Solve 4+5z-6z^2

$$4+5z-6 { z }^{ 2 }$$

Factor

$$\left(4-3z\right)\left(2z+1\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-6z^{2}+5z+4$$

Factor the expression by grouping. First, the expression needs to be rewritten as -6z^{2}+az+bz+4. To find a and b, set up a system to be solved.

$$a+b=5 ab=-6\times 4=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=8 b=-3$$

Rewrite -6z^{2}+5z+4 as \left(-6z^{2}+8z\right)+\left(-3z+4\right).

$$\left(-6z^{2}+8z\right)+\left(-3z+4\right)$$

Factor out 2z in -6z^{2}+8z.

$$2z\left(-3z+4\right)-3z+4$$

Factor out common term -3z+4 by using distributive property.

$$\left(-3z+4\right)\left(2z+1\right)$$

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.

$$-6z^{2}+5z+4=0$$

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.

$$z=\frac{-5±\sqrt{5^{2}-4\left(-6\right)\times 4}}{2\left(-6\right)}$$

Square 5.

$$z=\frac{-5±\sqrt{25-4\left(-6\right)\times 4}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$z=\frac{-5±\sqrt{25+24\times 4}}{2\left(-6\right)}$$

Multiply 24 times 4.

$$z=\frac{-5±\sqrt{25+96}}{2\left(-6\right)}$$

Add 25 to 96.

$$z=\frac{-5±\sqrt{121}}{2\left(-6\right)}$$

Take the square root of 121.

$$z=\frac{-5±11}{2\left(-6\right)}$$

Multiply 2 times -6.

$$z=\frac{-5±11}{-12}$$

Now solve the equation z=\frac{-5±11}{-12} when ± is plus. Add -5 to 11.

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

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

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

Now solve the equation z=\frac{-5±11}{-12} when ± is minus. Subtract 11 from -5.

$$z=-\frac{16}{-12}$$

Reduce the fraction \frac{-16}{-12} to lowest terms by extracting and canceling out 4.

$$z=\frac{4}{3}$$

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{4}{3} for x_{2}.

$$-6z^{2}+5z+4=-6\left(z-\left(-\frac{1}{2}\right)\right)\left(z-\frac{4}{3}\right)$$

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

$$-6z^{2}+5z+4=-6\left(z+\frac{1}{2}\right)\left(z-\frac{4}{3}\right)$$

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

$$-6z^{2}+5z+4=-6\times \frac{-2z-1}{-2}\left(z-\frac{4}{3}\right)$$

Subtract \frac{4}{3} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6z^{2}+5z+4=-6\times \frac{-2z-1}{-2}\times \frac{-3z+4}{-3}$$

Multiply \frac{-2z-1}{-2} times \frac{-3z+4}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6z^{2}+5z+4=-6\times \frac{\left(-2z-1\right)\left(-3z+4\right)}{-2\left(-3\right)}$$

Multiply -2 times -3.

$$-6z^{2}+5z+4=-6\times \frac{\left(-2z-1\right)\left(-3z+4\right)}{6}$$

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

$$-6z^{2}+5z+4=-\left(-2z-1\right)\left(-3z+4\right)$$

Evaluate

$$\left(4-3z\right)\left(2z+1\right)$$