Solve 6y^2+25y+11

$$6 y ^ { 2 } + 25 y + 11$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by+11. To find a and b, set up a system to be solved.

$$a+b=25 ab=6\times 11=66$$

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 66.

$$1,66 2,33 3,22 6,11$$

Calculate the sum for each pair.

$$1+66=67 2+33=35 3+22=25 6+11=17$$

The solution is the pair that gives sum 25.

$$a=3 b=22$$

Rewrite 6y^{2}+25y+11 as \left(6y^{2}+3y\right)+\left(22y+11\right).

$$\left(6y^{2}+3y\right)+\left(22y+11\right)$$

Factor out 3y in the first and 11 in the second group.

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

Factor out common term 2y+1 by using distributive property.

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

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.

$$6y^{2}+25y+11=0$$

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.

$$y=\frac{-25±\sqrt{25^{2}-4\times 6\times 11}}{2\times 6}$$

Square 25.

$$y=\frac{-25±\sqrt{625-4\times 6\times 11}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-25±\sqrt{625-24\times 11}}{2\times 6}$$

Multiply -24 times 11.

$$y=\frac{-25±\sqrt{625-264}}{2\times 6}$$

Add 625 to -264.

$$y=\frac{-25±\sqrt{361}}{2\times 6}$$

Take the square root of 361.

$$y=\frac{-25±19}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-25±19}{12}$$

Now solve the equation y=\frac{-25±19}{12} when ± is plus. Add -25 to 19.

$$y=-\frac{6}{12}$$

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

$$y=-\frac{1}{2}$$

Now solve the equation y=\frac{-25±19}{12} when ± is minus. Subtract 19 from -25.

$$y=-\frac{44}{12}$$

Reduce the fraction \frac{-44}{12} to lowest terms by extracting and canceling out 4.

$$y=-\frac{11}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{2} for x_{1} and -\frac{11}{3} for x_{2}.

$$6y^{2}+25y+11=6\left(y-\left(-\frac{1}{2}\right)\right)\left(y-\left(-\frac{11}{3}\right)\right)$$

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

$$6y^{2}+25y+11=6\left(y+\frac{1}{2}\right)\left(y+\frac{11}{3}\right)$$

Add \frac{1}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+25y+11=6\times \frac{2y+1}{2}\left(y+\frac{11}{3}\right)$$

Add \frac{11}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+25y+11=6\times \frac{2y+1}{2}\times \frac{3y+11}{3}$$

Multiply \frac{2y+1}{2} times \frac{3y+11}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+25y+11=6\times \frac{\left(2y+1\right)\left(3y+11\right)}{2\times 3}$$

Multiply 2 times 3.

$$6y^{2}+25y+11=6\times \frac{\left(2y+1\right)\left(3y+11\right)}{6}$$

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

$$6y^{2}+25y+11=\left(2y+1\right)\left(3y+11\right)$$

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

$$x ^ 2 +\frac{25}{6}x +\frac{11}{6} = 0$$

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

$$r + s = -\frac{25}{6} rs = \frac{11}{6}$$

Two numbers r and s sum up to -\frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{6} = -\frac{25}{12}. 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>

$$r = -\frac{25}{12} - u s = -\frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{11}{6}

$$(-\frac{25}{12} - u) (-\frac{25}{12} + u) = \frac{11}{6}$$

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

$$\frac{625}{144} - u^2 = \frac{11}{6}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = \frac{11}{6}-\frac{625}{144} = -\frac{361}{144}$$

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

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

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

$$r =-\frac{25}{12} - \frac{19}{12} = -3.667 s = -\frac{25}{12} + \frac{19}{12} = -0.500$$

Evaluate

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