Solve 5x^2+25x-250

$$5 x ^ { 2 } + 25 x - 250$$

Factor

$$5\left(x-5\right)\left(x+10\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}+5x-50\right)$$

Consider x^{2}+5x-50. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-50. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-50\right)=-50$$

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 x^{2}+5x-50 as \left(x^{2}-5x\right)+\left(10x-50\right).

$$\left(x^{2}-5x\right)+\left(10x-50\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-5\right)+10\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+10\right)$$

Rewrite the complete factored expression.

$$5\left(x-5\right)\left(x+10\right)$$

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.

$$5x^{2}+25x-250=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5\left(-250\right)}}{2\times 5}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 5\left(-250\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-25±\sqrt{625-20\left(-250\right)}}{2\times 5}$$

Multiply -20 times -250.

$$x=\frac{-25±\sqrt{625+5000}}{2\times 5}$$

Add 625 to 5000.

$$x=\frac{-25±\sqrt{5625}}{2\times 5}$$

Take the square root of 5625.

$$x=\frac{-25±75}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-25±75}{10}$$

Now solve the equation x=\frac{-25±75}{10} when ± is plus. Add -25 to 75.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{-25±75}{10} when ± is minus. Subtract 75 from -25.

$$x=-\frac{100}{10}$$

Divide -100 by 10.

$$x=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -10 for x_{2}.

$$5x^{2}+25x-250=5\left(x-5\right)\left(x-\left(-10\right)\right)$$

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

$$5x^{2}+25x-250=5\left(x-5\right)\left(x+10\right)$$

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

$$x ^ 2 +5x -50 = 0$$

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

$$r + s = -5 rs = -50$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -50

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -50$$

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

$$\frac{25}{4} - u^2 = -50$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -50-\frac{25}{4} = -\frac{225}{4}$$

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

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

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

$$r =-\frac{5}{2} - \frac{15}{2} = -10 s = -\frac{5}{2} + \frac{15}{2} = 5$$

Evaluate

$$5\left(x-5\right)\left(x+10\right)$$