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.

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

Evaluate

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