Solve 5x^2-23x-10

$$5 { x }^{ 2 } -23x-10$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-23 ab=5\left(-10\right)=-50$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -50.

$$1,-50 2,-25 5,-10$$

Calculate the sum for each pair.

$$1-50=-49 2-25=-23 5-10=-5$$

The solution is the pair that gives sum -23.

$$a=-25 b=2$$

Rewrite 5x^{2}-23x-10 as \left(5x^{2}-25x\right)+\left(2x-10\right).

$$\left(5x^{2}-25x\right)+\left(2x-10\right)$$

Factor out 5x in the first and 2 in the second group.

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

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

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

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.

$$5x^{2}-23x-10=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 5\left(-10\right)}}{2\times 5}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 5\left(-10\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-23\right)±\sqrt{529-20\left(-10\right)}}{2\times 5}$$

Multiply -20 times -10.

$$x=\frac{-\left(-23\right)±\sqrt{529+200}}{2\times 5}$$

Add 529 to 200.

$$x=\frac{-\left(-23\right)±\sqrt{729}}{2\times 5}$$

Take the square root of 729.

$$x=\frac{-\left(-23\right)±27}{2\times 5}$$

The opposite of -23 is 23.

$$x=\frac{23±27}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{23±27}{10}$$

Now solve the equation x=\frac{23±27}{10} when ± is plus. Add 23 to 27.

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

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{23±27}{10} when ± is minus. Subtract 27 from 23.

$$x=-\frac{4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{5}$$

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 -\frac{2}{5} for x_{2}.

$$5x^{2}-23x-10=5\left(x-5\right)\left(x-\left(-\frac{2}{5}\right)\right)$$

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

$$5x^{2}-23x-10=5\left(x-5\right)\left(x+\frac{2}{5}\right)$$

Add \frac{2}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-23x-10=5\left(x-5\right)\times \frac{5x+2}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}-23x-10=\left(x-5\right)\left(5x+2\right)$$

Evaluate

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