Solving quadratic equations of the form $$x^2+bx+c=0.$$ You can do this by completing the square, applying the quadratic formula, or in some (rare) cases by factoring. For example, the equation $$x^2-2x-15=0$$ has two real solutions, the larger one of which is and the smaller of which is .
Quadratic equations do not always occur in the form listed above. If $x^2$ is multiplied by a factor as in $$4x^2+12x+6 = 0$$ you can still apply the quadratic formula, but if you want to complete the square you first need to divide by the leading coefficient first on both sides of the equation. Be careful if you decide to rely on the quadratic formula, in my experience few people are able to remember it dependably and to apply it correctly. The solution of the above equation is $x=$ $\pm$
Sometimes equations do not look like quadratic equations but can be converted to such. For example, in the equation $$x^4 -13x^2 +36=0 \quad(*)$$ you would think of $z=x^2$ as the variable and obtain a quadratic equation in $z$: $$z^2-13z+36=0.$$ Once you know $z$ you can find $x$ by computing the positive and negative square roots of $z$. For example, the largest solution of $(*)$ is . In the equation $$x-13\sqrt{x}+36 = 0$$ you think of $z=\sqrt{x}$ as the variable, solve for $z$, square $z$ and get the largest solution . Sometimes you have rational expressions and you obtain a quadratic equation after multiplying with the appropriate denominators, as in the equation $$\frac{x-1}{x}=x$$ which has a pair of conjugate complex solutions: $x=$ $\pm$ $i$.
In radical equations you get rid of the radicals by isolating them and taking them to the appropriate power. For example, the solution of the equation $$\sqrt{x} + \sqrt{x+1} = 3$$ is $x=$
Polynomials, like any functions, can be evaluated, at numbers and also at algebraic expressions. For example if $$p(x)=3x^2-4x+5$$ then $p(x-3)=$ $x^2$ - $x+$ .
Understand how to use synthetic division to divide a polynomial by a linear factor.
Understand how to draw the graph of a quadratic function. It is always a parabola.
Understand how to obtain the standard form of a rational function (as a ratio of two polynomials). Consider for example the function $$r(x) = \frac{1}{x+1} - \frac{1}{1+\frac{1}{x}}.$$ It can be simplified to $$r(x) = \frac{1-x}{1+x},$$ the ratio of two polynomials. To get rid of the ratio in the denominator of the second term multiply the numerator and denominator in the second term with $x$. As an exercise simplify the following expression $\frac{1}{x+1} + \frac{1}{1+\frac{1}{x}}=$ .