Solve 10x^2+3x-1

$$10 { x }^{ 2 } +3x-1$$

Factor

$$\left(5x-1\right)\left(2x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=3 ab=10\left(-1\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite 10x^{2}+3x-1 as \left(10x^{2}-2x\right)+\left(5x-1\right).

$$\left(10x^{2}-2x\right)+\left(5x-1\right)$$

Factor out 2x in 10x^{2}-2x.

$$2x\left(5x-1\right)+5x-1$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(2x+1\right)$$

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.

$$10x^{2}+3x-1=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 10\left(-1\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-3±\sqrt{9-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$x=\frac{-3±\sqrt{9+40}}{2\times 10}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\times 10}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-3±7}{20}$$

Now solve the equation x=\frac{-3±7}{20} when ± is plus. Add -3 to 7.

$$x=\frac{4}{20}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-3±7}{20} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{20}$$

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

$$x=-\frac{1}{2}$$

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

$$10x^{2}+3x-1=10\left(x-\frac{1}{5}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$10x^{2}+3x-1=10\left(x-\frac{1}{5}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+3x-1=10\times \frac{5x-1}{5}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+3x-1=10\times \frac{5x-1}{5}\times \frac{2x+1}{2}$$

Multiply \frac{5x-1}{5} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+3x-1=10\times \frac{\left(5x-1\right)\left(2x+1\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+3x-1=10\times \frac{\left(5x-1\right)\left(2x+1\right)}{10}$$

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

$$10x^{2}+3x-1=\left(5x-1\right)\left(2x+1\right)$$

Evaluate

$$\left(5x-1\right)\left(2x+1\right)$$