Solve 4n^2-2n-90

$$4 n ^ { 2 } - 2 n - 90$$

Factor

$$2\left(n-5\right)\left(2n+9\right)$$

Solution Steps

Factor out 2.

$$2\left(2n^{2}-n-45\right)$$

Consider 2n^{2}-n-45. Factor the expression by grouping. First, the expression needs to be rewritten as 2n^{2}+an+bn-45. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-45\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -1.

$$a=-10 b=9$$

Rewrite 2n^{2}-n-45 as \left(2n^{2}-10n\right)+\left(9n-45\right).

$$\left(2n^{2}-10n\right)+\left(9n-45\right)$$

Factor out 2n in the first and 9 in the second group.

$$2n\left(n-5\right)+9\left(n-5\right)$$

Factor out common term n-5 by using distributive property.

$$\left(n-5\right)\left(2n+9\right)$$

Rewrite the complete factored expression.

$$2\left(n-5\right)\left(2n+9\right)$$

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.

$$4n^{2}-2n-90=0$$

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.

$$n=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 4\left(-90\right)}}{2\times 4}$$

Square -2.

$$n=\frac{-\left(-2\right)±\sqrt{4-4\times 4\left(-90\right)}}{2\times 4}$$

Multiply -4 times 4.

$$n=\frac{-\left(-2\right)±\sqrt{4-16\left(-90\right)}}{2\times 4}$$

Multiply -16 times -90.

$$n=\frac{-\left(-2\right)±\sqrt{4+1440}}{2\times 4}$$

Add 4 to 1440.

$$n=\frac{-\left(-2\right)±\sqrt{1444}}{2\times 4}$$

Take the square root of 1444.

$$n=\frac{-\left(-2\right)±38}{2\times 4}$$

The opposite of -2 is 2.

$$n=\frac{2±38}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{2±38}{8}$$

Now solve the equation n=\frac{2±38}{8} when ± is plus. Add 2 to 38.

$$n=\frac{40}{8}$$

Divide 40 by 8.

$$n=5$$

Now solve the equation n=\frac{2±38}{8} when ± is minus. Subtract 38 from 2.

$$n=-\frac{36}{8}$$

Reduce the fraction \frac{-36}{8} to lowest terms by extracting and canceling out 4.

$$n=-\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -\frac{9}{2} for x_{2}.

$$4n^{2}-2n-90=4\left(n-5\right)\left(n-\left(-\frac{9}{2}\right)\right)$$

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

$$4n^{2}-2n-90=4\left(n-5\right)\left(n+\frac{9}{2}\right)$$

Add \frac{9}{2} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4n^{2}-2n-90=4\left(n-5\right)\times \frac{2n+9}{2}$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4n^{2}-2n-90=2\left(n-5\right)\left(2n+9\right)$$

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 4

$$x ^ 2 -\frac{1}{2}x -\frac{45}{2} = 0$$

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

$$r + s = \frac{1}{2} rs = -\frac{45}{2}$$

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

$$r = \frac{1}{4} - u s = \frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{45}{2}

$$(\frac{1}{4} - u) (\frac{1}{4} + u) = -\frac{45}{2}$$

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

$$\frac{1}{16} - u^2 = -\frac{45}{2}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{45}{2}-\frac{1}{16} = -\frac{361}{16}$$

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

$$u^2 = \frac{361}{16} u = \pm\sqrt{\frac{361}{16}} = \pm \frac{19}{4}$$

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

$$r =\frac{1}{4} - \frac{19}{4} = -4.500 s = \frac{1}{4} + \frac{19}{4} = 5$$

Evaluate

$$2\left(n-5\right)\left(2n+9\right)$$