Solve 3v^2+28v+60

$$3 v ^ { 2 } + 28 v + 60$$

Factor

$$\left(v+6\right)\left(3v+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3v^{2}+av+bv+60. To find a and b, set up a system to be solved.

$$a+b=28 ab=3\times 60=180$$

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 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 28.

$$a=10 b=18$$

Rewrite 3v^{2}+28v+60 as \left(3v^{2}+10v\right)+\left(18v+60\right).

$$\left(3v^{2}+10v\right)+\left(18v+60\right)$$

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

$$v\left(3v+10\right)+6\left(3v+10\right)$$

Factor out common term 3v+10 by using distributive property.

$$\left(3v+10\right)\left(v+6\right)$$

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.

$$3v^{2}+28v+60=0$$

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.

$$v=\frac{-28±\sqrt{28^{2}-4\times 3\times 60}}{2\times 3}$$

Square 28.

$$v=\frac{-28±\sqrt{784-4\times 3\times 60}}{2\times 3}$$

Multiply -4 times 3.

$$v=\frac{-28±\sqrt{784-12\times 60}}{2\times 3}$$

Multiply -12 times 60.

$$v=\frac{-28±\sqrt{784-720}}{2\times 3}$$

Add 784 to -720.

$$v=\frac{-28±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$v=\frac{-28±8}{2\times 3}$$

Multiply 2 times 3.

$$v=\frac{-28±8}{6}$$

Now solve the equation v=\frac{-28±8}{6} when ± is plus. Add -28 to 8.

$$v=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation v=\frac{-28±8}{6} when ± is minus. Subtract 8 from -28.

$$v=-\frac{36}{6}$$

Divide -36 by 6.

$$v=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{10}{3} for x_{1} and -6 for x_{2}.

$$3v^{2}+28v+60=3\left(v-\left(-\frac{10}{3}\right)\right)\left(v-\left(-6\right)\right)$$

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

$$3v^{2}+28v+60=3\left(v+\frac{10}{3}\right)\left(v+6\right)$$

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

$$3v^{2}+28v+60=3\times \frac{3v+10}{3}\left(v+6\right)$$

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

$$3v^{2}+28v+60=\left(3v+10\right)\left(v+6\right)$$

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 3

$$x ^ 2 +\frac{28}{3}x +20 = 0$$

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

$$r + s = -\frac{28}{3} rs = 20$$

Two numbers r and s sum up to -\frac{28}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{28}{3} = -\frac{14}{3}. 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>

$$r = -\frac{14}{3} - u s = -\frac{14}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 20

$$(-\frac{14}{3} - u) (-\frac{14}{3} + u) = 20$$

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

$$\frac{196}{9} - u^2 = 20$$

Simplify the expression by subtracting \frac{196}{9} on both sides

$$-u^2 = 20-\frac{196}{9} = -\frac{16}{9}$$

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.

$$r =-\frac{14}{3} - \frac{4}{3} = -6.000 s = -\frac{14}{3} + \frac{4}{3} = -3.333$$

Evaluate

$$\left(v+6\right)\left(3v+10\right)$$