Solve 25h^2+120h+144=

$$25 h ^ { 2 } + 120 h + 144 =$$

Factor

$$\left(5h+12\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25h^{2}+ah+bh+144. To find a and b, set up a system to be solved.

$$a+b=120 ab=25\times 144=3600$$

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 3600.

$$1,3600 2,1800 3,1200 4,900 5,720 6,600 8,450 9,400 10,360 12,300 15,240 16,225 18,200 20,180 24,150 25,144 30,120 36,100 40,90 45,80 48,75 50,72 60,60$$

Calculate the sum for each pair.

$$1+3600=3601 2+1800=1802 3+1200=1203 4+900=904 5+720=725 6+600=606 8+450=458 9+400=409 10+360=370 12+300=312 15+240=255 16+225=241 18+200=218 20+180=200 24+150=174 25+144=169 30+120=150 36+100=136 40+90=130 45+80=125 48+75=123 50+72=122 60+60=120$$

The solution is the pair that gives sum 120.

$$a=60 b=60$$

Rewrite 25h^{2}+120h+144 as \left(25h^{2}+60h\right)+\left(60h+144\right).

$$\left(25h^{2}+60h\right)+\left(60h+144\right)$$

Factor out 5h in the first and 12 in the second group.

$$5h\left(5h+12\right)+12\left(5h+12\right)$$

Factor out common term 5h+12 by using distributive property.

$$\left(5h+12\right)\left(5h+12\right)$$

Rewrite as a binomial square.

$$\left(5h+12\right)^{2}$$

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.

$$factor(25h^{2}+120h+144)$$

Find the greatest common factor of the coefficients.

$$gcf(25,120,144)=1$$

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

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

Find the square root of the trailing term, 144.

$$\sqrt{144}=12$$

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.

$$\left(5h+12\right)^{2}$$

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.

$$25h^{2}+120h+144=0$$

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.

$$h=\frac{-120±\sqrt{120^{2}-4\times 25\times 144}}{2\times 25}$$

Square 120.

$$h=\frac{-120±\sqrt{14400-4\times 25\times 144}}{2\times 25}$$

Multiply -4 times 25.

$$h=\frac{-120±\sqrt{14400-100\times 144}}{2\times 25}$$

Multiply -100 times 144.

$$h=\frac{-120±\sqrt{14400-14400}}{2\times 25}$$

Add 14400 to -14400.

$$h=\frac{-120±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$h=\frac{-120±0}{2\times 25}$$

Multiply 2 times 25.

$$h=\frac{-120±0}{50}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{12}{5} for x_{1} and -\frac{12}{5} for x_{2}.

$$25h^{2}+120h+144=25\left(h-\left(-\frac{12}{5}\right)\right)\left(h-\left(-\frac{12}{5}\right)\right)$$

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

$$25h^{2}+120h+144=25\left(h+\frac{12}{5}\right)\left(h+\frac{12}{5}\right)$$

Add \frac{12}{5} to h by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25h^{2}+120h+144=25\times \frac{5h+12}{5}\left(h+\frac{12}{5}\right)$$

Add \frac{12}{5} to h by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25h^{2}+120h+144=25\times \frac{5h+12}{5}\times \frac{5h+12}{5}$$

Multiply \frac{5h+12}{5} times \frac{5h+12}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25h^{2}+120h+144=25\times \frac{\left(5h+12\right)\left(5h+12\right)}{5\times 5}$$

Multiply 5 times 5.

$$25h^{2}+120h+144=25\times \frac{\left(5h+12\right)\left(5h+12\right)}{25}$$

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

$$25h^{2}+120h+144=\left(5h+12\right)\left(5h+12\right)$$

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

$$x ^ 2 +\frac{24}{5}x +\frac{144}{25} = 0$$

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

$$r + s = -\frac{24}{5} rs = \frac{144}{25}$$

Two numbers r and s sum up to -\frac{24}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{24}{5} = -\frac{12}{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>

$$r = -\frac{12}{5} - u s = -\frac{12}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{144}{25}

$$(-\frac{12}{5} - u) (-\frac{12}{5} + u) = \frac{144}{25}$$

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

$$\frac{144}{25} - u^2 = \frac{144}{25}$$

Simplify the expression by subtracting \frac{144}{25} on both sides

$$-u^2 = \frac{144}{25}-\frac{144}{25} = 0$$

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.

$$r = s = -\frac{12}{5} = -2.400$$

Evaluate

$$\left(5h+12\right)^{2}$$