Solve 45b^2+180b+100

$$45 b ^ { 2 } + 180 b + 100$$

Factor

$$5\left(3b+2\right)\left(3b+10\right)$$

Solution Steps

Factor out 5.

$$5\left(9b^{2}+36b+20\right)$$

Consider 9b^{2}+36b+20. Factor the expression by grouping. First, the expression needs to be rewritten as 9b^{2}+pb+qb+20. To find p and q, set up a system to be solved.

$$p+q=36 pq=9\times 20=180$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 36.

$$p=6 q=30$$

Rewrite 9b^{2}+36b+20 as \left(9b^{2}+6b\right)+\left(30b+20\right).

$$\left(9b^{2}+6b\right)+\left(30b+20\right)$$

Factor out 3b in the first and 10 in the second group.

$$3b\left(3b+2\right)+10\left(3b+2\right)$$

Factor out common term 3b+2 by using distributive property.

$$\left(3b+2\right)\left(3b+10\right)$$

Rewrite the complete factored expression.

$$5\left(3b+2\right)\left(3b+10\right)$$

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.

$$45b^{2}+180b+100=0$$

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.

$$b=\frac{-180±\sqrt{180^{2}-4\times 45\times 100}}{2\times 45}$$

Square 180.

$$b=\frac{-180±\sqrt{32400-4\times 45\times 100}}{2\times 45}$$

Multiply -4 times 45.

$$b=\frac{-180±\sqrt{32400-180\times 100}}{2\times 45}$$

Multiply -180 times 100.

$$b=\frac{-180±\sqrt{32400-18000}}{2\times 45}$$

Add 32400 to -18000.

$$b=\frac{-180±\sqrt{14400}}{2\times 45}$$

Take the square root of 14400.

$$b=\frac{-180±120}{2\times 45}$$

Multiply 2 times 45.

$$b=\frac{-180±120}{90}$$

Now solve the equation b=\frac{-180±120}{90} when ± is plus. Add -180 to 120.

$$b=-\frac{60}{90}$$

Reduce the fraction \frac{-60}{90} to lowest terms by extracting and canceling out 30.

$$b=-\frac{2}{3}$$

Now solve the equation b=\frac{-180±120}{90} when ± is minus. Subtract 120 from -180.

$$b=-\frac{300}{90}$$

Reduce the fraction \frac{-300}{90} to lowest terms by extracting and canceling out 30.

$$b=-\frac{10}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3} for x_{1} and -\frac{10}{3} for x_{2}.

$$45b^{2}+180b+100=45\left(b-\left(-\frac{2}{3}\right)\right)\left(b-\left(-\frac{10}{3}\right)\right)$$

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

$$45b^{2}+180b+100=45\left(b+\frac{2}{3}\right)\left(b+\frac{10}{3}\right)$$

Add \frac{2}{3} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$45b^{2}+180b+100=45\times \frac{3b+2}{3}\left(b+\frac{10}{3}\right)$$

Add \frac{10}{3} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$45b^{2}+180b+100=45\times \frac{3b+2}{3}\times \frac{3b+10}{3}$$

Multiply \frac{3b+2}{3} times \frac{3b+10}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$45b^{2}+180b+100=45\times \frac{\left(3b+2\right)\left(3b+10\right)}{3\times 3}$$

Multiply 3 times 3.

$$45b^{2}+180b+100=45\times \frac{\left(3b+2\right)\left(3b+10\right)}{9}$$

Cancel out 9, the greatest common factor in 45 and 9.

$$45b^{2}+180b+100=5\left(3b+2\right)\left(3b+10\right)$$

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 45

$$x ^ 2 +4x +\frac{20}{9} = 0$$

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

$$r + s = -4 rs = \frac{20}{9}$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. 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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{9}

$$(-2 - u) (-2 + u) = \frac{20}{9}$$

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

$$4 - u^2 = \frac{20}{9}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{20}{9}-4 = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-2 - \frac{4}{3} = -3.333 s = -2 + \frac{4}{3} = -0.667$$

Evaluate

$$45b^{2}+180b+100$$