Algebra II: Polynomials & Complex Numbers

How to do it:

1. Add or subtract like terms (same variable, same exponent).

2. Multiply using distribution or an area model.

3. The degree of a product is the sum of the degrees.

Worked example: $(x^2+1)(x-3)=x^3-3x^2+x-3\text{.}$ See remainder theorem.

Check: $(x^2+1)+(x-3)$ is $x^2+x-2\text{,}$ not a product. Polynomials stay polynomials under $+$ and $\times\text{,}$ like integers.

How to do it:

1. If $p(a)=0\text{,}$ then $(x-a)$ is a factor (factor theorem).

2. Plot the zeros. Odd multiplicity crosses; even multiplicity bounces.

3. Use the leading term for end behavior as $x\to\pm\infty\text{.}$

Worked example: $p(x)=(x+2)(x-1)^2$ has zeros $-2$ (simple, crosses) and $1$ (double, bounce). Leading term $x^3\text{,}$ so as $x\to\infty\text{,}$ $p(x)\to\infty\text{,}$ and as $x\to-\infty\text{,}$ $p(x)\to-\infty\text{.}$

Check: a double zero on the axis looks like it "touches and turns." Expanding to check: $(x+2)(x^2-2x+1)=x^3-3x+2\text{.}$

How to do it:

1. Define $i$ by $i^2=-1\text{.}$ Write $z=a+bi\text{.}$

2. Add real parts and imaginary parts separately.

3. Multiply with distribution, then replace $i^2$ with $-1\text{.}$

Worked example: $(2+3i)(1-i)=2-2i+3i-3i^2=2+i+3=5+i\text{.}$

Check: $x^2+1=0$ → $x=\pm i\text{.}$ The quadratic formula still works; a negative discriminant becomes $bi\text{.}$ Plot $5+i$ as the point $(5,1)$ on the plane.

How to do it:

1. Write one factor across the top of a grid and the other down the side, one term per cell.

2. Multiply into every cell.

3. Add the cells and combine like terms; the degree of the product is the sum of the degrees.

Worked example: $(x+2)(x^2-3x+4)\text{.}$ Row $x\text{:}$ $x^3\text{,}$ $-3x^2\text{,}$ $4x\text{.}$ Row $2\text{:}$ $2x^2\text{,}$ $-6x\text{,}$ $8\text{.}$ Add: $x^3-x^2-2x+8\text{.}$

Check: a degree-$1$ times a degree-$2$ polynomial has degree $3\text{,}$ and a quick test at $x=1$ works: $(3)(2)=6$ and $1-1-2+8=6\text{.}$

How to do it:

1. Divide as with long division: divide leading terms, multiply back, subtract, bring down, repeat.

2. Remainder theorem: dividing $p(x)$ by $(x-a)$ leaves remainder $p(a)\text{,}$ so you can find the remainder by substituting.

3. Remainder $0$ means $(x-a)$ is a factor and $a$ is a zero (factor theorem).

Worked example: $(x^3-3x+2)\div(x-1)\text{.}$ $x^3\div x=x^2\text{;}$ subtract $x^2(x-1)=x^3-x^2$ to get $x^2-3x\text{;}$ then $+x$ leaves $-2x+2\text{;}$ then $-2$ leaves $0\text{.}$ Quotient $x^2+x-2\text{,}$ remainder $0\text{,}$ so $x-1$ is a factor. For the remainder on dividing by $(x-2)\text{,}$ just substitute: $p(2)=8-6+2=4\text{.}$

Check: $(x-1)(x^2+x-2)=x^3-3x+2\text{.}$ And $p(1)=1-3+2=0$ confirms the zero at $1\text{.}$

How to do it:

1. Difference of squares: $a^2-b^2=(a-b)(a+b)\text{.}$

2. Perfect square trinomial: $a^2\pm 2ab+b^2=(a\pm b)^2\text{.}$

3. Sum and difference of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)\text{.}$ The sign in the binomial matches the original; the middle sign in the trinomial is opposite.

Worked example: $4x^2-25=(2x-5)(2x+5)\text{.}$ $x^2+6x+9=(x+3)^2\text{.}$ $x^3-8=x^3-2^3=(x-2)(x^2+2x+4)\text{.}$

Check: expand $(x-2)(x^2+2x+4)=x^3+2x^2+4x-2x^2-4x-8=x^3-8\text{.}$ The middle terms cancel in pairs, which is the whole point of the pattern.

How to do it:

1. The conjugate of $a+bi$ is $a-bi\text{.}$ Their product is real: $(a+bi)(a-bi)=a^2+b^2\text{.}$

2. To divide, multiply the top and bottom by the conjugate of the bottom.

3. Simplify with $i^2=-1$ and write the answer as $a+bi\text{.}$

Worked example: $\dfrac{2+3i}{1-i}\text{.}$ Multiply by $\dfrac{1+i}{1+i}\text{:}$ top $(2+3i)(1+i)=2+2i+3i+3i^2=-1+5i\text{;}$ bottom $(1-i)(1+i)=1-i^2=2\text{.}$ So $\dfrac{-1+5i}{2}=-\dfrac{1}{2}+\dfrac{5}{2}i\text{.}$

Check: multiply back: $(1-i)\left(-\tfrac{1}{2}+\tfrac{5}{2}i\right)=-\tfrac{1}{2}+\tfrac{5}{2}i+\tfrac{1}{2}i-\tfrac{5}{2}i^2=2+3i\text{.}$

How to do it:

1. The powers cycle every four: $i^1=i\text{,}$ $i^2=-1\text{,}$ $i^3=-i\text{,}$ $i^4=1\text{,}$ then repeat.

2. For a big power, divide the exponent by $4$ and use the remainder.

3. $x^2+k=0$ with $k>0$ has the two solutions $x=\pm i\sqrt{k}\text{;}$ a quadratic with a negative discriminant has two complex solutions that are conjugates.

Worked example: $i^{23}\text{:}$ $23=4(5)+3\text{,}$ so $i^{23}=i^3=-i\text{.}$ $x^2+9=0\text{:}$ $x^2=-9\text{,}$ so $x=\pm 3i\text{.}$ $x^2-2x+5=0\text{:}$ the discriminant is $4-20=-16\text{,}$ so $x=\dfrac{2\pm\sqrt{-16}}{2}=\dfrac{2\pm 4i}{2}=1\pm 2i\text{.}$

Check: $(3i)^2=9i^2=-9\text{,}$ so $x^2+9=0$ holds. The two solutions $1+2i$ and $1-2i$ are conjugates, as they must be when the coefficients are real.