Solve 5x^2-35x+50

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

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}.

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

Evaluate

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