Polynomial Functions and Complex Zeros · 多项式函数与复零点
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| degree/dɪˈɡriː/ | 次数 | cì shù |
| real zeros/rɪəl ˈzɪərəʊz/ | 实零点 | shí líng diǎn |
| multiplicity/ˌmʌltɪˈplɪsɪti/ | 重数 | chóng shù |
| factored form/ˈfæktəd fɔːm/ | 因式分解形式 | yīn shì fēn jiě xíng shì |
| complex zeros/ˈkɒmpleks ˈzɪərəʊz/ | 复零点 | fù líng diǎn |
| conjugate pairs/ˈkɒndʒuːɡeɪt peəz/ | 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.
其他的根去哪了?
- $y = x^2 + 1$ 的图像是一条完全浮在 x 轴上方的抛物线。
- 它从不穿过 x 轴,所以看起来 $x^2 + 1 = 0$ 没有解。
- 可是代数坚持:一个二次方程应该有两个根。
- 那些"消失"的根,藏在复数里。
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.
实零点与重数
- 多项式的实零点(real zeros)是它的图像穿过或接触 x 轴的那些 x 值。
- 从因式分解形式(factored form)可以直接读出:$(x-2)(x+3)$ 的零点在 $2$ 和 $-3$。
- 像 $(x-2)^2$ 这样的重复因式,给出一个重数(multiplicity)为二的零点。
- 偶数重数 → 图像接触并弹回;奇数重数 → 图像穿过。
A zero where the graph just touches the x-axis (instead of crossing) has an even . · 图像只是接触而不穿过 x 轴的那种零点,具有偶数。
Even multiplicity (like $(x-3)^2$) makes the curve bounce off the axis; odd multiplicity makes it cross. · 偶数重数(比如 $(x-3)^2$)让曲线在轴上弹回;奇数重数则让它穿过。
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.
当图像错过 x 轴时
- 有些多项式从不触及 x 轴,所以它们的实零点比次数要少。
- 那些"消失"的零点就是复零点(complex zeros)——用 $i = \sqrt{-1}$ 构成的值。
- 作为方程的解,它们同样真实;只是不以 x 轴交点的形式出现。
A parabola that never meets the x-axis has how many real zeros? · 一条从不与 x 轴相交的抛物线,有多少个实零点?
No x-intercepts means no real zeros. But it still has two zeros — they are complex. · 没有 x 轴交点就意味着没有实零点。但它仍然有两个零点——它们是复数。
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.
复零点成对出现
- 如果一个多项式系数为实数,且 $a + bi$ 是它的零点,那么 $a - bi$ 也是零点。
- 这一对镜像伙伴被称为共轭对(conjugate pairs)。
- 这就是复零点总是一次来两个、从不单独出现的原因。
- 所以一个实系数多项式可以有 $0, 2, 4, \dots$ 个复零点,但绝不会是奇数个。
A complex zero and its mirror partner · 一个复零点和它的镜像伙伴
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. · 画出一个零点 a + bi。对实系数多项式,它的共轭 a − bi(关于实轴的镜像)也一定是零点。
For a polynomial with real coefficients, complex zeros always come in conjugate pairs. · 对实系数多项式,复零点总是成共轭对出现。
If $a + bi$ is a zero, then $a - bi$ must be one too. That is why complex zeros arrive two at a time. · 如果 $a + bi$ 是零点,那么 $a - bi$ 也一定是。这就是复零点总是成对出现的原因。
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.
数清每一个零点
- 一个 $n$ 次多项式恰好有 $n$ 个零点——只要你把实零点和复零点都按重数一起数。
- 这就是代数基本定理,它从不失效。
- 所以三次函数永远有 $3$ 个零点;四次函数永远有 $4$ 个。
- 如果你看到的 x 轴交点比次数(degree)少,剩下的就是复数。
Counted with multiplicity, how many zeros does a degree 4 polynomial have in total? · 按重数计算,一个4 次多项式总共有多少个零点?
A degree-$n$ polynomial has exactly $n$ zeros in total, real and complex counted with multiplicity. · 一个 $n$ 次多项式总共恰好有 $n$ 个零点,实的和复的按重数一起算。
A cubic (degree 3) has exactly 1 real zero. How many complex (non-real) zeros must it have? · 一个三次(3 次)多项式恰好有1个实零点。它一定有多少个复(非实)零点?
Total zeros $= 3$. With $1$ real, the other $2$ are a complex conjugate pair. · 零点总数 $= 3$。其中 $1$ 个是实的,另外 $2$ 个是一对共轭复数。
Select all · 所有 true statements. · 选出所有正确的说法。
A missing x-intercept does not · 不 reduce the zero count — the missing zeros are complex. The others are all true. · 缺少交点不会减少零点数量——缺的那些零点是复数。其余都对。
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.
一个"消失"的 x 轴交点并不意味着多项式的零点更少。零点总数永远等于次数——看不见的零点只是复数而已。不要因为只看到两个交点,就把一个四次函数说成"有两个零点"。
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.
解 $x^2 + 1 = 0$。
- 移项:$x^2 = -1$,所以 $x = \pm\sqrt{-1} = \pm i$。
- 两个零点是 $i$ 和 $-i$——一对共轭,正如所承诺的。
- 抛物线 $y = x^2 + 1$ 浮在轴上方,因为两个零点都不是实数。
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.
实零点就是 x 轴交点,偶数/奇数重数决定是接触还是穿过。剩下的零点是复零点,它们成共轭对出现。按重数计算,零点总数永远等于次数。