Q1。验证下面三次多项式旁边给出的数字是否为零。另外，在以下每种情况下，验证零和系数之间的关系：
(i)f(x)= 2x 3 + x 2 – 5x + 2； 1/2，1，-2
解决方案：
For the followings numbers to be the zeros of polynomial they must satisfy the following equation i.e. f(x)=0 so now checking:
When x = 1/2
f(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2
f(1/2) = 1/4 + 1/4 – 5/2 + 2 = 0
f(1/2) = 0,
Hence, x = 1/2 is a zero of the given polynomial.
When x = 1
f(1) = 2(1)3 + (1)2 – 5(1) + 2
f(1) = 2 + 1 – 5 + 2 = 0
f(1) = 0,
Hence, x = 1 is also a zero of the given polynomial.
When x = -2
f(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
f(-2) = -16 + 4 + 10 + 2 = 0
f(-2) = 0,
Hence, x = -2 is also a zero of the given polynomial.
We know that Sum of zeros = -b/a;
Sum of the products of the zeros taken two at a time = c/a;
Product of zeros = -d/a;
Here sum=1/2 + 1 – 2 =-1/2 and -b/a=-1/2
Here sum of products=(1/2 * 1) + (1 * -2) + (1/2 * -2) =-5/2 and c/a=-5/2
Here product =1/2 x 1 x (- 2) = -1 and -d/a=-1
Hence, the relationship between the zeros and coefficients is verified.
(ii)g(x)= x 3 – 4x 2 + 5x – 2； 2 1 1
解决方案：
For the followings numbers to be the zeros of polynomial they must satisfy the following equation ie. g(x)=0 so now checking:
When x = 2
g(2) = (2)3 – 4(2)2 + 5(2) – 2
g(2) = 8 – 16 + 10 – 2 = 0
g(2) = 0,
Hence, x = 2 is a zero of the given polynomial.
Now we have two same roots, so we will check only once.
When x = 1
g(1) = (1)3 – 4(1)2 + 5(1) – 2
g(1) = 1 – 4 + 5 – 2 = 0
g(1) = 0,
Hence, x = 1 is also a zero of the given polynomial.
We know that Sum of zeros = -b/a;
Sum of the products of the zeros taken two at a time = c/a;
Product of zeros = – d/a;
Here sum=2+1+1= 4 and -b/a=4
Here sum of products=(1 * 1) + (1 * 2) + (2 * 1) =5 and c/a =5
Here product = 2*1*1=2 and -d/a=2
Hence, the relationship between the zeros and coefficients is verified.
问题2。找到三次多项式，其总和，一次取零的乘积之和和一次分别为3，-1和-3的零乘积。
解决方案：
A cubic polynomial say, f(x) is of the form ax3 + bx2 + cx + d.
We know that f(x) = k [x3 – (sum of roots)x2 + (sum of products of roots taken two at a time)x -(product of roots)]
Sum of roots = 3;
Sum of products of roots taken two at a time=-1;
Product of roots=-3
f(x) = k [x3 – (3)x2 + (-1)x – (-3)]
∴ f(x) = k [x3 – 3x2 – x + 3)]
Required polynomial f(x) = k [x3 – (3)x2 + (-1)x – (-3)]
∴ f(x) = k[x3 – 3x2 – x + 3)]
问题3.如果多项式f(x)= 2x 3 – 15x 2 + 37x – 30的零在AP中，请找到它们。
解决方案：
Let the roots be α = a – d, β = a and γ = a +d, Where, a is the first term and d is the common difference.
From given f(x), a= 2, b= -15, c= 37 and d= 30
=> Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2
So, calculating for a, we get 3a = 15/2 i.e. a = 5/2
=> Product of roots = (a – d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15 i.e. a(a2 –d2) = 15
Substituting ‘a’ we get ∴ d = 1/2 or -1/2
When d=1/2
Roots are α = 5/2-1/2, β = 5/2 and γ = 5/2 +1/2 i.e. α = 2, β = 2.5 and γ = 3
When d=-1/2
Roots are α = 5/2-(-1/2), β = 5/2 and γ = 5/2 +(-1/2) i.e. α = 3, β = 2.5 and γ = 2
问题4.确定多项式f(x)= x 3 + 3px 2 + 3qx + r的零点可能在AP中的条件
解决方案：
Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.
sum = A-D+A+A+D = -b/a i.e. 3A=-3b so, A=-b
Since β = A is a root so f(A)=0;
a3+3pa2+3qa+r=0
Now put a=-p; so, we get
2p2-3pq+r=0 is the required condition.
问题5.如果多项式的零点f(x)= ax 3 + 3bx 2 + 3cx + d在AP中，则证明2b 3 -3abc + a 2 d = 0。
解决方案：
Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.
sum = A-D+A+A+D = -b/a i.e. 3A=-3b/a so, A=-b/a
Now f(A)=0 so,
f(x)=aA3+3bA2+3cA+d
Now put A=-b/a; So, we get:
2b3-3abc+a2d =0
Hence, proved.
问题6.如果多项式f(x)= x 3 -12x 2 + 39x + k的零在AP中，则求出k的值。
解决方案：
Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.
sum = A-D+A+A+D = -b/a i.e., 3A=12 so, A=4
f(β)=0 i.e. f(A)=0
(4)3-12(4)2+39(4)+k=0
64-192+156+k=0
28+k=0
k=-28