Solve 3y^2+11y+10

$$3 y ^ { 2 } + 11 y + 10$$

Factor

$$\left(y+2\right)\left(3y+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+10. To find a and b, set up a system to be solved.

$$a+b=11 ab=3\times 10=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

Rewrite 3y^{2}+11y+10 as \left(3y^{2}+5y\right)+\left(6y+10\right).

$$\left(3y^{2}+5y\right)+\left(6y+10\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(3y+5\right)+2\left(3y+5\right)$$

Factor out common term 3y+5 by using distributive property.

$$\left(3y+5\right)\left(y+2\right)$$

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.

$$3y^{2}+11y+10=0$$

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.

$$y=\frac{-11±\sqrt{11^{2}-4\times 3\times 10}}{2\times 3}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-11±\sqrt{121-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$y=\frac{-11±\sqrt{121-120}}{2\times 3}$$

Add 121 to -120.

$$y=\frac{-11±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$y=\frac{-11±1}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-11±1}{6}$$

Now solve the equation y=\frac{-11±1}{6} when ± is plus. Add -11 to 1.

$$y=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$y=-\frac{5}{3}$$

Now solve the equation y=\frac{-11±1}{6} when ± is minus. Subtract 1 from -11.

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

Divide -12 by 6.

$$y=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{3} for x_{1} and -2 for x_{2}.

$$3y^{2}+11y+10=3\left(y-\left(-\frac{5}{3}\right)\right)\left(y-\left(-2\right)\right)$$

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

$$3y^{2}+11y+10=3\left(y+\frac{5}{3}\right)\left(y+2\right)$$

Add \frac{5}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}+11y+10=3\times \frac{3y+5}{3}\left(y+2\right)$$

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

$$3y^{2}+11y+10=\left(3y+5\right)\left(y+2\right)$$

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

$$x ^ 2 +\frac{11}{3}x +\frac{10}{3} = 0$$

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

$$r + s = -\frac{11}{3} rs = \frac{10}{3}$$

Two numbers r and s sum up to -\frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{3} = -\frac{11}{6}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{3}

$$(-\frac{11}{6} - u) (-\frac{11}{6} + u) = \frac{10}{3}$$

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

$$\frac{121}{36} - u^2 = \frac{10}{3}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = \frac{10}{3}-\frac{121}{36} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =-\frac{11}{6} - \frac{1}{6} = -2.000 s = -\frac{11}{6} + \frac{1}{6} = -1.667$$

Evaluate

$$\left(y+2\right)\left(3y+5\right)$$