Solve 3x^2-65x+250

$$3 x ^ { 2 } - 65 x + 250$$

Factor

$$\left(x-5\right)\left(3x-50\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+250. To find a and b, set up a system to be solved.

$$a+b=-65 ab=3\times 250=750$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 750.

$$-1,-750 -2,-375 -3,-250 -5,-150 -6,-125 -10,-75 -15,-50 -25,-30$$

Calculate the sum for each pair.

$$-1-750=-751 -2-375=-377 -3-250=-253 -5-150=-155 -6-125=-131 -10-75=-85 -15-50=-65 -25-30=-55$$

The solution is the pair that gives sum -65.

$$a=-50 b=-15$$

Rewrite 3x^{2}-65x+250 as \left(3x^{2}-50x\right)+\left(-15x+250\right).

$$\left(3x^{2}-50x\right)+\left(-15x+250\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(3x-50\right)-5\left(3x-50\right)$$

Factor out common term 3x-50 by using distributive property.

$$\left(3x-50\right)\left(x-5\right)$$

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.

$$3x^{2}-65x+250=0$$

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.

$$x=\frac{-\left(-65\right)±\sqrt{\left(-65\right)^{2}-4\times 3\times 250}}{2\times 3}$$

Square -65.

$$x=\frac{-\left(-65\right)±\sqrt{4225-4\times 3\times 250}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-65\right)±\sqrt{4225-12\times 250}}{2\times 3}$$

Multiply -12 times 250.

$$x=\frac{-\left(-65\right)±\sqrt{4225-3000}}{2\times 3}$$

Add 4225 to -3000.

$$x=\frac{-\left(-65\right)±\sqrt{1225}}{2\times 3}$$

Take the square root of 1225.

$$x=\frac{-\left(-65\right)±35}{2\times 3}$$

The opposite of -65 is 65.

$$x=\frac{65±35}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{65±35}{6}$$

Now solve the equation x=\frac{65±35}{6} when ± is plus. Add 65 to 35.

$$x=\frac{100}{6}$$

Reduce the fraction \frac{100}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{50}{3}$$

Now solve the equation x=\frac{65±35}{6} when ± is minus. Subtract 35 from 65.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{50}{3} for x_{1} and 5 for x_{2}.

$$3x^{2}-65x+250=3\left(x-\frac{50}{3}\right)\left(x-5\right)$$

Subtract \frac{50}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-65x+250=3\times \frac{3x-50}{3}\left(x-5\right)$$

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

$$3x^{2}-65x+250=\left(3x-50\right)\left(x-5\right)$$

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 3

$$x ^ 2 -\frac{65}{3}x +\frac{250}{3} = 0$$

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

$$r + s = \frac{65}{3} rs = \frac{250}{3}$$

Two numbers r and s sum up to \frac{65}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{65}{3} = \frac{65}{6}. 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>

$$r = \frac{65}{6} - u s = \frac{65}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{250}{3}

$$(\frac{65}{6} - u) (\frac{65}{6} + u) = \frac{250}{3}$$

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

$$\frac{4225}{36} - u^2 = \frac{250}{3}$$

Simplify the expression by subtracting \frac{4225}{36} on both sides

$$-u^2 = \frac{250}{3}-\frac{4225}{36} = -\frac{1225}{36}$$

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

$$u^2 = \frac{1225}{36} u = \pm\sqrt{\frac{1225}{36}} = \pm \frac{35}{6}$$

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

$$r =\frac{65}{6} - \frac{35}{6} = 5.000 s = \frac{65}{6} + \frac{35}{6} = 16.667$$

Evaluate

$$\left(x-5\right)\left(3x-50\right)$$