Solve 200n^2+30n+1

$$200 n ^ { 2 } + 30 n + 1$$

Factor

$$\left(10n+1\right)\left(20n+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 200n^{2}+an+bn+1. To find a and b, set up a system to be solved.

$$a+b=30 ab=200\times 1=200$$

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 200.

$$1,200 2,100 4,50 5,40 8,25 10,20$$

Calculate the sum for each pair.

$$1+200=201 2+100=102 4+50=54 5+40=45 8+25=33 10+20=30$$

The solution is the pair that gives sum 30.

$$a=10 b=20$$

Rewrite 200n^{2}+30n+1 as \left(200n^{2}+10n\right)+\left(20n+1\right).

$$\left(200n^{2}+10n\right)+\left(20n+1\right)$$

Factor out 10n in 200n^{2}+10n.

$$10n\left(20n+1\right)+20n+1$$

Factor out common term 20n+1 by using distributive property.

$$\left(20n+1\right)\left(10n+1\right)$$

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.

$$200n^{2}+30n+1=0$$

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.

$$n=\frac{-30±\sqrt{30^{2}-4\times 200}}{2\times 200}$$

Square 30.

$$n=\frac{-30±\sqrt{900-4\times 200}}{2\times 200}$$

Multiply -4 times 200.

$$n=\frac{-30±\sqrt{900-800}}{2\times 200}$$

Add 900 to -800.

$$n=\frac{-30±\sqrt{100}}{2\times 200}$$

Take the square root of 100.

$$n=\frac{-30±10}{2\times 200}$$

Multiply 2 times 200.

$$n=\frac{-30±10}{400}$$

Now solve the equation n=\frac{-30±10}{400} when ± is plus. Add -30 to 10.

$$n=-\frac{20}{400}$$

Reduce the fraction \frac{-20}{400} to lowest terms by extracting and canceling out 20.

$$n=-\frac{1}{20}$$

Now solve the equation n=\frac{-30±10}{400} when ± is minus. Subtract 10 from -30.

$$n=-\frac{40}{400}$$

Reduce the fraction \frac{-40}{400} to lowest terms by extracting and canceling out 40.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{20} for x_{1} and -\frac{1}{10} for x_{2}.

$$200n^{2}+30n+1=200\left(n-\left(-\frac{1}{20}\right)\right)\left(n-\left(-\frac{1}{10}\right)\right)$$

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

$$200n^{2}+30n+1=200\left(n+\frac{1}{20}\right)\left(n+\frac{1}{10}\right)$$

Add \frac{1}{20} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$200n^{2}+30n+1=200\times \frac{20n+1}{20}\left(n+\frac{1}{10}\right)$$

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

$$200n^{2}+30n+1=200\times \frac{20n+1}{20}\times \frac{10n+1}{10}$$

Multiply \frac{20n+1}{20} times \frac{10n+1}{10} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$200n^{2}+30n+1=200\times \frac{\left(20n+1\right)\left(10n+1\right)}{20\times 10}$$

Multiply 20 times 10.

$$200n^{2}+30n+1=200\times \frac{\left(20n+1\right)\left(10n+1\right)}{200}$$

Cancel out 200, the greatest common factor in 200 and 200.

$$200n^{2}+30n+1=\left(20n+1\right)\left(10n+1\right)$$

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 200

$$x ^ 2 +\frac{3}{20}x +\frac{1}{200} = 0$$

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

$$r + s = -\frac{3}{20} rs = \frac{1}{200}$$

Two numbers r and s sum up to -\frac{3}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{20} = -\frac{3}{40}. 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>

$$r = -\frac{3}{40} - u s = -\frac{3}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{200}

$$(-\frac{3}{40} - u) (-\frac{3}{40} + u) = \frac{1}{200}$$

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

$$\frac{9}{1600} - u^2 = \frac{1}{200}$$

Simplify the expression by subtracting \frac{9}{1600} on both sides

$$-u^2 = \frac{1}{200}-\frac{9}{1600} = -\frac{1}{1600}$$

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

$$u^2 = \frac{1}{1600} u = \pm\sqrt{\frac{1}{1600}} = \pm \frac{1}{40}$$

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

$$r =-\frac{3}{40} - \frac{1}{40} = -0.100 s = -\frac{3}{40} + \frac{1}{40} = -0.050$$

Evaluate

$$\left(10n+1\right)\left(20n+1\right)$$