Solve 25x^2+25x-150

$$25 x ^ { 2 } + 25 x - 150$$

Factor

$$25\left(x-2\right)\left(x+3\right)$$

Solution Steps

Factor out 25.

$$25\left(x^{2}+x-6\right)$$

Consider x^{2}+x-6. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite x^{2}+x-6 as \left(x^{2}-2x\right)+\left(3x-6\right).

$$\left(x^{2}-2x\right)+\left(3x-6\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-2\right)+3\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$25\left(x-2\right)\left(x+3\right)$$

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.

$$25x^{2}+25x-150=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 25\left(-150\right)}}{2\times 25}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 25\left(-150\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-25±\sqrt{625-100\left(-150\right)}}{2\times 25}$$

Multiply -100 times -150.

$$x=\frac{-25±\sqrt{625+15000}}{2\times 25}$$

Add 625 to 15000.

$$x=\frac{-25±\sqrt{15625}}{2\times 25}$$

Take the square root of 15625.

$$x=\frac{-25±125}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-25±125}{50}$$

Now solve the equation x=\frac{-25±125}{50} when ± is plus. Add -25 to 125.

$$x=\frac{100}{50}$$

Divide 100 by 50.

$$x=2$$

Now solve the equation x=\frac{-25±125}{50} when ± is minus. Subtract 125 from -25.

$$x=-\frac{150}{50}$$

Divide -150 by 50.

$$x=-3$$

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

$$25x^{2}+25x-150=25\left(x-2\right)\left(x-\left(-3\right)\right)$$

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

$$25x^{2}+25x-150=25\left(x-2\right)\left(x+3\right)$$

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 25

$$x ^ 2 +1x -6 = 0$$

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

$$r + s = -1 rs = -6$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -6

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -6$$

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

$$\frac{1}{4} - u^2 = -6$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -6-\frac{1}{4} = -\frac{25}{4}$$

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.

$$r =-\frac{1}{2} - \frac{5}{2} = -3 s = -\frac{1}{2} + \frac{5}{2} = 2$$

Evaluate

$$25\left(x-2\right)\left(x+3\right)$$