Solve 100w^2+115w+30=

$$100 w ^ { 2 } + 115 w + 30 =$$

Factor

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

Solution Steps

Factor out 5.

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

Consider 20w^{2}+23w+6. Factor the expression by grouping. First, the expression needs to be rewritten as 20w^{2}+aw+bw+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 20w^{2}+23w+6 as \left(20w^{2}+8w\right)+\left(15w+6\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square 115.

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

Multiply -4 times 100.

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

Multiply -400 times 30.

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

Add 13225 to -12000.

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

Take the square root of 1225.

$$w=\frac{-115±35}{2\times 100}$$

Multiply 2 times 100.

$$w=\frac{-115±35}{200}$$

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

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

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

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

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

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

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

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

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}.

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

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

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

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

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

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

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

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

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

Multiply 5 times 4.

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

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

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

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 100

$$x ^ 2 +\frac{23}{20}x +\frac{3}{10} = 0$$

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

$$r + s = -\frac{23}{20} rs = \frac{3}{10}$$

Two numbers r and s sum up to -\frac{23}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{20} = -\frac{23}{40}. 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>

$$r = -\frac{23}{40} - u s = -\frac{23}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{10}

$$(-\frac{23}{40} - u) (-\frac{23}{40} + u) = \frac{3}{10}$$

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

$$\frac{529}{1600} - u^2 = \frac{3}{10}$$

Simplify the expression by subtracting \frac{529}{1600} on both sides

$$-u^2 = \frac{3}{10}-\frac{529}{1600} = -\frac{49}{1600}$$

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

$$u^2 = \frac{49}{1600} u = \pm\sqrt{\frac{49}{1600}} = \pm \frac{7}{40}$$

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

$$r =-\frac{23}{40} - \frac{7}{40} = -0.750 s = -\frac{23}{40} + \frac{7}{40} = -0.400$$

Evaluate

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