Solve 39x^2+200x+209

$$39 x ^ { 2 } + 200 x + 209$$

Factor

$$\left(3x+11\right)\left(13x+19\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 39x^{2}+ax+bx+209. To find a and b, set up a system to be solved.

$$a+b=200 ab=39\times 209=8151$$

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 8151.

$$1,8151 3,2717 11,741 13,627 19,429 33,247 39,209 57,143$$

Calculate the sum for each pair.

$$1+8151=8152 3+2717=2720 11+741=752 13+627=640 19+429=448 33+247=280 39+209=248 57+143=200$$

The solution is the pair that gives sum 200.

$$a=57 b=143$$

Rewrite 39x^{2}+200x+209 as \left(39x^{2}+57x\right)+\left(143x+209\right).

$$\left(39x^{2}+57x\right)+\left(143x+209\right)$$

Factor out 3x in the first and 11 in the second group.

$$3x\left(13x+19\right)+11\left(13x+19\right)$$

Factor out common term 13x+19 by using distributive property.

$$\left(13x+19\right)\left(3x+11\right)$$

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.

$$39x^{2}+200x+209=0$$

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.

$$x=\frac{-200±\sqrt{200^{2}-4\times 39\times 209}}{2\times 39}$$

Square 200.

$$x=\frac{-200±\sqrt{40000-4\times 39\times 209}}{2\times 39}$$

Multiply -4 times 39.

$$x=\frac{-200±\sqrt{40000-156\times 209}}{2\times 39}$$

Multiply -156 times 209.

$$x=\frac{-200±\sqrt{40000-32604}}{2\times 39}$$

Add 40000 to -32604.

$$x=\frac{-200±\sqrt{7396}}{2\times 39}$$

Take the square root of 7396.

$$x=\frac{-200±86}{2\times 39}$$

Multiply 2 times 39.

$$x=\frac{-200±86}{78}$$

Now solve the equation x=\frac{-200±86}{78} when ± is plus. Add -200 to 86.

$$x=-\frac{114}{78}$$

Reduce the fraction \frac{-114}{78} to lowest terms by extracting and canceling out 6.

$$x=-\frac{19}{13}$$

Now solve the equation x=\frac{-200±86}{78} when ± is minus. Subtract 86 from -200.

$$x=-\frac{286}{78}$$

Reduce the fraction \frac{-286}{78} to lowest terms by extracting and canceling out 26.

$$x=-\frac{11}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{19}{13} for x_{1} and -\frac{11}{3} for x_{2}.

$$39x^{2}+200x+209=39\left(x-\left(-\frac{19}{13}\right)\right)\left(x-\left(-\frac{11}{3}\right)\right)$$

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

$$39x^{2}+200x+209=39\left(x+\frac{19}{13}\right)\left(x+\frac{11}{3}\right)$$

Add \frac{19}{13} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$39x^{2}+200x+209=39\times \frac{13x+19}{13}\left(x+\frac{11}{3}\right)$$

Add \frac{11}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$39x^{2}+200x+209=39\times \frac{13x+19}{13}\times \frac{3x+11}{3}$$

Multiply \frac{13x+19}{13} times \frac{3x+11}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$39x^{2}+200x+209=39\times \frac{\left(13x+19\right)\left(3x+11\right)}{13\times 3}$$

Multiply 13 times 3.

$$39x^{2}+200x+209=39\times \frac{\left(13x+19\right)\left(3x+11\right)}{39}$$

Cancel out 39, the greatest common factor in 39 and 39.

$$39x^{2}+200x+209=\left(13x+19\right)\left(3x+11\right)$$

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 39

$$x ^ 2 +\frac{200}{39}x +\frac{209}{39} = 0$$

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

$$r + s = -\frac{200}{39} rs = \frac{209}{39}$$

Two numbers r and s sum up to -\frac{200}{39} exactly when the average of the two numbers is \frac{1}{2}*-\frac{200}{39} = -\frac{100}{39}. 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>

$$r = -\frac{100}{39} - u s = -\frac{100}{39} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{209}{39}

$$(-\frac{100}{39} - u) (-\frac{100}{39} + u) = \frac{209}{39}$$

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

$$\frac{10000}{1521} - u^2 = \frac{209}{39}$$

Simplify the expression by subtracting \frac{10000}{1521} on both sides

$$-u^2 = \frac{209}{39}-\frac{10000}{1521} = -\frac{1849}{1521}$$

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

$$u^2 = \frac{1849}{1521} u = \pm\sqrt{\frac{1849}{1521}} = \pm \frac{43}{39}$$

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

$$r =-\frac{100}{39} - \frac{43}{39} = -3.667 s = -\frac{100}{39} + \frac{43}{39} = -1.462$$

Evaluate

$$\left(3x+11\right)\left(13x+19\right)$$