Solve 1+z-2z^2

$$1 + z - 2 z ^ { 2 }$$

Factor

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

Solution Steps

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

$$-2z^{2}+z+1$$

Factor the expression by grouping. First, the expression needs to be rewritten as -2z^{2}+az+bz+1. To find a and b, set up a system to be solved.

$$a+b=1 ab=-2=-2$$

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. The only such pair is the system solution.

$$a=2 b=-1$$

Rewrite -2z^{2}+z+1 as \left(-2z^{2}+2z\right)+\left(-z+1\right).

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

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

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

Factor out common term -z+1 by using distributive property.

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

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.

$$-2z^{2}+z+1=0$$

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.

$$z=\frac{-1±\sqrt{1^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Square 1.

$$z=\frac{-1±\sqrt{1-4\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$z=\frac{-1±\sqrt{1+8}}{2\left(-2\right)}$$

Add 1 to 8.

$$z=\frac{-1±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of 9.

$$z=\frac{-1±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$z=\frac{-1±3}{-4}$$

Now solve the equation z=\frac{-1±3}{-4} when ± is plus. Add -1 to 3.

$$z=\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation z=\frac{-1±3}{-4} when ± is minus. Subtract 3 from -1.

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

Divide -4 by -4.

$$z=1$$

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 1 for x_{2}.

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

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

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

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

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

Cancel out 2, the greatest common factor in -2 and 2.

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

Evaluate

$$1+z-2z^{2}$$