Solve 12x^2+80x+100

$$12 { x }^{ 2 } +80x+100$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(3x^{2}+20x+25\right)$$

Consider 3x^{2}+20x+25. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=20 ab=3\times 25=75$$

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 75.

$$1,75 3,25 5,15$$

Calculate the sum for each pair.

$$1+75=76 3+25=28 5+15=20$$

The solution is the pair that gives sum 20.

$$a=5 b=15$$

Rewrite 3x^{2}+20x+25 as \left(3x^{2}+5x\right)+\left(15x+25\right).

$$\left(3x^{2}+5x\right)+\left(15x+25\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$12x^{2}+80x+100=0$$

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.

$$x=\frac{-80±\sqrt{80^{2}-4\times 12\times 100}}{2\times 12}$$

Square 80.

$$x=\frac{-80±\sqrt{6400-4\times 12\times 100}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-80±\sqrt{6400-48\times 100}}{2\times 12}$$

Multiply -48 times 100.

$$x=\frac{-80±\sqrt{6400-4800}}{2\times 12}$$

Add 6400 to -4800.

$$x=\frac{-80±\sqrt{1600}}{2\times 12}$$

Take the square root of 1600.

$$x=\frac{-80±40}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-80±40}{24}$$

Now solve the equation x=\frac{-80±40}{24} when ± is plus. Add -80 to 40.

$$x=-\frac{40}{24}$$

Reduce the fraction \frac{-40}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-80±40}{24} when ± is minus. Subtract 40 from -80.

$$x=-\frac{120}{24}$$

Divide -120 by 24.

$$x=-5$$

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 -5 for x_{2}.

$$12x^{2}+80x+100=12\left(x-\left(-\frac{5}{3}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$12x^{2}+80x+100=12\left(x+\frac{5}{3}\right)\left(x+5\right)$$

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

$$12x^{2}+80x+100=12\times \frac{3x+5}{3}\left(x+5\right)$$

Cancel out 3, the greatest common factor in 12 and 3.

$$12x^{2}+80x+100=4\left(3x+5\right)\left(x+5\right)$$

Evaluate

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