Solve 10x^2-x-2=

$$10 x ^ { 2 } - x - 2 =$$

Factor

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

Solution Steps

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

$$a+b=-1 ab=10\left(-2\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite 10x^{2}-x-2 as \left(10x^{2}-5x\right)+\left(4x-2\right).

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

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

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

Factor out common term 2x-1 by using distributive property.

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

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.

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

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.

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

Multiply -4 times 10.

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

Multiply -40 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 10}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\times 10}$$

Take the square root of 81.

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

The opposite of -1 is 1.

$$x=\frac{1±9}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{1±9}{20}$$

Now solve the equation x=\frac{1±9}{20} when ± is plus. Add 1 to 9.

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

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

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

Now solve the equation x=\frac{1±9}{20} when ± is minus. Subtract 9 from 1.

$$x=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

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

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

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

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

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

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

Multiply 2 times 5.

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

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

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 10

$$x ^ 2 -\frac{1}{10}x -\frac{1}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{1}{10} rs = -\frac{1}{5}$$

Two numbers r and s sum up to \frac{1}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{10} = \frac{1}{20}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{20} - u s = \frac{1}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{5}

$$(\frac{1}{20} - u) (\frac{1}{20} + u) = -\frac{1}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{400} - u^2 = -\frac{1}{5}$$

Simplify the expression by subtracting \frac{1}{400} on both sides

$$-u^2 = -\frac{1}{5}-\frac{1}{400} = -\frac{81}{400}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{81}{400} u = \pm\sqrt{\frac{81}{400}} = \pm \frac{9}{20}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{1}{20} - \frac{9}{20} = -0.400 s = \frac{1}{20} + \frac{9}{20} = 0.500$$

Evaluate

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