多项式用于模拟现实生活中发生的某些物理现象，它们在数学上描述情况非常有用。它们几乎用于科学的每个领域，甚至包括科学之外的领域，例如在经济学和其他相关领域。这些多项式的零或根是其性质的一个非常重要的方面，在描述它们或将其绘制在图形上时非常有用。让我们看一下它们的定义和详细找出根源的方法。
多项式的零/根
我们说如果p(x)= 0，则x = a是多项式的根。找出零的过程基本上就是找出任何多项式方程的解的过程。让我们看一些有关为二次多项式找到零的示例。
问题1：找出P(x)= x 2 + 2x – 15的零点。
回答：
x2 + 2x – 15 = 0
⇒ x2 + 5x – 3x – 15 = 0
⇒ x(x + 5) – 3(x + 5) = 0
⇒ (x – 3) (x + 5) = 0
⇒ x = 3, -5
问题2：找出P(x)= x 2 – 16x + 64的零点。
回答：
x2 – 16x + 64 = 0
⇒ x2 – 8x – 8x + 64 = 0
⇒ x(x – 8) – 8(x – 8) = 0
⇒ (x – 8) (x – 8) = 0
⇒ (x – 8)2 = 0
x = 8, 8
This is called a double root.
假设我们有一个多项式P(x)= 0分解为
P(x)=(x – r) k (x – a) m
如果r是多项式的零，并且其项的产生根的指数是k，那么我们说r具有多重性k 。复数为1的零通常称为简单零。
问题3：P(x)是5级多项式，已为您分解。列出根及其多样性。
P(x)= 5x 5 −20x 4 + 5x 3 + 50x 2 −20x−40 = 5(x + 1) 2 (x−2) 3
回答：
Given, P(x) = 5(x+1)2(x−2)3
Putting this polynomial equal to zero we get the root,
x = -1, -1, 2, 2, 2
Notice that -1 occurs two times as a root. So its multiplicity is 2 while the multiplicity of the root “2” is 3.
线性代数的基本定理
If P(x) is a polynomial of degree “n” then P(x) will have exactly n zeros, some of which may repeat.
这意味着，如果我们列出所有零，并在k是其多重性时列出k一次，则将其列出k次。我们将在列表中恰好有n个数字。这很有用，因为它可以使我们了解多项式中应该有多少个零。因此，一旦达到所需的零数，我们就可以停止寻找零。
因子定理
For the polynomial P(x),
- If r is a zero of P(x) then x−r will be a factor of P(x).
- If x−r is a factor of P(x) then r will be a zero of P(x).
可以通过查看前面的示例来验证。这个因子定理可以得出一些有趣的结果，
Result 1: If P(x) is a polynomial of degree “n”, and “r” is a zero of P(x) then P(x) can be written in the following form,
P(x) = (x – r) Q(x)
Where Q(x) is a polynomial of degree “n-1” and can be found out by dividing P(x) with (x – r).
Result 2: If P(x) = (x-r)Q(x) and x = t is a zero of Q(x) then x = t will also be zero of P(x).
为了验证上述事实，
假设“ t”是根Q(x)，这意味着Q(t)= 0。
我们知道“ r”是多项式P(x)的根，其中P(x)=(x – r)Q(x)，
因此，我们需要检查x = t是否也是P(x)的根，让我们将x = t放入P(x)
P(t)=(t – r)Q(t)= 0
因此，x = t也是根P(x)。
因此，证明。
样本问题
问题1：假设x = 2是P(x)= x 3 + 2x 2 -5x-6的零。找到其他两个零。
解决方案：
From the fundamental theorem we studied earlier, we can say that P(x) will have 3 roots because it is a three degree polynomial. One of them is x = 2.
So we can rewrite P(x),
P(x) = (x – 2) Q(x)
For finding the other two roots, we need to find out the Q(x).
Q(x) can be found out by dividing P(x) by (x-2).
After dividing, the Q(x) comes out to be,
Q(x) = x2 + 4x + 3
The remaining two roots can be found out from this,
Q(x) = x2 + 3x + x + 3
⇒ x(x + 3) + 1(x + 3)
⇒ (x + 1) (x + 3)
Q(x) = 0,
x = -1, -3
Thus, the other two roots are x = -1 and x = -3.
问题2：假设x = r是多项式的根，请找出多项式的其他根。
P(x)＝ x 3 -6x 2 -16x； r = −2
解决方案：
We know that x = -2 is a root,
So, P(x) can be rewritten as, P(x) = (x + 2) Q(x).
Now to find Q(x), we do the same thing as we did in the previous question, we divide P(x) with (x + 2).
We get,
Q(x) = x2 – 8x
Now to find the other two roots, factorize Q(x)
Q(x) = x (x – 8) = 0
So, the roots are x = 0, 8.
Thus, we have three roots, x = -2, 0, 8.
SO, this polynomial can also be written in factored form,
P(x) = (x + 2) (x) (x – 8)
问题3：找出多项式的根，即4x 3 -3x 2 -25x-6 = 0
解决方案：
Trick to solve polynomial equations with degree 3,
Find the smallest integer that can make the polynomial value 0, start with 1,-1,2, and so on…
Here we can see -2 can make the polynomial value 0.
Write (x+2) at 3 places and then write the coefficients accordingly to make the complete polynomial
4x2 (x+2) -11x(x+2) -3(x+2) =0
Now, notice carefully, the first coefficient is 4x2, because when it is multiplied with the x inside the bracket, it gives 4x3
When 4x2 is multiplied with 2, it gives 8x2, but the second term must be -3x2, hence the coefficient added next is -11x
Now, we know how to adjust the terms so that when we simplify it gives back the original polynomial.
We get a quadratic equation and a root is already there,
(4x2-11x-3)(x+2) = 0
Factorize the quadratic equation,
(4x2-12x+x-3)(x+2) = 0
(4x(x-3)+1(x-3))(x+2) = 0
(4x+1)(x-3)(x+2) = 0
x = -2, x = 3, x = -1/4
问题4：找出多项式的零点4x 6 – 16x 4 = 0
解决方案：
The Polynomial has up to degree 6, hence, there exist 6 roots of the polynomial.
4x4(x2-4) = 0
4x4(x2-22) = 0
4x4[(x+2)(x-2)] = 0
Therefore, x= 0, 0, 0, 0, 2, -2