Solve 75x^2+200x-400

$$75 x ^ { 2 } + 200 x - 400$$

Factor

$$25\left(3x-4\right)\left(x+4\right)$$

Solution Steps

Factor out 25.

$$25\left(3x^{2}+8x-16\right)$$

Consider 3x^{2}+8x-16. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=3\left(-16\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite 3x^{2}+8x-16 as \left(3x^{2}-4x\right)+\left(12x-16\right).

$$\left(3x^{2}-4x\right)+\left(12x-16\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(3x-4\right)+4\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$25\left(3x-4\right)\left(x+4\right)$$

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.

$$75x^{2}+200x-400=0$$

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.

$$x=\frac{-200±\sqrt{200^{2}-4\times 75\left(-400\right)}}{2\times 75}$$

Square 200.

$$x=\frac{-200±\sqrt{40000-4\times 75\left(-400\right)}}{2\times 75}$$

Multiply -4 times 75.

$$x=\frac{-200±\sqrt{40000-300\left(-400\right)}}{2\times 75}$$

Multiply -300 times -400.

$$x=\frac{-200±\sqrt{40000+120000}}{2\times 75}$$

Add 40000 to 120000.

$$x=\frac{-200±\sqrt{160000}}{2\times 75}$$

Take the square root of 160000.

$$x=\frac{-200±400}{2\times 75}$$

Multiply 2 times 75.

$$x=\frac{-200±400}{150}$$

Now solve the equation x=\frac{-200±400}{150} when ± is plus. Add -200 to 400.

$$x=\frac{200}{150}$$

Reduce the fraction \frac{200}{150} to lowest terms by extracting and canceling out 50.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{-200±400}{150} when ± is minus. Subtract 400 from -200.

$$x=-\frac{600}{150}$$

Divide -600 by 150.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -4 for x_{2}.

$$75x^{2}+200x-400=75\left(x-\frac{4}{3}\right)\left(x-\left(-4\right)\right)$$

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

$$75x^{2}+200x-400=75\left(x-\frac{4}{3}\right)\left(x+4\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$75x^{2}+200x-400=75\times \frac{3x-4}{3}\left(x+4\right)$$

Cancel out 3, the greatest common factor in 75 and 3.

$$75x^{2}+200x-400=25\left(3x-4\right)\left(x+4\right)$$

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 75

$$x ^ 2 +\frac{8}{3}x -\frac{16}{3} = 0$$

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

$$r + s = -\frac{8}{3} rs = -\frac{16}{3}$$

Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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>

$$r = -\frac{4}{3} - u s = -\frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{3}

$$(-\frac{4}{3} - u) (-\frac{4}{3} + u) = -\frac{16}{3}$$

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

$$\frac{16}{9} - u^2 = -\frac{16}{3}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -\frac{16}{3}-\frac{16}{9} = -\frac{64}{9}$$

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

$$u^2 = \frac{64}{9} u = \pm\sqrt{\frac{64}{9}} = \pm \frac{8}{3}$$

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

$$r =-\frac{4}{3} - \frac{8}{3} = -4 s = -\frac{4}{3} + \frac{8}{3} = 1.333$$

Evaluate

$$25\left(3x-4\right)\left(x+4\right)$$