Solve 25x^2+25x+4

$$25 x ^ { 2 } + 25 x + 4$$

Factor

$$\left(5x+1\right)\left(5x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=25 ab=25\times 4=100$$

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 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 25.

$$a=5 b=20$$

Rewrite 25x^{2}+25x+4 as \left(25x^{2}+5x\right)+\left(20x+4\right).

$$\left(25x^{2}+5x\right)+\left(20x+4\right)$$

Factor out 5x in the first and 4 in the second group.

$$5x\left(5x+1\right)+4\left(5x+1\right)$$

Factor out common term 5x+1 by using distributive property.

$$\left(5x+1\right)\left(5x+4\right)$$

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.

$$25x^{2}+25x+4=0$$

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.

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

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 25\times 4}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-25±\sqrt{625-100\times 4}}{2\times 25}$$

Multiply -100 times 4.

$$x=\frac{-25±\sqrt{625-400}}{2\times 25}$$

Add 625 to -400.

$$x=\frac{-25±\sqrt{225}}{2\times 25}$$

Take the square root of 225.

$$x=\frac{-25±15}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-25±15}{50}$$

Now solve the equation x=\frac{-25±15}{50} when ± is plus. Add -25 to 15.

$$x=-\frac{10}{50}$$

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

$$x=-\frac{1}{5}$$

Now solve the equation x=\frac{-25±15}{50} when ± is minus. Subtract 15 from -25.

$$x=-\frac{40}{50}$$

Reduce the fraction \frac{-40}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{4}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{5} for x_{1} and -\frac{4}{5} for x_{2}.

$$25x^{2}+25x+4=25\left(x-\left(-\frac{1}{5}\right)\right)\left(x-\left(-\frac{4}{5}\right)\right)$$

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

$$25x^{2}+25x+4=25\left(x+\frac{1}{5}\right)\left(x+\frac{4}{5}\right)$$

Add \frac{1}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+25x+4=25\times \frac{5x+1}{5}\left(x+\frac{4}{5}\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+25x+4=25\times \frac{5x+1}{5}\times \frac{5x+4}{5}$$

Multiply \frac{5x+1}{5} times \frac{5x+4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+25x+4=25\times \frac{\left(5x+1\right)\left(5x+4\right)}{5\times 5}$$

Multiply 5 times 5.

$$25x^{2}+25x+4=25\times \frac{\left(5x+1\right)\left(5x+4\right)}{25}$$

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

$$25x^{2}+25x+4=\left(5x+1\right)\left(5x+4\right)$$

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

$$x ^ 2 +1x +\frac{4}{25} = 0$$

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

$$r + s = -1 rs = \frac{4}{25}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{25}

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = \frac{4}{25}$$

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

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

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = \frac{4}{25}-\frac{1}{4} = -\frac{9}{100}$$

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

$$u^2 = \frac{9}{100} u = \pm\sqrt{\frac{9}{100}} = \pm \frac{3}{10}$$

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

$$r =-\frac{1}{2} - \frac{3}{10} = -0.800 s = -\frac{1}{2} + \frac{3}{10} = -0.200$$

Evaluate

$$\left(5x+1\right)\left(5x+4\right)$$