Solve f(x)=25x^2+100x+99

$$f ( x ) = 25 x ^ { 2 } + 100 x + 99$$

Factor

$$\left(5x+9\right)\left(5x+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25x^{2}+ax+bx+99. To find a and b, set up a system to be solved.

$$a+b=100 ab=25\times 99=2475$$

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 2475.

$$1,2475 3,825 5,495 9,275 11,225 15,165 25,99 33,75 45,55$$

Calculate the sum for each pair.

$$1+2475=2476 3+825=828 5+495=500 9+275=284 11+225=236 15+165=180 25+99=124 33+75=108 45+55=100$$

The solution is the pair that gives sum 100.

$$a=45 b=55$$

Rewrite 25x^{2}+100x+99 as \left(25x^{2}+45x\right)+\left(55x+99\right).

$$\left(25x^{2}+45x\right)+\left(55x+99\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(5x+9\right)+11\left(5x+9\right)$$

Factor out common term 5x+9 by using distributive property.

$$\left(5x+9\right)\left(5x+11\right)$$

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.

$$25x^{2}+100x+99=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\times 25\times 99}}{2\times 25}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\times 25\times 99}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-100±\sqrt{10000-100\times 99}}{2\times 25}$$

Multiply -100 times 99.

$$x=\frac{-100±\sqrt{10000-9900}}{2\times 25}$$

Add 10000 to -9900.

$$x=\frac{-100±\sqrt{100}}{2\times 25}$$

Take the square root of 100.

$$x=\frac{-100±10}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-100±10}{50}$$

Now solve the equation x=\frac{-100±10}{50} when ± is plus. Add -100 to 10.

$$x=-\frac{90}{50}$$

Reduce the fraction \frac{-90}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{9}{5}$$

Now solve the equation x=\frac{-100±10}{50} when ± is minus. Subtract 10 from -100.

$$x=-\frac{110}{50}$$

Reduce the fraction \frac{-110}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{11}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{9}{5} for x_{1} and -\frac{11}{5} for x_{2}.

$$25x^{2}+100x+99=25\left(x-\left(-\frac{9}{5}\right)\right)\left(x-\left(-\frac{11}{5}\right)\right)$$

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

$$25x^{2}+100x+99=25\left(x+\frac{9}{5}\right)\left(x+\frac{11}{5}\right)$$

Add \frac{9}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+100x+99=25\times \frac{5x+9}{5}\left(x+\frac{11}{5}\right)$$

Add \frac{11}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+100x+99=25\times \frac{5x+9}{5}\times \frac{5x+11}{5}$$

Multiply \frac{5x+9}{5} times \frac{5x+11}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+100x+99=25\times \frac{\left(5x+9\right)\left(5x+11\right)}{5\times 5}$$

Multiply 5 times 5.

$$25x^{2}+100x+99=25\times \frac{\left(5x+9\right)\left(5x+11\right)}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25x^{2}+100x+99=\left(5x+9\right)\left(5x+11\right)$$

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 25

$$x ^ 2 +4x +\frac{99}{25} = 0$$

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

$$r + s = -4 rs = \frac{99}{25}$$

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

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{99}{25}

$$(-2 - u) (-2 + u) = \frac{99}{25}$$

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

$$4 - u^2 = \frac{99}{25}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{99}{25}-4 = -\frac{1}{25}$$

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

$$u^2 = \frac{1}{25} u = \pm\sqrt{\frac{1}{25}} = \pm \frac{1}{5}$$

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

$$r =-2 - \frac{1}{5} = -2.200 s = -2 + \frac{1}{5} = -1.800$$

Evaluate

$$\left(5x+9\right)\left(5x+11\right)$$