Solve 24x^2+100x+100

$$24 x ^ { 2 } + 100 x + 100$$

Factor

$$4\left(2x+5\right)\left(3x+5\right)$$

Solution Steps

Factor out 4.

$$4\left(6x^{2}+25x+25\right)$$

Consider 6x^{2}+25x+25. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=25 ab=6\times 25=150$$

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 150.

$$1,150 2,75 3,50 5,30 6,25 10,15$$

Calculate the sum for each pair.

$$1+150=151 2+75=77 3+50=53 5+30=35 6+25=31 10+15=25$$

The solution is the pair that gives sum 25.

$$a=10 b=15$$

Rewrite 6x^{2}+25x+25 as \left(6x^{2}+10x\right)+\left(15x+25\right).

$$\left(6x^{2}+10x\right)+\left(15x+25\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(3x+5\right)+5\left(3x+5\right)$$

Factor out common term 3x+5 by using distributive property.

$$\left(3x+5\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$4\left(3x+5\right)\left(2x+5\right)$$

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.

$$24x^{2}+100x+100=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\times 24\times 100}}{2\times 24}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\times 24\times 100}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-100±\sqrt{10000-96\times 100}}{2\times 24}$$

Multiply -96 times 100.

$$x=\frac{-100±\sqrt{10000-9600}}{2\times 24}$$

Add 10000 to -9600.

$$x=\frac{-100±\sqrt{400}}{2\times 24}$$

Take the square root of 400.

$$x=\frac{-100±20}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-100±20}{48}$$

Now solve the equation x=\frac{-100±20}{48} when ± is plus. Add -100 to 20.

$$x=-\frac{80}{48}$$

Reduce the fraction \frac{-80}{48} to lowest terms by extracting and canceling out 16.

$$x=-\frac{5}{3}$$

Now solve the equation x=\frac{-100±20}{48} when ± is minus. Subtract 20 from -100.

$$x=-\frac{120}{48}$$

Reduce the fraction \frac{-120}{48} to lowest terms by extracting and canceling out 24.

$$x=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{3} for x_{1} and -\frac{5}{2} for x_{2}.

$$24x^{2}+100x+100=24\left(x-\left(-\frac{5}{3}\right)\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$24x^{2}+100x+100=24\left(x+\frac{5}{3}\right)\left(x+\frac{5}{2}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+100x+100=24\times \frac{3x+5}{3}\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+100x+100=24\times \frac{3x+5}{3}\times \frac{2x+5}{2}$$

Multiply \frac{3x+5}{3} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+100x+100=24\times \frac{\left(3x+5\right)\left(2x+5\right)}{3\times 2}$$

Multiply 3 times 2.

$$24x^{2}+100x+100=24\times \frac{\left(3x+5\right)\left(2x+5\right)}{6}$$

Cancel out 6, the greatest common factor in 24 and 6.

$$24x^{2}+100x+100=4\left(3x+5\right)\left(2x+5\right)$$

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 24

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

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

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

Two numbers r and s sum up to -\frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{6} = -\frac{25}{12}. 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>

$$r = -\frac{25}{12} - u s = -\frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{6}

$$(-\frac{25}{12} - u) (-\frac{25}{12} + u) = \frac{25}{6}$$

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

$$\frac{625}{144} - u^2 = \frac{25}{6}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = \frac{25}{6}-\frac{625}{144} = -\frac{25}{144}$$

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

$$u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}$$

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

$$r =-\frac{25}{12} - \frac{5}{12} = -2.500 s = -\frac{25}{12} + \frac{5}{12} = -1.667$$

Evaluate

$$24x^{2}+100x+100$$