Solve 20x^2-7x-40

$$20 { x }^{ 2 } -7x-40$$

Factor

$$\left(5x-8\right)\left(4x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-7 ab=20\left(-40\right)=-800$$

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 -800.

$$1,-800 2,-400 4,-200 5,-160 8,-100 10,-80 16,-50 20,-40 25,-32$$

Calculate the sum for each pair.

$$1-800=-799 2-400=-398 4-200=-196 5-160=-155 8-100=-92 10-80=-70 16-50=-34 20-40=-20 25-32=-7$$

The solution is the pair that gives sum -7.

$$a=-32 b=25$$

Rewrite 20x^{2}-7x-40 as \left(20x^{2}-32x\right)+\left(25x-40\right).

$$\left(20x^{2}-32x\right)+\left(25x-40\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(5x-8\right)+5\left(5x-8\right)$$

Factor out common term 5x-8 by using distributive property.

$$\left(5x-8\right)\left(4x+5\right)$$

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.

$$20x^{2}-7x-40=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 20\left(-40\right)}}{2\times 20}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 20\left(-40\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-7\right)±\sqrt{49-80\left(-40\right)}}{2\times 20}$$

Multiply -80 times -40.

$$x=\frac{-\left(-7\right)±\sqrt{49+3200}}{2\times 20}$$

Add 49 to 3200.

$$x=\frac{-\left(-7\right)±\sqrt{3249}}{2\times 20}$$

Take the square root of 3249.

$$x=\frac{-\left(-7\right)±57}{2\times 20}$$

The opposite of -7 is 7.

$$x=\frac{7±57}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{7±57}{40}$$

Now solve the equation x=\frac{7±57}{40} when ± is plus. Add 7 to 57.

$$x=\frac{64}{40}$$

Reduce the fraction \frac{64}{40} to lowest terms by extracting and canceling out 8.

$$x=\frac{8}{5}$$

Now solve the equation x=\frac{7±57}{40} when ± is minus. Subtract 57 from 7.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{5} for x_{1} and -\frac{5}{4} for x_{2}.

$$20x^{2}-7x-40=20\left(x-\frac{8}{5}\right)\left(x-\left(-\frac{5}{4}\right)\right)$$

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

$$20x^{2}-7x-40=20\left(x-\frac{8}{5}\right)\left(x+\frac{5}{4}\right)$$

Subtract \frac{8}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-7x-40=20\times \frac{5x-8}{5}\left(x+\frac{5}{4}\right)$$

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

$$20x^{2}-7x-40=20\times \frac{5x-8}{5}\times \frac{4x+5}{4}$$

Multiply \frac{5x-8}{5} times \frac{4x+5}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-7x-40=20\times \frac{\left(5x-8\right)\left(4x+5\right)}{5\times 4}$$

Multiply 5 times 4.

$$20x^{2}-7x-40=20\times \frac{\left(5x-8\right)\left(4x+5\right)}{20}$$

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

$$20x^{2}-7x-40=\left(5x-8\right)\left(4x+5\right)$$

Evaluate

$$\left(5x-8\right)\left(4x+5\right)$$