Solve 115x+100x^2+30

$$115 x + 100 x ^ { 2 } + 30$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(23x+20x^{2}+6\right)$$

Consider 23x+20x^{2}+6. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$20x^{2}+23x+6$$

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=23 ab=20\times 6=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=8 b=15$$

Rewrite 20x^{2}+23x+6 as \left(20x^{2}+8x\right)+\left(15x+6\right).

$$\left(20x^{2}+8x\right)+\left(15x+6\right)$$

Factor out 4x in the first and 3 in the second group.

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

Factor out common term 5x+2 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$100x^{2}+115x+30=0$$

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.

$$x=\frac{-115±\sqrt{115^{2}-4\times 100\times 30}}{2\times 100}$$

Square 115.

$$x=\frac{-115±\sqrt{13225-4\times 100\times 30}}{2\times 100}$$

Multiply -4 times 100.

$$x=\frac{-115±\sqrt{13225-400\times 30}}{2\times 100}$$

Multiply -400 times 30.

$$x=\frac{-115±\sqrt{13225-12000}}{2\times 100}$$

Add 13225 to -12000.

$$x=\frac{-115±\sqrt{1225}}{2\times 100}$$

Take the square root of 1225.

$$x=\frac{-115±35}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{-115±35}{200}$$

Now solve the equation x=\frac{-115±35}{200} when ± is plus. Add -115 to 35.

$$x=-\frac{80}{200}$$

Reduce the fraction \frac{-80}{200} to lowest terms by extracting and canceling out 40.

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

Now solve the equation x=\frac{-115±35}{200} when ± is minus. Subtract 35 from -115.

$$x=-\frac{150}{200}$$

Reduce the fraction \frac{-150}{200} to lowest terms by extracting and canceling out 50.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{5} for x_{1} and -\frac{3}{4} for x_{2}.

$$100x^{2}+115x+30=100\left(x-\left(-\frac{2}{5}\right)\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$100x^{2}+115x+30=100\left(x+\frac{2}{5}\right)\left(x+\frac{3}{4}\right)$$

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

$$100x^{2}+115x+30=100\times \frac{5x+2}{5}\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$100x^{2}+115x+30=100\times \frac{5x+2}{5}\times \frac{4x+3}{4}$$

Multiply \frac{5x+2}{5} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$100x^{2}+115x+30=100\times \frac{\left(5x+2\right)\left(4x+3\right)}{5\times 4}$$

Multiply 5 times 4.

$$100x^{2}+115x+30=100\times \frac{\left(5x+2\right)\left(4x+3\right)}{20}$$

Cancel out 20, the greatest common factor in 100 and 20.

$$100x^{2}+115x+30=5\left(5x+2\right)\left(4x+3\right)$$

Evaluate

$$100x^{2}+115x+30$$