Solve 50g^2-15g-2

$$50 g ^ { 2 } - 15 g - 2$$

Factor

$$\left(5g-2\right)\left(10g+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 50g^{2}+ag+bg-2. To find a and b, set up a system to be solved.

$$a+b=-15 ab=50\left(-2\right)=-100$$

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 -100.

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

Calculate the sum for each pair.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

The solution is the pair that gives sum -15.

$$a=-20 b=5$$

Rewrite 50g^{2}-15g-2 as \left(50g^{2}-20g\right)+\left(5g-2\right).

$$\left(50g^{2}-20g\right)+\left(5g-2\right)$$

Factor out 10g in 50g^{2}-20g.

$$10g\left(5g-2\right)+5g-2$$

Factor out common term 5g-2 by using distributive property.

$$\left(5g-2\right)\left(10g+1\right)$$

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.

$$50g^{2}-15g-2=0$$

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.

$$g=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 50\left(-2\right)}}{2\times 50}$$

Square -15.

$$g=\frac{-\left(-15\right)±\sqrt{225-4\times 50\left(-2\right)}}{2\times 50}$$

Multiply -4 times 50.

$$g=\frac{-\left(-15\right)±\sqrt{225-200\left(-2\right)}}{2\times 50}$$

Multiply -200 times -2.

$$g=\frac{-\left(-15\right)±\sqrt{225+400}}{2\times 50}$$

Add 225 to 400.

$$g=\frac{-\left(-15\right)±\sqrt{625}}{2\times 50}$$

Take the square root of 625.

$$g=\frac{-\left(-15\right)±25}{2\times 50}$$

The opposite of -15 is 15.

$$g=\frac{15±25}{2\times 50}$$

Multiply 2 times 50.

$$g=\frac{15±25}{100}$$

Now solve the equation g=\frac{15±25}{100} when ± is plus. Add 15 to 25.

$$g=\frac{40}{100}$$

Reduce the fraction \frac{40}{100} to lowest terms by extracting and canceling out 20.

$$g=\frac{2}{5}$$

Now solve the equation g=\frac{15±25}{100} when ± is minus. Subtract 25 from 15.

$$g=-\frac{10}{100}$$

Reduce the fraction \frac{-10}{100} to lowest terms by extracting and canceling out 10.

$$g=-\frac{1}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{1}{10} for x_{2}.

$$50g^{2}-15g-2=50\left(g-\frac{2}{5}\right)\left(g-\left(-\frac{1}{10}\right)\right)$$

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

$$50g^{2}-15g-2=50\left(g-\frac{2}{5}\right)\left(g+\frac{1}{10}\right)$$

Subtract \frac{2}{5} from g by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$50g^{2}-15g-2=50\times \frac{5g-2}{5}\left(g+\frac{1}{10}\right)$$

Add \frac{1}{10} to g by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$50g^{2}-15g-2=50\times \frac{5g-2}{5}\times \frac{10g+1}{10}$$

Multiply \frac{5g-2}{5} times \frac{10g+1}{10} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$50g^{2}-15g-2=50\times \frac{\left(5g-2\right)\left(10g+1\right)}{5\times 10}$$

Multiply 5 times 10.

$$50g^{2}-15g-2=50\times \frac{\left(5g-2\right)\left(10g+1\right)}{50}$$

Cancel out 50, the greatest common factor in 50 and 50.

$$50g^{2}-15g-2=\left(5g-2\right)\left(10g+1\right)$$

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 50

$$x ^ 2 -\frac{3}{10}x -\frac{1}{25} = 0$$

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

$$r + s = \frac{3}{10} rs = -\frac{1}{25}$$

Two numbers r and s sum up to \frac{3}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{10} = \frac{3}{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>

$$r = \frac{3}{20} - u s = \frac{3}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{25}

$$(\frac{3}{20} - u) (\frac{3}{20} + u) = -\frac{1}{25}$$

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

$$\frac{9}{400} - u^2 = -\frac{1}{25}$$

Simplify the expression by subtracting \frac{9}{400} on both sides

$$-u^2 = -\frac{1}{25}-\frac{9}{400} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{3}{20} - \frac{1}{4} = -0.100 s = \frac{3}{20} + \frac{1}{4} = 0.400$$

Evaluate

$$\left(5g-2\right)\left(10g+1\right)$$