Solve 8y^2-13y+5

$$8 y ^ { 2 } - 13 y + 5$$

Factor

$$\left(y-1\right)\left(8y-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8y^{2}+ay+by+5. To find a and b, set up a system to be solved.

$$a+b=-13 ab=8\times 5=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite 8y^{2}-13y+5 as \left(8y^{2}-8y\right)+\left(-5y+5\right).

$$\left(8y^{2}-8y\right)+\left(-5y+5\right)$$

Factor out 8y in the first and -5 in the second group.

$$8y\left(y-1\right)-5\left(y-1\right)$$

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

$$\left(y-1\right)\left(8y-5\right)$$

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.

$$8y^{2}-13y+5=0$$

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.

$$y=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 8\times 5}}{2\times 8}$$

Square -13.

$$y=\frac{-\left(-13\right)±\sqrt{169-4\times 8\times 5}}{2\times 8}$$

Multiply -4 times 8.

$$y=\frac{-\left(-13\right)±\sqrt{169-32\times 5}}{2\times 8}$$

Multiply -32 times 5.

$$y=\frac{-\left(-13\right)±\sqrt{169-160}}{2\times 8}$$

Add 169 to -160.

$$y=\frac{-\left(-13\right)±\sqrt{9}}{2\times 8}$$

Take the square root of 9.

$$y=\frac{-\left(-13\right)±3}{2\times 8}$$

The opposite of -13 is 13.

$$y=\frac{13±3}{2\times 8}$$

Multiply 2 times 8.

$$y=\frac{13±3}{16}$$

Now solve the equation y=\frac{13±3}{16} when ± is plus. Add 13 to 3.

$$y=\frac{16}{16}$$

Divide 16 by 16.

$$y=1$$

Now solve the equation y=\frac{13±3}{16} when ± is minus. Subtract 3 from 13.

$$y=\frac{10}{16}$$

Reduce the fraction \frac{10}{16} to lowest terms by extracting and canceling out 2.

$$y=\frac{5}{8}$$

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{5}{8} for x_{2}.

$$8y^{2}-13y+5=8\left(y-1\right)\left(y-\frac{5}{8}\right)$$

Subtract \frac{5}{8} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8y^{2}-13y+5=8\left(y-1\right)\times \frac{8y-5}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8y^{2}-13y+5=\left(y-1\right)\left(8y-5\right)$$

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 8

$$x ^ 2 -\frac{13}{8}x +\frac{5}{8} = 0$$

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

$$r + s = \frac{13}{8} rs = \frac{5}{8}$$

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{5}{8}

$$(\frac{13}{16} - u) (\frac{13}{16} + u) = \frac{5}{8}$$

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

$$\frac{169}{256} - u^2 = \frac{5}{8}$$

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

$$-u^2 = \frac{5}{8}-\frac{169}{256} = -\frac{9}{256}$$

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

$$u^2 = \frac{9}{256} u = \pm\sqrt{\frac{9}{256}} = \pm \frac{3}{16}$$

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

$$r =\frac{13}{16} - \frac{3}{16} = 0.625 s = \frac{13}{16} + \frac{3}{16} = 1$$

Evaluate

$$\left(y-1\right)\left(8y-5\right)$$