10类NCERT解决方案-第2章多项式-练习2.3

问题1.将多项式p(x)除以多项式g(x)，然后在以下每一项中找到商和余数：

(i)p(x)= x ³ – 3x ² + 5x – 3，g(x)= x ² – 2

(ii)p(x)= x ⁴ – 3x ² + 4x + 5，g(x)= x ² +1 – x

(iii)p(x)= x ⁴ – 5x + 6，g(x)= 2 – x ²

解决方案：

i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

R = 7x-9

Q = x-3

ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

R = 8

Q = x²+x-3

iii) p(x) = x⁴– 5x + 6, g(x) = 2 – x²

Q = -x²-2

R = -5x+10

问题2.通过将第二多项式除以第一多项式，检查第一多项式是否为第二多项式的因数。

(i)t ² – 3，2t ⁴ + 3t ³ – 2t ² – 9t – 12

(ii)x ² + 3x + 1，3x ⁴ + 5x ³ – 7x ² + 2x + 2

(iii)x ³ – 3x + 1，x ⁵ – 4x ³ + x ² + 3x + 1

解决方案：

i) t² – 3, 2t⁴ + 3t³ – 2t²– 9t – 12

Q = 2t³+3t+4

R = 0

Yes 1^st polynomial is factor of 2^nd polynomial. 

ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

R = 0

Q = 3x²-4x+2

iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

R = x²-1

Q = 2

问题3.如果其他两个零分别为√(5/3)和-√(5/3)，则获得所有其他零，分别为3x ⁴ + 6x ³ – 2x ^{2 – 10x – 5。}

解决方案：

$(x- \sqrt{\frac{5}{3}}) (x + \sqrt{\frac{5}{3}}) = x^{2}-(\sqrt{\frac{5}{3}})^{2} = x^{2}-\frac{5}{3}$

R = 0

Q = 3x²+6x+3

∴ we are factorizing

3x²+6x+3

x²+2x+1

(x+1)²

(x+1)  (x+1) = 0

∴ x = -1 and x = -1

$\sqrt{\frac{5}{3}}, - \sqrt{\frac{5}{3}}, -1, -1$

问题4.将x ³ – 3x ² + x + 2除以多项式g(x)时，商和余数分别为x – 2和-2x + 4。求g(x)

解决方案：

Dividend = Divisor * Quotient + Remainder

x³-3x²+3x-2/x-2

R = 0

Q = x² -x +1

Answer: g(x)=x²-x+1

问题5.给出满足除法算法的多项式p(x)，g(x)，q(x)和r(x)的示例，并：

(i)度p(x)=度q(x)

(ii)度q(x)=度r(x)

(iii)度r(x)= 0

解决方案：

i) deg p(x) = deg q(x)

p(x)=2x²-2x+14, g(x)=2

p(x)/g(x)=2x²-2x+14/2=(x²-x+7)

=x²-x+7=q(x)

=q(x)=x²-x+7

r(x)=0

ii) deg q(x)=deg r(x)

p(x)=4x²+4x+4, g(x)=x²+x+1

q(x) = 4

r(x) = 0

∴Here deg q(x)=deg r(x)

iii) deg r(x)=0

p(x)=x³+2x²-x+2 ,g(x)=x²-1

q(x) = x+2

r(x) = 4

deg of r(x) = 0