Solve 5y^2-16y+3

$$5 y ^ { 2 } - 16 y + 3$$

Factor

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

Solution Steps

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

$$a+b=-16 ab=5\times 3=15$$

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 -16.

$$a=-15 b=-1$$

Rewrite 5y^{2}-16y+3 as \left(5y^{2}-15y\right)+\left(-y+3\right).

$$\left(5y^{2}-15y\right)+\left(-y+3\right)$$

Factor out 5y in the first and -1 in the second group.

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

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

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

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.

$$5y^{2}-16y+3=0$$

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.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Square -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-16\right)±\sqrt{256-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$y=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 5}$$

Add 256 to -60.

$$y=\frac{-\left(-16\right)±\sqrt{196}}{2\times 5}$$

Take the square root of 196.

$$y=\frac{-\left(-16\right)±14}{2\times 5}$$

The opposite of -16 is 16.

$$y=\frac{16±14}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{16±14}{10}$$

Now solve the equation y=\frac{16±14}{10} when ± is plus. Add 16 to 14.

$$y=\frac{30}{10}$$

Divide 30 by 10.

$$y=3$$

Now solve the equation y=\frac{16±14}{10} when ± is minus. Subtract 14 from 16.

$$y=\frac{2}{10}$$

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

$$y=\frac{1}{5}$$

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

$$5y^{2}-16y+3=5\left(y-3\right)\left(y-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5y^{2}-16y+3=5\left(y-3\right)\times \frac{5y-1}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5y^{2}-16y+3=\left(y-3\right)\left(5y-1\right)$$

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 5

$$x ^ 2 -\frac{16}{5}x +\frac{3}{5} = 0$$

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

$$r + s = \frac{16}{5} rs = \frac{3}{5}$$

Two numbers r and s sum up to \frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{5} = \frac{8}{5}. 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>

$$r = \frac{8}{5} - u s = \frac{8}{5} + u$$

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

$$(\frac{8}{5} - u) (\frac{8}{5} + u) = \frac{3}{5}$$

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

$$\frac{64}{25} - u^2 = \frac{3}{5}$$

Simplify the expression by subtracting \frac{64}{25} on both sides

$$-u^2 = \frac{3}{5}-\frac{64}{25} = -\frac{49}{25}$$

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

$$u^2 = \frac{49}{25} u = \pm\sqrt{\frac{49}{25}} = \pm \frac{7}{5}$$

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

$$r =\frac{8}{5} - \frac{7}{5} = 0.200 s = \frac{8}{5} + \frac{7}{5} = 3$$

Evaluate

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