Solve 3z^2+16z+20

$$3 z ^ { 2 } + 16 z + 20$$

Factor

$$\left(z+2\right)\left(3z+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3z^{2}+az+bz+20. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 20=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=6 b=10$$

Rewrite 3z^{2}+16z+20 as \left(3z^{2}+6z\right)+\left(10z+20\right).

$$\left(3z^{2}+6z\right)+\left(10z+20\right)$$

Factor out 3z in the first and 10 in the second group.

$$3z\left(z+2\right)+10\left(z+2\right)$$

Factor out common term z+2 by using distributive property.

$$\left(z+2\right)\left(3z+10\right)$$

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.

$$3z^{2}+16z+20=0$$

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.

$$z=\frac{-16±\sqrt{16^{2}-4\times 3\times 20}}{2\times 3}$$

Square 16.

$$z=\frac{-16±\sqrt{256-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$z=\frac{-16±\sqrt{256-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$z=\frac{-16±\sqrt{256-240}}{2\times 3}$$

Add 256 to -240.

$$z=\frac{-16±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

$$z=\frac{-16±4}{2\times 3}$$

Multiply 2 times 3.

$$z=\frac{-16±4}{6}$$

Now solve the equation z=\frac{-16±4}{6} when ± is plus. Add -16 to 4.

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

Divide -12 by 6.

$$z=-2$$

Now solve the equation z=\frac{-16±4}{6} when ± is minus. Subtract 4 from -16.

$$z=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$z=-\frac{10}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and -\frac{10}{3} for x_{2}.

$$3z^{2}+16z+20=3\left(z-\left(-2\right)\right)\left(z-\left(-\frac{10}{3}\right)\right)$$

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

$$3z^{2}+16z+20=3\left(z+2\right)\left(z+\frac{10}{3}\right)$$

Add \frac{10}{3} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3z^{2}+16z+20=3\left(z+2\right)\times \frac{3z+10}{3}$$

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

$$3z^{2}+16z+20=\left(z+2\right)\left(3z+10\right)$$

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

$$x ^ 2 +\frac{16}{3}x +\frac{20}{3} = 0$$

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

$$r + s = -\frac{16}{3} rs = \frac{20}{3}$$

Two numbers r and s sum up to -\frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{3} = -\frac{8}{3}. 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>

$$r = -\frac{8}{3} - u s = -\frac{8}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{3}

$$(-\frac{8}{3} - u) (-\frac{8}{3} + u) = \frac{20}{3}$$

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

$$\frac{64}{9} - u^2 = \frac{20}{3}$$

Simplify the expression by subtracting \frac{64}{9} on both sides

$$-u^2 = \frac{20}{3}-\frac{64}{9} = -\frac{4}{9}$$

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

$$u^2 = \frac{4}{9} u = \pm\sqrt{\frac{4}{9}} = \pm \frac{2}{3}$$

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

$$r =-\frac{8}{3} - \frac{2}{3} = -3.333 s = -\frac{8}{3} + \frac{2}{3} = -2.000$$

Evaluate

$$\left(z+2\right)\left(3z+10\right)$$