Solve 6x^2+39x+33

$$6 x ^ { 2 } + 39 x + 33$$

Factor

$$3\left(x+1\right)\left(2x+11\right)$$

Solution Steps

Factor out 3.

$$3\left(2x^{2}+13x+11\right)$$

Consider 2x^{2}+13x+11. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=13 ab=2\times 11=22$$

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 22.

$$1,22 2,11$$

Calculate the sum for each pair.

$$1+22=23 2+11=13$$

The solution is the pair that gives sum 13.

$$a=2 b=11$$

Rewrite 2x^{2}+13x+11 as \left(2x^{2}+2x\right)+\left(11x+11\right).

$$\left(2x^{2}+2x\right)+\left(11x+11\right)$$

Factor out 2x in the first and 11 in the second group.

$$2x\left(x+1\right)+11\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(2x+11\right)$$

Rewrite the complete factored expression.

$$3\left(x+1\right)\left(2x+11\right)$$

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.

$$6x^{2}+39x+33=0$$

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.

$$x=\frac{-39±\sqrt{39^{2}-4\times 6\times 33}}{2\times 6}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\times 6\times 33}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-39±\sqrt{1521-24\times 33}}{2\times 6}$$

Multiply -24 times 33.

$$x=\frac{-39±\sqrt{1521-792}}{2\times 6}$$

Add 1521 to -792.

$$x=\frac{-39±\sqrt{729}}{2\times 6}$$

Take the square root of 729.

$$x=\frac{-39±27}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-39±27}{12}$$

Now solve the equation x=\frac{-39±27}{12} when ± is plus. Add -39 to 27.

$$x=-\frac{12}{12}$$

Divide -12 by 12.

$$x=-1$$

Now solve the equation x=\frac{-39±27}{12} when ± is minus. Subtract 27 from -39.

$$x=-\frac{66}{12}$$

Reduce the fraction \frac{-66}{12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{11}{2}$$

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 -\frac{11}{2} for x_{2}.

$$6x^{2}+39x+33=6\left(x-\left(-1\right)\right)\left(x-\left(-\frac{11}{2}\right)\right)$$

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

$$6x^{2}+39x+33=6\left(x+1\right)\left(x+\frac{11}{2}\right)$$

Add \frac{11}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+39x+33=6\left(x+1\right)\times \frac{2x+11}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}+39x+33=3\left(x+1\right)\left(2x+11\right)$$

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 6

$$x ^ 2 +\frac{13}{2}x +\frac{11}{2} = 0$$

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

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

Two numbers r and s sum up to -\frac{13}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{2} = -\frac{13}{4}. 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>

$$r = -\frac{13}{4} - u s = -\frac{13}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{11}{2}

$$(-\frac{13}{4} - u) (-\frac{13}{4} + u) = \frac{11}{2}$$

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

$$\frac{169}{16} - u^2 = \frac{11}{2}$$

Simplify the expression by subtracting \frac{169}{16} on both sides

$$-u^2 = \frac{11}{2}-\frac{169}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{13}{4} - \frac{9}{4} = -5.500 s = -\frac{13}{4} + \frac{9}{4} = -1$$

Evaluate

$$3\left(x+1\right)\left(2x+11\right)$$