Solve 5x^2-35x+50

$$5 x ^ { 2 } - 35 x + 50$$

Factor

$$5\left(x-5\right)\left(x-2\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}-7x+10\right)$$

Consider x^{2}-7x+10. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite x^{2}-7x+10 as \left(x^{2}-5x\right)+\left(-2x+10\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-5\right)-2\left(x-5\right)$$

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

$$\left(x-5\right)\left(x-2\right)$$

Rewrite the complete factored expression.

$$5\left(x-5\right)\left(x-2\right)$$

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.

$$5x^{2}-35x+50=0$$

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.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 5\times 50}}{2\times 5}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 5\times 50}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-35\right)±\sqrt{1225-20\times 50}}{2\times 5}$$

Multiply -20 times 50.

$$x=\frac{-\left(-35\right)±\sqrt{1225-1000}}{2\times 5}$$

Add 1225 to -1000.

$$x=\frac{-\left(-35\right)±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$x=\frac{-\left(-35\right)±15}{2\times 5}$$

The opposite of -35 is 35.

$$x=\frac{35±15}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{35±15}{10}$$

Now solve the equation x=\frac{35±15}{10} when ± is plus. Add 35 to 15.

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

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{35±15}{10} when ± is minus. Subtract 15 from 35.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

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 2 for x_{2}.

$$5x^{2}-35x+50=5\left(x-5\right)\left(x-2\right)$$

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

$$x ^ 2 -7x +10 = 0$$

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

$$r + s = 7 rs = 10$$

Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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>

$$r = \frac{7}{2} - u s = \frac{7}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(\frac{7}{2} - u) (\frac{7}{2} + u) = 10$$

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

$$\frac{49}{4} - u^2 = 10$$

Simplify the expression by subtracting \frac{49}{4} on both sides

$$-u^2 = 10-\frac{49}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =\frac{7}{2} - \frac{3}{2} = 2 s = \frac{7}{2} + \frac{3}{2} = 5$$

Evaluate

$$5\left(x-5\right)\left(x-2\right)$$