Solve 2r^2+5r-25

$$2 r ^ { 2 } + 5 r - 25$$

Factor

$$\left(2r-5\right)\left(r+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2r^{2}+ar+br-25. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-25\right)=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 5.

$$a=-5 b=10$$

Rewrite 2r^{2}+5r-25 as \left(2r^{2}-5r\right)+\left(10r-25\right).

$$\left(2r^{2}-5r\right)+\left(10r-25\right)$$

Factor out r in the first and 5 in the second group.

$$r\left(2r-5\right)+5\left(2r-5\right)$$

Factor out common term 2r-5 by using distributive property.

$$\left(2r-5\right)\left(r+5\right)$$

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.

$$2r^{2}+5r-25=0$$

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.

$$r=\frac{-5±\sqrt{5^{2}-4\times 2\left(-25\right)}}{2\times 2}$$

Square 5.

$$r=\frac{-5±\sqrt{25-4\times 2\left(-25\right)}}{2\times 2}$$

Multiply -4 times 2.

$$r=\frac{-5±\sqrt{25-8\left(-25\right)}}{2\times 2}$$

Multiply -8 times -25.

$$r=\frac{-5±\sqrt{25+200}}{2\times 2}$$

Add 25 to 200.

$$r=\frac{-5±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$r=\frac{-5±15}{2\times 2}$$

Multiply 2 times 2.

$$r=\frac{-5±15}{4}$$

Now solve the equation r=\frac{-5±15}{4} when ± is plus. Add -5 to 15.

$$r=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$r=\frac{5}{2}$$

Now solve the equation r=\frac{-5±15}{4} when ± is minus. Subtract 15 from -5.

$$r=-\frac{20}{4}$$

Divide -20 by 4.

$$r=-5$$

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

$$2r^{2}+5r-25=2\left(r-\frac{5}{2}\right)\left(r-\left(-5\right)\right)$$

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

$$2r^{2}+5r-25=2\left(r-\frac{5}{2}\right)\left(r+5\right)$$

Subtract \frac{5}{2} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2r^{2}+5r-25=2\times \frac{2r-5}{2}\left(r+5\right)$$

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

$$2r^{2}+5r-25=\left(2r-5\right)\left(r+5\right)$$

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 2

$$x ^ 2 +\frac{5}{2}x -\frac{25}{2} = 0$$

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

$$r + s = -\frac{5}{2} rs = -\frac{25}{2}$$

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

$$r = -\frac{5}{4} - u s = -\frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{25}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = -\frac{25}{2}$$

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

$$\frac{25}{16} - u^2 = -\frac{25}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{25}{2}-\frac{25}{16} = -\frac{225}{16}$$

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

$$u^2 = \frac{225}{16} u = \pm\sqrt{\frac{225}{16}} = \pm \frac{15}{4}$$

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

$$r =-\frac{5}{4} - \frac{15}{4} = -5 s = -\frac{5}{4} + \frac{15}{4} = 2.500$$

Evaluate

$$\left(2r-5\right)\left(r+5\right)$$