Polynomial Functions and Complex Zeros
| English | Chinese | Pinyin |
|---|---|---|
| degree | 次数 | cì shù |
| real zeros | 实零点 | shí líng diǎn |
| multiplicity | 重数 | chóng shù |
| factored form | 因式分解形式 | yīn shì fēn jiě xíng shì |
| complex zeros | 复零点 | fù líng diǎn |
| conjugate pairs | 共轭对 | gòng è duì |
Where did the other roots go?
- The graph of $y = x^2 + 1$ is a parabola floating entirely above the x-axis.
- It never crosses, so it seems to have no solutions to $x^2 + 1 = 0$.
- Yet algebra insists a degree-2 equation should have two roots.
- The missing roots are hiding in the complex numbers.
Real zeros and multiplicity
- The real zeros 实零点 of a polynomial are the x-values where its graph crosses or touches the x-axis.
- Read them straight from the factored form 因式分解形式: $(x-2)(x+3)$ has zeros at $2$ and $-3$.
- A repeated factor like $(x-2)^2$ gives a zero of multiplicity 重数 two.
- Even multiplicity → the graph touches and bounces; odd multiplicity → it crosses.
A zero where the graph just touches the x-axis (instead of crossing) has an even ____.
Even multiplicity (like $(x-3)^2$) makes the curve bounce off the axis; odd multiplicity makes it cross.
When the graph misses the axis
- Some polynomials never reach the x-axis, so they have fewer real zeros than their degree.
- The zeros that are "missing" are complex zeros 复零点 — values built with $i = \sqrt{-1}$.
- They are just as real as solutions; they simply do not appear as x-intercepts.
A parabola that never meets the x-axis has how many real zeros?
No x-intercepts means no real zeros. But it still has two zeros — they are complex.
Complex zeros travel in pairs
- If a polynomial has real coefficients and $a + bi$ is a zero, then $a - bi$ is a zero too.
- These mirror-image partners are called conjugate pairs 共轭对.
- That is why complex zeros always arrive two at a time — never alone.
- So a real polynomial can have $0, 2, 4, \dots$ complex zeros, but never an odd number.
A complex zero and its mirror partner
Plot a zero a + bi. For a real polynomial its conjugate a − bi (the mirror image across the real axis) must be a zero too.
For a polynomial with real coefficients, complex zeros always come in conjugate pairs.
If $a + bi$ is a zero, then $a - bi$ must be one too. That is why complex zeros arrive two at a time.
Counting every zero
- A polynomial of degree $n$ has exactly $n$ zeros, once you count real and complex zeros with multiplicity.
- This is the Fundamental Theorem of Algebra, and it never fails.
- So a cubic always has $3$ zeros; a quartic always has $4$.
- If you can see fewer x-intercepts than the degree 次数, the rest are complex.
Counted with multiplicity, how many zeros does a degree 4 polynomial have in total?
A degree-$n$ polynomial has exactly $n$ zeros in total, real and complex counted with multiplicity.
A cubic (degree 3) has exactly 1 real zero. How many complex (non-real) zeros must it have?
Total zeros $= 3$. With $1$ real, the other $2$ are a complex conjugate pair.
Select all true statements.
A missing x-intercept does not reduce the zero count — the missing zeros are complex. The others are all true.
A "missing" x-intercept does not mean the polynomial has fewer zeros. The zero count always equals the degree — the unseen zeros are simply complex. Do not report a quartic as having "two zeros" just because you see two x-intercepts.
Solve $x^2 + 1 = 0$.
- Rearranged: $x^2 = -1$, so $x = \pm\sqrt{-1} = \pm i$.
- The two zeros are $i$ and $-i$ — a conjugate pair, exactly as promised.
- The parabola $y = x^2 + 1$ floats above the axis because neither zero is real.
The real zeros are the x-intercepts, with even/odd multiplicity deciding touch vs cross. Any remaining zeros are complex zeros, which occur in conjugate pairs. Counted with multiplicity, the total number of zeros always equals the degree.