任何具有常数和变量的代数表达式都称为多项式。多项式是“ Poly”和“ nominal”两个词的组合,其中poly表示“许多”而名义上表示“项”,因此,多项式可以包含尽可能多的项,但绝不能是无限的。在多项式中,零和系数存在于已经提供给我们的系数中,但是,我们需要借助系数来获取零的值。
零和系数之间的关系
我们知道,任何多项式的零点都是多项式图切割x轴的点。也可以使用多项式中不同项的系数找出这些零。让我们看一下零与多项式系数之间的关系。
线性多项式
通常,线性多项式由下式定义:
y =轴+ b
我们知道,对于零,我们需要找到y = 0的点。求解y = 0的一般方程。
y =轴+ b
⇒0=斧+ b
⇒x=
这为我们提供了零和线性多项式系数之间的关系。
In general for a linear equation y = ax + b, a ≠ 0, the graph of ax + b is a straight line that cuts the x-axis at (, 0)
问题:使用上述公式和图形方法验证线性多项式的零。
y = 4x + 2
解决方案:
We are given with the equation y = 4x + 2,
Here a = 4 and b = 2
So, by the formula mentioned above the zero will occur at (, 0) that is
Let’s verify this zero with graphical method. We need to plot the graph of this equation.
y = 4x + 2
Let’s bring it to the intercept form.
Now we know the intercepts on the x and y-axis.
二次多项式
二次多项式具有最高的2阶数,并且伴随着因式分解,还有其他一些方法可以找到二次多项式的零点,例如Dharacharya方法。由于它具有最高的2级,因此在二次多项式中存在2个零。
让我们得出零与二次多项式系数之间的关系。假设一个多项式,
P(x)= 2x 2 – 8x + 6
该多项式可以分解如下,
P(x)= 2x 2 -8x + 6
= 2x 2 – 6x – 2x + 6
= 2x(x – 3)-2(x – 3)
= 2(x – 1)(x – 3)
所以这个方程的根为x = 1和x = 3
请注意,
Sum of zeros = 1 + 3 = 4 =
Product of zeros = 1 × 3 = 3 =
这是零和二阶系数的系数之间的关系。
因此,以一般形式
For a polynomial, p(x) = ax2 + bx + c which has m and n as roots
m + n =
m × n =
问题:验证上述方程式的属性6x 2 – 10x + 4
解决方案:
6x2 – 10x + 4
⇒ 6x2 – 6x -4x + 4
⇒ 6x(x – 1) – 4(x – 1)
⇒ (6x – 4) (x – 1)
Thus, the roots for this equation come out to be x = and x = 1
Now we know according to above properties,
Sum of zeros =
Product of zeros =
Let’s verify it
Sun of zeros = 1 + =
Both the values come to be equal. Hence, verified.
Let’s verify the product of roots property
Product of zeros =
In this case also, both values are equal. Hence, verified.
三次多项式
对于三次多项式可以得出类似的关系。三次多项式是次数为3的多项式,并且由于其最高次数为3,所以三次多项式存在三个零。假设获得的多项式的根/零为p,q,r,则零与多项式之间的关系为:
For a cubic polynomial,
ax3 + bx2 + cx + d
Which has roots x = p, q and r
p + q+ r =
pq + qr + pr =
pqr =
让我们看一些有关这些属性的示例。
样本问题
问题1:验证-1,1,2是多项式x 3 -2x 2 – x + 2的根。还要验证上述属性。
解决方案:
x = -1, 1 and 2 are the roots of polynomial that is P(x) = 0 at all these points. Let’s plug in the values of x one by one in the polynomial.
P(-1) = (-1)3 – 2(-1)2 -(-1) + 2
= -1 – 2 + 1 + 2
= -3 + 3
= 0
P(1) = (1)3 – 2(1)2 – 1 + 2
= 1 – 2 – 1 + 2
= 0
P(2) = (2)3 – 2(2)2 – 2 + 2
= 8 – 8 – 2 + 2
= 0
Thus, all these points are roots of the polynomial
Let’s verify the properties
(1) p + q+ r =
⇒ -1 + 1+ 2 =
⇒ 2 = 2
L.H.S = R.H.S
Hence Verified
(2) pq + qr + pr =
⇒ (-1)(1) + (1)(2) + (-1)(2) =
⇒ -1+ 2 -2 = -1
⇒ -1 = -1
L.H.S = R.H.S
Hence Verified
(3) pqr =
⇒ (-1)(1)(2) =
⇒ -2 = -2
L.H.S = R.H.S
Hence Verified
问题2:评估二次多项式6x 2 + 18的零的总和。
解决方案:
General form of a quadratic polynomial is ax2 + bx + c = 0.
Given polynomial 6x2 + 18 can be rewritten as,
6x2 + 0.x + 18
As studied above the sum and product of roots of quadratic polynomial is given by,
m + n =
m.n =
Where m and n are the roots of the polynomial
In our case a = 6, b = 0 and c = 18. Plugging the values in the formulas
m + n =
m.n =
Thus, the sum of roots is 0 and product is given by 3.
问题3:给定多项式ax 2 + bx +1。其根为-1和3。找到a和b的值。
解决方案:
We know the formula for sum and product of the root of a quadratic polynomial .
Let m and n be the roots,
Here, m = -1 and n =3
So, by the formula
m + n = …(1)
m.n = …..(2)
From the equation (1)
-1 + 3 =
⇒ 2 = \frac{-b}{a}
⇒ 2a = -b
From equation (2)
(-1)(3) =
⇒ -3a = 1
⇒ a =
Putting this value of “a” in the above equation
b =