Solve 3y^2+13y+4

$$3 y ^ { 2 } + 13 y + 4$$

Factor

$$\left(y+4\right)\left(3y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\times 4=12$$

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.

$$a=1 b=12$$

Rewrite 3y^{2}+13y+4 as \left(3y^{2}+y\right)+\left(12y+4\right).

$$\left(3y^{2}+y\right)+\left(12y+4\right)$$

Factor out y in the first and 4 in the second group.

$$y\left(3y+1\right)+4\left(3y+1\right)$$

Factor out common term 3y+1 by using distributive property.

$$\left(3y+1\right)\left(y+4\right)$$

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.

$$3y^{2}+13y+4=0$$

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.

$$y=\frac{-13±\sqrt{13^{2}-4\times 3\times 4}}{2\times 3}$$

Square 13.

$$y=\frac{-13±\sqrt{169-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-13±\sqrt{169-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$y=\frac{-13±\sqrt{169-48}}{2\times 3}$$

Add 169 to -48.

$$y=\frac{-13±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$y=\frac{-13±11}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-13±11}{6}$$

Now solve the equation y=\frac{-13±11}{6} when ± is plus. Add -13 to 11.

$$y=-\frac{2}{6}$$

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

$$y=-\frac{1}{3}$$

Now solve the equation y=\frac{-13±11}{6} when ± is minus. Subtract 11 from -13.

$$y=-\frac{24}{6}$$

Divide -24 by 6.

$$y=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and -4 for x_{2}.

$$3y^{2}+13y+4=3\left(y-\left(-\frac{1}{3}\right)\right)\left(y-\left(-4\right)\right)$$

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

$$3y^{2}+13y+4=3\left(y+\frac{1}{3}\right)\left(y+4\right)$$

Add \frac{1}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}+13y+4=3\times \frac{3y+1}{3}\left(y+4\right)$$

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

$$3y^{2}+13y+4=\left(3y+1\right)\left(y+4\right)$$

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

$$x ^ 2 +\frac{13}{3}x +\frac{4}{3} = 0$$

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

$$r + s = -\frac{13}{3} rs = \frac{4}{3}$$

Two numbers r and s sum up to -\frac{13}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{3} = -\frac{13}{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>

$$r = -\frac{13}{6} - u s = -\frac{13}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{3}

$$(-\frac{13}{6} - u) (-\frac{13}{6} + u) = \frac{4}{3}$$

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

$$\frac{169}{36} - u^2 = \frac{4}{3}$$

Simplify the expression by subtracting \frac{169}{36} on both sides

$$-u^2 = \frac{4}{3}-\frac{169}{36} = -\frac{121}{36}$$

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

$$u^2 = \frac{121}{36} u = \pm\sqrt{\frac{121}{36}} = \pm \frac{11}{6}$$

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

$$r =-\frac{13}{6} - \frac{11}{6} = -4 s = -\frac{13}{6} + \frac{11}{6} = -0.333$$

Evaluate

$$\left(y+4\right)\left(3y+1\right)$$