Solve 3y^2-8y+5

$$3 y ^ { 2 } - 8 y + 5$$

Factor

$$\left(y-1\right)\left(3y-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+5. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite 3y^{2}-8y+5 as \left(3y^{2}-5y\right)+\left(-3y+5\right).

$$\left(3y^{2}-5y\right)+\left(-3y+5\right)$$

Factor out y in the first and -1 in the second group.

$$y\left(3y-5\right)-\left(3y-5\right)$$

Factor out common term 3y-5 by using distributive property.

$$\left(3y-5\right)\left(y-1\right)$$

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.

$$3y^{2}-8y+5=0$$

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.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-8\right)±\sqrt{64-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$y=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 3}$$

Add 64 to -60.

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

Take the square root of 4.

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

The opposite of -8 is 8.

$$y=\frac{8±2}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{8±2}{6}$$

Now solve the equation y=\frac{8±2}{6} when ± is plus. Add 8 to 2.

$$y=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{5}{3}$$

Now solve the equation y=\frac{8±2}{6} when ± is minus. Subtract 2 from 8.

$$y=\frac{6}{6}$$

Divide 6 by 6.

$$y=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{3} for x_{1} and 1 for x_{2}.

$$3y^{2}-8y+5=3\left(y-\frac{5}{3}\right)\left(y-1\right)$$

Subtract \frac{5}{3} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}-8y+5=3\times \frac{3y-5}{3}\left(y-1\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3y^{2}-8y+5=\left(3y-5\right)\left(y-1\right)$$

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 3

$$x ^ 2 -\frac{8}{3}x +\frac{5}{3} = 0$$

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

$$r + s = \frac{8}{3} rs = \frac{5}{3}$$

Two numbers r and s sum up to \frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{3} = \frac{4}{3}. 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>

$$r = \frac{4}{3} - u s = \frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(\frac{4}{3} - u) (\frac{4}{3} + u) = \frac{5}{3}$$

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

$$\frac{16}{9} - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = \frac{5}{3}-\frac{16}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =\frac{4}{3} - \frac{1}{3} = 1.000 s = \frac{4}{3} + \frac{1}{3} = 1.667$$

Evaluate

$$\left(y-1\right)\left(3y-5\right)$$