Solve 20x^2-17x-10

$$20 { x }^{ 2 } -17x-10$$

Factor

$$\left(4x-5\right)\left(5x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-17 ab=20\left(-10\right)=-200$$

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 -200.

$$1,-200 2,-100 4,-50 5,-40 8,-25 10,-20$$

Calculate the sum for each pair.

$$1-200=-199 2-100=-98 4-50=-46 5-40=-35 8-25=-17 10-20=-10$$

The solution is the pair that gives sum -17.

$$a=-25 b=8$$

Rewrite 20x^{2}-17x-10 as \left(20x^{2}-25x\right)+\left(8x-10\right).

$$\left(20x^{2}-25x\right)+\left(8x-10\right)$$

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

$$5x\left(4x-5\right)+2\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(5x+2\right)$$

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.

$$20x^{2}-17x-10=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 20\left(-10\right)}}{2\times 20}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 20\left(-10\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-17\right)±\sqrt{289-80\left(-10\right)}}{2\times 20}$$

Multiply -80 times -10.

$$x=\frac{-\left(-17\right)±\sqrt{289+800}}{2\times 20}$$

Add 289 to 800.

$$x=\frac{-\left(-17\right)±\sqrt{1089}}{2\times 20}$$

Take the square root of 1089.

$$x=\frac{-\left(-17\right)±33}{2\times 20}$$

The opposite of -17 is 17.

$$x=\frac{17±33}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{17±33}{40}$$

Now solve the equation x=\frac{17±33}{40} when ± is plus. Add 17 to 33.

$$x=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{17±33}{40} when ± is minus. Subtract 33 from 17.

$$x=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and -\frac{2}{5} for x_{2}.

$$20x^{2}-17x-10=20\left(x-\frac{5}{4}\right)\left(x-\left(-\frac{2}{5}\right)\right)$$

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

$$20x^{2}-17x-10=20\left(x-\frac{5}{4}\right)\left(x+\frac{2}{5}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-17x-10=20\times \frac{4x-5}{4}\left(x+\frac{2}{5}\right)$$

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

$$20x^{2}-17x-10=20\times \frac{4x-5}{4}\times \frac{5x+2}{5}$$

Multiply \frac{4x-5}{4} times \frac{5x+2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-17x-10=20\times \frac{\left(4x-5\right)\left(5x+2\right)}{4\times 5}$$

Multiply 4 times 5.

$$20x^{2}-17x-10=20\times \frac{\left(4x-5\right)\left(5x+2\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20x^{2}-17x-10=\left(4x-5\right)\left(5x+2\right)$$

Evaluate

$$\left(4x-5\right)\left(5x+2\right)$$