Solve 10x^2+17x-20

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

Factor

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

Solution Steps

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

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

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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.

$$a=-8 b=25$$

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

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

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

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

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

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

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.

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

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.

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

Square 17.

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

Multiply -4 times 10.

$$x=\frac{-17±\sqrt{289-40\left(-20\right)}}{2\times 10}$$

Multiply -40 times -20.

$$x=\frac{-17±\sqrt{289+800}}{2\times 10}$$

Add 289 to 800.

$$x=\frac{-17±\sqrt{1089}}{2\times 10}$$

Take the square root of 1089.

$$x=\frac{-17±33}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-17±33}{20}$$

Now solve the equation x=\frac{-17±33}{20} when ± is plus. Add -17 to 33.

$$x=\frac{16}{20}$$

Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-17±33}{20} when ± is minus. Subtract 33 from -17.

$$x=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

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

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

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

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

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

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

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

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

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

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

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

Multiply 5 times 2.

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

Cancel out 10, the greatest common factor in 10 and 10.

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

Evaluate

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