第10类NCERT解决方案–第2章多项式–练习2.4

问题1.验证以下三次多项式旁边给出的数字是否为零。另外，在每种情况下，验证零和系数之间的关系：

(i)2x ³ + x ² – 5x + 2； 1/2，1，-2

解决方案：

p(x) = 2x³+x²-5x+2

p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 

           = (1/4)+(1/4)-(5/2)+2 

           = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Therefore, 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression

ax³+bx²+cx+d = 2x³+x²-5x+2

a=2, b=1, c= -5 and d = 2

α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

α+β+γ = ½+1+(-2) 

            = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) 

                  = -5/2 = c/a

α β γ = ½×1×(-2) 

         = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii)x ³ – 4x ² + 5x – 2 ； 2 1 1

解决方案：

p(x) = x³-4x²+5x-2

Zeroes are 2,1,1.

p(2)= 2³-4(2)²+5(2)-2 

      = 0

p(1) = 1³-(4)(1² )+(5)(1)-2 = 0

Therefore, proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

On comparing the given polynomial with general expression

ax³+bx²+cx+d = x³-4x²+5x-2

a = 1, b = -4, c = 5 and d = -2

Therefore,

α + β + γ = –b/a

                = 2+1+1 

                = 4 

         –b/a = -(-4)/1

αβ + βγ + γα = c/a

                      = 2×1+1×1+1×2 

                      = 5 

                c/a = 5/1 

α β γ = – d/a.

          = 2×1×1 

          = 2 

 -d/a = -(-2)/1

Hence, the relationship between the zeroes and the coefficients is satisfied.

问题2。找到一个三次多项式，其总和为一次取两个零的乘积之和，而其零的乘积分别为2 – –7 – –14。

解决方案：

Let us consider the cubic polynomial as ax³+bx²+cx+d and zeroes of the polynomials be α, β, γ.

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

On comparing

a = 1, b = -2, c = -7, d = 14

Therefore, the cubic polynomial is x³-2x²-7x+14

问题3。如果多项式x ³ – 3x ² + x + 1的零为a – b，a，a + b，则找到a和b。

解决方案：

p(x) = x³-3x²+x+1

Zeroes are given as a – b, a, a + b

px³+qx²+rx+s = x³-3x²+x+1

On comparing

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Therefore, zeroes are 1-b, 1, 1+b.

Product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = √2

Therefore,1-√2, 1,1+√2 are the zeroes of x³-3x²+x+1.

问题4.如果多项式^×4 -6x ³ -26x ² + 138X-35的两个零是2±√3，找到其他零。

解决方案：

Degree of polynomial is 4

Therefore, it has four roots

f(x) = x⁴-6x³-26x²+138x-35

As 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

Therefore, [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

Therefore, x²-4x+1 is a factor of polynomial f(x).

Let it be g(x) = x²-4x+1

By dividing f(x) by g(x) we get another factor of f(x)

x⁴-6x³-26x²+138x-35 = (x²-4x+1)(x² –2x−35)

On factorizing (x²–2x−35) by splitting the middle term

x²–(7−5)x −35 = x²– 7x+5x-35

                         =x(x −7)+5(x−7)

(x+5)(x−7) = 0

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3, 2-√3, −5 and 7.