Solve 25x^2+150x+216

$$25 x ^ { 2 } + 150 x + 216$$

Factor

$$\left(5x+12\right)\left(5x+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25x^{2}+ax+bx+216. To find a and b, set up a system to be solved.

$$a+b=150 ab=25\times 216=5400$$

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 5400.

$$1,5400 2,2700 3,1800 4,1350 5,1080 6,900 8,675 9,600 10,540 12,450 15,360 18,300 20,270 24,225 25,216 27,200 30,180 36,150 40,135 45,120 50,108 54,100 60,90 72,75$$

Calculate the sum for each pair.

$$1+5400=5401 2+2700=2702 3+1800=1803 4+1350=1354 5+1080=1085 6+900=906 8+675=683 9+600=609 10+540=550 12+450=462 15+360=375 18+300=318 20+270=290 24+225=249 25+216=241 27+200=227 30+180=210 36+150=186 40+135=175 45+120=165 50+108=158 54+100=154 60+90=150 72+75=147$$

The solution is the pair that gives sum 150.

$$a=60 b=90$$

Rewrite 25x^{2}+150x+216 as \left(25x^{2}+60x\right)+\left(90x+216\right).

$$\left(25x^{2}+60x\right)+\left(90x+216\right)$$

Factor out 5x in the first and 18 in the second group.

$$5x\left(5x+12\right)+18\left(5x+12\right)$$

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

$$\left(5x+12\right)\left(5x+18\right)$$

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.

$$25x^{2}+150x+216=0$$

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.

$$x=\frac{-150±\sqrt{150^{2}-4\times 25\times 216}}{2\times 25}$$

Square 150.

$$x=\frac{-150±\sqrt{22500-4\times 25\times 216}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-150±\sqrt{22500-100\times 216}}{2\times 25}$$

Multiply -100 times 216.

$$x=\frac{-150±\sqrt{22500-21600}}{2\times 25}$$

Add 22500 to -21600.

$$x=\frac{-150±\sqrt{900}}{2\times 25}$$

Take the square root of 900.

$$x=\frac{-150±30}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-150±30}{50}$$

Now solve the equation x=\frac{-150±30}{50} when ± is plus. Add -150 to 30.

$$x=-\frac{120}{50}$$

Reduce the fraction \frac{-120}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{12}{5}$$

Now solve the equation x=\frac{-150±30}{50} when ± is minus. Subtract 30 from -150.

$$x=-\frac{180}{50}$$

Reduce the fraction \frac{-180}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{18}{5}$$

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{18}{5} for x_{2}.

$$25x^{2}+150x+216=25\left(x-\left(-\frac{12}{5}\right)\right)\left(x-\left(-\frac{18}{5}\right)\right)$$

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

$$25x^{2}+150x+216=25\left(x+\frac{12}{5}\right)\left(x+\frac{18}{5}\right)$$

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

$$25x^{2}+150x+216=25\times \frac{5x+12}{5}\left(x+\frac{18}{5}\right)$$

Add \frac{18}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+150x+216=25\times \frac{5x+12}{5}\times \frac{5x+18}{5}$$

Multiply \frac{5x+12}{5} times \frac{5x+18}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+150x+216=25\times \frac{\left(5x+12\right)\left(5x+18\right)}{5\times 5}$$

Multiply 5 times 5.

$$25x^{2}+150x+216=25\times \frac{\left(5x+12\right)\left(5x+18\right)}{25}$$

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

$$25x^{2}+150x+216=\left(5x+12\right)\left(5x+18\right)$$

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

$$x ^ 2 +6x +\frac{216}{25} = 0$$

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

$$r + s = -6 rs = \frac{216}{25}$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{216}{25}

$$(-3 - u) (-3 + u) = \frac{216}{25}$$

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

$$9 - u^2 = \frac{216}{25}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{216}{25}-9 = -\frac{9}{25}$$

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

$$u^2 = \frac{9}{25} u = \pm\sqrt{\frac{9}{25}} = \pm \frac{3}{5}$$

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

$$r =-3 - \frac{3}{5} = -3.600 s = -3 + \frac{3}{5} = -2.400$$

Evaluate

$$\left(5x+12\right)\left(5x+18\right)$$