Solve 5r^2+32r+12

$$5 r ^ { 2 } + 32 r + 12$$

Factor

$$\left(r+6\right)\left(5r+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5r^{2}+ar+br+12. To find a and b, set up a system to be solved.

$$a+b=32 ab=5\times 12=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 32.

$$a=2 b=30$$

Rewrite 5r^{2}+32r+12 as \left(5r^{2}+2r\right)+\left(30r+12\right).

$$\left(5r^{2}+2r\right)+\left(30r+12\right)$$

Factor out r in the first and 6 in the second group.

$$r\left(5r+2\right)+6\left(5r+2\right)$$

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

$$\left(5r+2\right)\left(r+6\right)$$

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.

$$5r^{2}+32r+12=0$$

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.

$$r=\frac{-32±\sqrt{32^{2}-4\times 5\times 12}}{2\times 5}$$

Square 32.

$$r=\frac{-32±\sqrt{1024-4\times 5\times 12}}{2\times 5}$$

Multiply -4 times 5.

$$r=\frac{-32±\sqrt{1024-20\times 12}}{2\times 5}$$

Multiply -20 times 12.

$$r=\frac{-32±\sqrt{1024-240}}{2\times 5}$$

Add 1024 to -240.

$$r=\frac{-32±\sqrt{784}}{2\times 5}$$

Take the square root of 784.

$$r=\frac{-32±28}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{-32±28}{10}$$

Now solve the equation r=\frac{-32±28}{10} when ± is plus. Add -32 to 28.

$$r=-\frac{4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation r=\frac{-32±28}{10} when ± is minus. Subtract 28 from -32.

$$r=-\frac{60}{10}$$

Divide -60 by 10.

$$r=-6$$

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 -6 for x_{2}.

$$5r^{2}+32r+12=5\left(r-\left(-\frac{2}{5}\right)\right)\left(r-\left(-6\right)\right)$$

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

$$5r^{2}+32r+12=5\left(r+\frac{2}{5}\right)\left(r+6\right)$$

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

$$5r^{2}+32r+12=5\times \frac{5r+2}{5}\left(r+6\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5r^{2}+32r+12=\left(5r+2\right)\left(r+6\right)$$

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 5

$$x ^ 2 +\frac{32}{5}x +\frac{12}{5} = 0$$

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

$$r + s = -\frac{32}{5} rs = \frac{12}{5}$$

Two numbers r and s sum up to -\frac{32}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{32}{5} = -\frac{16}{5}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{16}{5} - u s = -\frac{16}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{12}{5}

$$(-\frac{16}{5} - u) (-\frac{16}{5} + u) = \frac{12}{5}$$

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

$$\frac{256}{25} - u^2 = \frac{12}{5}$$

Simplify the expression by subtracting \frac{256}{25} on both sides

$$-u^2 = \frac{12}{5}-\frac{256}{25} = -\frac{196}{25}$$

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

$$u^2 = \frac{196}{25} u = \pm\sqrt{\frac{196}{25}} = \pm \frac{14}{5}$$

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

$$r =-\frac{16}{5} - \frac{14}{5} = -6 s = -\frac{16}{5} + \frac{14}{5} = -0.400$$

Evaluate

$$\left(r+6\right)\left(5r+2\right)$$