Solve 3z^2-5z-2

$$3 { z }^{ 2 } -5z-2$$

Factor

$$\left(z-2\right)\left(3z+1\right)$$

Solution Steps

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

$$a+b=-5 ab=3\left(-2\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 3z^{2}-5z-2 as \left(3z^{2}-6z\right)+\left(z-2\right).

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

Factor out 3z in 3z^{2}-6z.

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

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

$$\left(z-2\right)\left(3z+1\right)$$

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.

$$3z^{2}-5z-2=0$$

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.

$$z=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\left(-2\right)}}{2\times 3}$$

Square -5.

$$z=\frac{-\left(-5\right)±\sqrt{25-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$z=\frac{-\left(-5\right)±\sqrt{25-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$z=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 3}$$

Add 25 to 24.

$$z=\frac{-\left(-5\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$z=\frac{-\left(-5\right)±7}{2\times 3}$$

The opposite of -5 is 5.

$$z=\frac{5±7}{2\times 3}$$

Multiply 2 times 3.

$$z=\frac{5±7}{6}$$

Now solve the equation z=\frac{5±7}{6} when ± is plus. Add 5 to 7.

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

Divide 12 by 6.

$$z=2$$

Now solve the equation z=\frac{5±7}{6} when ± is minus. Subtract 7 from 5.

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

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

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

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{1}{3} for x_{2}.

$$3z^{2}-5z-2=3\left(z-2\right)\left(z-\left(-\frac{1}{3}\right)\right)$$

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

$$3z^{2}-5z-2=3\left(z-2\right)\left(z+\frac{1}{3}\right)$$

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

$$3z^{2}-5z-2=3\left(z-2\right)\times \frac{3z+1}{3}$$

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

$$3z^{2}-5z-2=\left(z-2\right)\left(3z+1\right)$$

Evaluate

$$\left(z-2\right)\left(3z+1\right)$$