Solve -3v^2+39v-36

$$- 3 v ^ { 2 } + 39 v - 36$$

Factor

$$3\left(1-v\right)\left(v-12\right)$$

Solution Steps

Factor out 3.

$$3\left(-v^{2}+13v-12\right)$$

Consider -v^{2}+13v-12. Factor the expression by grouping. First, the expression needs to be rewritten as -v^{2}+av+bv-12. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=12 b=1$$

Rewrite -v^{2}+13v-12 as \left(-v^{2}+12v\right)+\left(v-12\right).

$$\left(-v^{2}+12v\right)+\left(v-12\right)$$

Factor out -v in -v^{2}+12v.

$$-v\left(v-12\right)+v-12$$

Factor out common term v-12 by using distributive property.

$$\left(v-12\right)\left(-v+1\right)$$

Rewrite the complete factored expression.

$$3\left(v-12\right)\left(-v+1\right)$$

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.

$$-3v^{2}+39v-36=0$$

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.

$$v=\frac{-39±\sqrt{39^{2}-4\left(-3\right)\left(-36\right)}}{2\left(-3\right)}$$

Square 39.

$$v=\frac{-39±\sqrt{1521-4\left(-3\right)\left(-36\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$v=\frac{-39±\sqrt{1521+12\left(-36\right)}}{2\left(-3\right)}$$

Multiply 12 times -36.

$$v=\frac{-39±\sqrt{1521-432}}{2\left(-3\right)}$$

Add 1521 to -432.

$$v=\frac{-39±\sqrt{1089}}{2\left(-3\right)}$$

Take the square root of 1089.

$$v=\frac{-39±33}{2\left(-3\right)}$$

Multiply 2 times -3.

$$v=\frac{-39±33}{-6}$$

Now solve the equation v=\frac{-39±33}{-6} when ± is plus. Add -39 to 33.

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

Divide -6 by -6.

$$v=1$$

Now solve the equation v=\frac{-39±33}{-6} when ± is minus. Subtract 33 from -39.

$$v=-\frac{72}{-6}$$

Divide -72 by -6.

$$v=12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and 12 for x_{2}.

$$-3v^{2}+39v-36=-3\left(v-1\right)\left(v-12\right)$$

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.

$$x ^ 2 -13x +12 = 0$$

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

$$r + s = 13 rs = 12$$

Two numbers r and s sum up to 13 exactly when the average of the two numbers is \frac{1}{2}*13 = \frac{13}{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>

$$r = \frac{13}{2} - u s = \frac{13}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 12

$$(\frac{13}{2} - u) (\frac{13}{2} + u) = 12$$

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

$$\frac{169}{4} - u^2 = 12$$

Simplify the expression by subtracting \frac{169}{4} on both sides

$$-u^2 = 12-\frac{169}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =\frac{13}{2} - \frac{11}{2} = 1 s = \frac{13}{2} + \frac{11}{2} = 12$$

Evaluate

$$3\left(1-v\right)\left(v-12\right)$$