 9}}{2\times 26}$$

Multiply -104 times 9.

$$y=\frac{-\left(-35\right)±\sqrt{1225-936}}{2\times 26}$$

Add 1225 to -936.

$$y=\frac{-\left(-35\right)±\sqrt{289}}{2\times 26}$$

Take the square root of 289.

$$y=\frac{-\left(-35\right)±17}{2\times 26}$$

The opposite of -35 is 35.

$$y=\frac{35±17}{2\times 26}$$

Multiply 2 times 26.

$$y=\frac{35±17}{52}$$

Now solve the equation y=\frac{35±17}{52} when ± is plus. Add 35 to 17.

$$y=\frac{52}{52}$$

Divide 52 by 52.

$$y=1$$

Now solve the equation y=\frac{35±17}{52} when ± is minus. Subtract 17 from 35.

$$y=\frac{18}{52}$$

Reduce the fraction \frac{18}{52} to lowest terms by extracting and canceling out 2.

$$y=\frac{9}{26}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and \frac{9}{26} for x_{2}.

$$26y^{2}-35y+9=26\left(y-1\right)\left(y-\frac{9}{26}\right)$$

Subtract \frac{9}{26} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$26y^{2}-35y+9=26\left(y-1\right)\times \frac{26y-9}{26}$$

Cancel out 26, the greatest common factor in 26 and 26.

$$26y^{2}-35y+9=\left(y-1\right)\left(26y-9\right)$$

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 26

$$x ^ 2 -\frac{35}{26}x +\frac{9}{26} = 0$$

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

$$r + s = \frac{35}{26} rs = \frac{9}{26}$$

Two numbers r and s sum up to \frac{35}{26} exactly when the average of the two numbers is \frac{1}{2}*\frac{35}{26} = \frac{35}{52}. 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>

$$r = \frac{35}{52} - u s = \frac{35}{52} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{26}

$$(\frac{35}{52} - u) (\frac{35}{52} + u) = \frac{9}{26}$$

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

$$\frac{1225}{2704} - u^2 = \frac{9}{26}$$

Simplify the expression by subtracting \frac{1225}{2704} on both sides

$$-u^2 = \frac{9}{26}-\frac{1225}{2704} = -\frac{289}{2704}$$

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

$$u^2 = \frac{289}{2704} u = \pm\sqrt{\frac{289}{2704}} = \pm \frac{17}{52}$$

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

$$r =\frac{35}{52} - \frac{17}{52} = 0.346 s = \frac{35}{52} + \frac{17}{52} = 1.000$$

Evaluate

$$\left(y-1\right)\left(26y-9\right)$$