Solve 2n^2-3n-20

$$2 n ^ { 2 } - 3 n - 20$$

Factor

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

Solution Steps

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

$$a+b=-3 ab=2\left(-20\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite 2n^{2}-3n-20 as \left(2n^{2}-8n\right)+\left(5n-20\right).

$$\left(2n^{2}-8n\right)+\left(5n-20\right)$$

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

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

Factor out common term n-4 by using distributive property.

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

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.

$$2n^{2}-3n-20=0$$

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.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-\left(-3\right)±\sqrt{9-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$n=\frac{-\left(-3\right)±\sqrt{9+160}}{2\times 2}$$

Add 9 to 160.

$$n=\frac{-\left(-3\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$n=\frac{-\left(-3\right)±13}{2\times 2}$$

The opposite of -3 is 3.

$$n=\frac{3±13}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{3±13}{4}$$

Now solve the equation n=\frac{3±13}{4} when ± is plus. Add 3 to 13.

$$n=\frac{16}{4}$$

Divide 16 by 4.

$$n=4$$

Now solve the equation n=\frac{3±13}{4} when ± is minus. Subtract 13 from 3.

$$n=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

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

$$2n^{2}-3n-20=2\left(n-4\right)\left(n-\left(-\frac{5}{2}\right)\right)$$

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

$$2n^{2}-3n-20=2\left(n-4\right)\left(n+\frac{5}{2}\right)$$

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

$$2n^{2}-3n-20=2\left(n-4\right)\times \frac{2n+5}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2n^{2}-3n-20=\left(n-4\right)\left(2n+5\right)$$

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 2

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

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

$$r + s = \frac{3}{2} rs = -10$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -10$$

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

$$\frac{9}{16} - u^2 = -10$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -10-\frac{9}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =\frac{3}{4} - \frac{13}{4} = -2.500 s = \frac{3}{4} + \frac{13}{4} = 4$$

Evaluate

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