Solve y^2-35y+300

$$y ^ { 2 } - 35 y + 300$$

Factor

$$\left(y-20\right)\left(y-15\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+300. To find a and b, set up a system to be solved.

$$a+b=-35 ab=1\times 300=300$$

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 300.

$$-1,-300 -2,-150 -3,-100 -4,-75 -5,-60 -6,-50 -10,-30 -12,-25 -15,-20$$

Calculate the sum for each pair.

$$-1-300=-301 -2-150=-152 -3-100=-103 -4-75=-79 -5-60=-65 -6-50=-56 -10-30=-40 -12-25=-37 -15-20=-35$$

The solution is the pair that gives sum -35.

$$a=-20 b=-15$$

Rewrite y^{2}-35y+300 as \left(y^{2}-20y\right)+\left(-15y+300\right).

$$\left(y^{2}-20y\right)+\left(-15y+300\right)$$

Factor out y in the first and -15 in the second group.

$$y\left(y-20\right)-15\left(y-20\right)$$

Factor out common term y-20 by using distributive property.

$$\left(y-20\right)\left(y-15\right)$$

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.

$$y^{2}-35y+300=0$$

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.

$$y=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 300}}{2}$$

Square -35.

$$y=\frac{-\left(-35\right)±\sqrt{1225-4\times 300}}{2}$$

Multiply -4 times 300.

$$y=\frac{-\left(-35\right)±\sqrt{1225-1200}}{2}$$

Add 1225 to -1200.

$$y=\frac{-\left(-35\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-\left(-35\right)±5}{2}$$

The opposite of -35 is 35.

$$y=\frac{35±5}{2}$$

Now solve the equation y=\frac{35±5}{2} when ± is plus. Add 35 to 5.

$$y=\frac{40}{2}$$

Divide 40 by 2.

$$y=20$$

Now solve the equation y=\frac{35±5}{2} when ± is minus. Subtract 5 from 35.

$$y=\frac{30}{2}$$

Divide 30 by 2.

$$y=15$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and 15 for x_{2}.

$$y^{2}-35y+300=\left(y-20\right)\left(y-15\right)$$

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.

$$x ^ 2 -35x +300 = 0$$

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

$$r + s = 35 rs = 300$$

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

$$r = \frac{35}{2} - u s = \frac{35}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 300

$$(\frac{35}{2} - u) (\frac{35}{2} + u) = 300$$

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

$$\frac{1225}{4} - u^2 = 300$$

Simplify the expression by subtracting \frac{1225}{4} on both sides

$$-u^2 = 300-\frac{1225}{4} = -\frac{25}{4}$$

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

$$u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{2}$$

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

$$r =\frac{35}{2} - \frac{5}{2} = 15 s = \frac{35}{2} + \frac{5}{2} = 20$$

Evaluate

$$\left(y-20\right)\left(y-15\right)$$