Solve 25t^2+110t+121=

$$25 t ^ { 2 } + 110 t + 121 =$$

Factor

$$\left(5t+11\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25t^{2}+at+bt+121. To find a and b, set up a system to be solved.

$$a+b=110 ab=25\times 121=3025$$

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 3025.

$$1,3025 5,605 11,275 25,121 55,55$$

Calculate the sum for each pair.

$$1+3025=3026 5+605=610 11+275=286 25+121=146 55+55=110$$

The solution is the pair that gives sum 110.

$$a=55 b=55$$

Rewrite 25t^{2}+110t+121 as \left(25t^{2}+55t\right)+\left(55t+121\right).

$$\left(25t^{2}+55t\right)+\left(55t+121\right)$$

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

$$5t\left(5t+11\right)+11\left(5t+11\right)$$

Factor out common term 5t+11 by using distributive property.

$$\left(5t+11\right)\left(5t+11\right)$$

Rewrite as a binomial square.

$$\left(5t+11\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(25t^{2}+110t+121)$$

Find the greatest common factor of the coefficients.

$$gcf(25,110,121)=1$$

Find the square root of the leading term, 25t^{2}.

$$\sqrt{25t^{2}}=5t$$

Find the square root of the trailing term, 121.

$$\sqrt{121}=11$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$\left(5t+11\right)^{2}$$

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.

$$25t^{2}+110t+121=0$$

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.

$$t=\frac{-110±\sqrt{110^{2}-4\times 25\times 121}}{2\times 25}$$

Square 110.

$$t=\frac{-110±\sqrt{12100-4\times 25\times 121}}{2\times 25}$$

Multiply -4 times 25.

$$t=\frac{-110±\sqrt{12100-100\times 121}}{2\times 25}$$

Multiply -100 times 121.

$$t=\frac{-110±\sqrt{12100-12100}}{2\times 25}$$

Add 12100 to -12100.

$$t=\frac{-110±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$t=\frac{-110±0}{2\times 25}$$

Multiply 2 times 25.

$$t=\frac{-110±0}{50}$$

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

$$25t^{2}+110t+121=25\left(t-\left(-\frac{11}{5}\right)\right)\left(t-\left(-\frac{11}{5}\right)\right)$$

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

$$25t^{2}+110t+121=25\left(t+\frac{11}{5}\right)\left(t+\frac{11}{5}\right)$$

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

$$25t^{2}+110t+121=25\times \frac{5t+11}{5}\left(t+\frac{11}{5}\right)$$

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

$$25t^{2}+110t+121=25\times \frac{5t+11}{5}\times \frac{5t+11}{5}$$

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

$$25t^{2}+110t+121=25\times \frac{\left(5t+11\right)\left(5t+11\right)}{5\times 5}$$

Multiply 5 times 5.

$$25t^{2}+110t+121=25\times \frac{\left(5t+11\right)\left(5t+11\right)}{25}$$

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

$$25t^{2}+110t+121=\left(5t+11\right)\left(5t+11\right)$$

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

$$x ^ 2 +\frac{22}{5}x +\frac{121}{25} = 0$$

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

$$r + s = -\frac{22}{5} rs = \frac{121}{25}$$

Two numbers r and s sum up to -\frac{22}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{22}{5} = -\frac{11}{5}. 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>

$$r = -\frac{11}{5} - u s = -\frac{11}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{121}{25}

$$(-\frac{11}{5} - u) (-\frac{11}{5} + u) = \frac{121}{25}$$

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

$$\frac{121}{25} - u^2 = \frac{121}{25}$$

Simplify the expression by subtracting \frac{121}{25} on both sides

$$-u^2 = \frac{121}{25}-\frac{121}{25} = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = -\frac{11}{5} = -2.200$$

Evaluate

$$\left(5t+11\right)^{2}$$