Solve 10n^2+45n+50

$$10 n ^ { 2 } + 45 n + 50$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(2n^{2}+9n+10\right)$$

Consider 2n^{2}+9n+10. Factor the expression by grouping. First, the expression needs to be rewritten as 2n^{2}+an+bn+10. To find a and b, set up a system to be solved.

$$a+b=9 ab=2\times 10=20$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

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

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 9.

$$a=4 b=5$$

Rewrite 2n^{2}+9n+10 as \left(2n^{2}+4n\right)+\left(5n+10\right).

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

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

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

Factor out common term n+2 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$10n^{2}+45n+50=0$$

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.

$$n=\frac{-45±\sqrt{45^{2}-4\times 10\times 50}}{2\times 10}$$

Square 45.

$$n=\frac{-45±\sqrt{2025-4\times 10\times 50}}{2\times 10}$$

Multiply -4 times 10.

$$n=\frac{-45±\sqrt{2025-40\times 50}}{2\times 10}$$

Multiply -40 times 50.

$$n=\frac{-45±\sqrt{2025-2000}}{2\times 10}$$

Add 2025 to -2000.

$$n=\frac{-45±\sqrt{25}}{2\times 10}$$

Take the square root of 25.

$$n=\frac{-45±5}{2\times 10}$$

Multiply 2 times 10.

$$n=\frac{-45±5}{20}$$

Now solve the equation n=\frac{-45±5}{20} when ± is plus. Add -45 to 5.

$$n=-\frac{40}{20}$$

Divide -40 by 20.

$$n=-2$$

Now solve the equation n=\frac{-45±5}{20} when ± is minus. Subtract 5 from -45.

$$n=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

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

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

$$10n^{2}+45n+50=10\left(n-\left(-2\right)\right)\left(n-\left(-\frac{5}{2}\right)\right)$$

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

$$10n^{2}+45n+50=10\left(n+2\right)\left(n+\frac{5}{2}\right)$$

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

$$10n^{2}+45n+50=10\left(n+2\right)\times \frac{2n+5}{2}$$

Cancel out 2, the greatest common factor in 10 and 2.

$$10n^{2}+45n+50=5\left(n+2\right)\left(2n+5\right)$$

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 10

$$x ^ 2 +\frac{9}{2}x +5 = 0$$

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

$$r + s = -\frac{9}{2} rs = 5$$

Two numbers r and s sum up to -\frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{2} = -\frac{9}{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>

$$r = -\frac{9}{4} - u s = -\frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(-\frac{9}{4} - u) (-\frac{9}{4} + u) = 5$$

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

$$\frac{81}{16} - u^2 = 5$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = 5-\frac{81}{16} = -\frac{1}{16}$$

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.

$$r =-\frac{9}{4} - \frac{1}{4} = -2.500 s = -\frac{9}{4} + \frac{1}{4} = -2$$

Evaluate

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