Solve 2y^2+y-300=

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

Factor

$$\left(y-12\right)\left(2y+25\right)$$

Solution Steps

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

$$a+b=1 ab=2\left(-300\right)=-600$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 1.

$$a=-24 b=25$$

Rewrite 2y^{2}+y-300 as \left(2y^{2}-24y\right)+\left(25y-300\right).

$$\left(2y^{2}-24y\right)+\left(25y-300\right)$$

Factor out 2y in the first and 25 in the second group.

$$2y\left(y-12\right)+25\left(y-12\right)$$

Factor out common term y-12 by using distributive property.

$$\left(y-12\right)\left(2y+25\right)$$

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.

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

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.

$$y=\frac{-1±\sqrt{1^{2}-4\times 2\left(-300\right)}}{2\times 2}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 2\left(-300\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-1±\sqrt{1-8\left(-300\right)}}{2\times 2}$$

Multiply -8 times -300.

$$y=\frac{-1±\sqrt{1+2400}}{2\times 2}$$

Add 1 to 2400.

$$y=\frac{-1±\sqrt{2401}}{2\times 2}$$

Take the square root of 2401.

$$y=\frac{-1±49}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-1±49}{4}$$

Now solve the equation y=\frac{-1±49}{4} when ± is plus. Add -1 to 49.

$$y=\frac{48}{4}$$

Divide 48 by 4.

$$y=12$$

Now solve the equation y=\frac{-1±49}{4} when ± is minus. Subtract 49 from -1.

$$y=-\frac{50}{4}$$

Reduce the fraction \frac{-50}{4} to lowest terms by extracting and canceling out 2.

$$y=-\frac{25}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and -\frac{25}{2} for x_{2}.

$$2y^{2}+y-300=2\left(y-12\right)\left(y-\left(-\frac{25}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$2y^{2}+y-300=2\left(y-12\right)\left(y+\frac{25}{2}\right)$$

Add \frac{25}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2y^{2}+y-300=2\left(y-12\right)\times \frac{2y+25}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2y^{2}+y-300=\left(y-12\right)\left(2y+25\right)$$

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 2

$$x ^ 2 +\frac{1}{2}x -150 = 0$$

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

$$r + s = -\frac{1}{2} rs = -150$$

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

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -150

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -150$$

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

$$\frac{1}{16} - u^2 = -150$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -150-\frac{1}{16} = -\frac{2401}{16}$$

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

$$u^2 = \frac{2401}{16} u = \pm\sqrt{\frac{2401}{16}} = \pm \frac{49}{4}$$

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

$$r =-\frac{1}{4} - \frac{49}{4} = -12.500 s = -\frac{1}{4} + \frac{49}{4} = 12$$

Evaluate

$$\left(y-12\right)\left(2y+25\right)$$