8类RD Sharma解–第8章代数表达式的除法–练习8.4 |套装1

问题1.将5x ³ – 15x ² + 25x除以5x

解决方案：

We have to divide 5x³ – 15x² + 25x by 5x

So, (5x³ – 15x² + 25x)/5x

5x³/5x – 15x²/5x + 25x/5x  

x² – 3x + 5

问题2：将4z ³ + 6z ² – z除以−1/2 z

解决方案：

We have to divide 4z³ + 6z²-z by −1/2 z

So, (4z³ + 6z² – z) / (-1/2)z 

(4z³/(-1/2)z) + (6z²/(-1/2)z) + (-z/(-1/2)z)

(-4 × 2/1)z² + (6 × (-2)/1)z + (-1 × (-2/1))

-8z² – 12z +2

问题3.将9x ² y – 6xy + 12xy ²除以(−3/2)xy。

解决方案：

We have to divide 9y – 6xy + 12xy² by (-3/2)xy

So, (9x²y – 6xy + 12xy²) / (-3/2)xy

(9x²y/(-3/2)xy ) – (6xy/(-3/2)xy) + (12xy²/(-3/2)xy)

(9 × (-2/3)x) – (6 × (-2/3)) + (12 × (-2/3))y

 -6x -8y +4

问题4.将3x ² y ² + 2x ² y + 15xy除以3xy

解决方案：

We have to divide 3x²y² + 2x²y + 15xy by 3xy

So, (x²y² + 2x²y + 15xy) / 3xy

(3x²y²/3xy) + (2x²y/3xy) + (15xy/3xy)

x²y + (2/3)x +5

问题5：将x ² + 7x + 12除以x + 4

解决方案：

We have to divide x² + 7x + 12 by x + 4

So by using long division method we get

Quotient = x + 3

Remainder = 0

问题6.用4y ² + 3y + 1/2除以2y + 1

解决方案：

We have to divide 4y² + 3y + 12 by 2y + 1

So by using long division method we get

Quotient = 2y + 1/2

Remainder = 0

问题7.将3x ³ + 4x ² + 5x + 18除以x + 2

解决方案：

We have to divide 3x³ + 4x² + 5x + 18 by x + 2

 -2x² – 4x + 18

So by using long division method we get

Quotient = 3x² – 2x + 9

Remainder = 0

问题8.将14x ² – 53x + 45除以7x – 9

解决方案：

We have to divide 14x² – 53x + 45 by 7x – 9

So by using long division method we get

Quotient = 2x – 5

Remainder = 0

问题9.将-21 + 71x – 31x ² – 24x ³除以3 – 8x

解决方案：

We have to divide -21 + 71x – 31x² – 24x³ by 3 – 8x

So by using long division method we get

Quotient = 3x² + 5x – 7

Remainder = 0

问题10.将3y ⁴ – 3y ³ – 4y ² – 4y除以y ² – 2y

解决方案：

We have to divide 3y⁴ – 3y³ – 4y² – 4y by y² – 2y

2y² – 4y

So by using long division method we get

Quotient = 3x² + 3y + 2 

Remainder = 0

问题11.将2y ⁵ + 10y ⁴ + 6y ³ + y ² + 5y + 3除以2y ³ +1

解决方案：

We have to divide 2y⁵ + 10y⁴ + 6y³ + y² + 5y + 3 by 2y³ + 1

So by using long division method we get

Quotient = y² + 5y + 3 

Remainder = 0 

问题12.将x ⁴ – 2x ³ + 2x ² + x + 4除以x ² + x + 1。

解决方案：

We have to divide x⁴ – 2x³ + 2x² + x + 4 by x² + x + 1

4x² + 4x + 4

So by using long division method we get

Quotient = x²– 3x + 4

Remainder = 0 

问题13.将m ³ – 14m ² + 37m – 26除以m ² – 12m + 13。

解决方案：

We have to divide m³ – 14m² + 37m – 26 by m² – 12m + 13

-2m² + 24m – 26

So by using long division method we get

Quotient = m – 2

Remainder = 0

问题14.将x ⁴ + x ² +1除以x ² + x +1

解决方案：

We have to divide x⁴ + x² + 1 by x² + x + 1

So by using long division method we get

Quotient = x² – x + 1

Remainder = 0

问题15：将x ⁵ + x ⁴ + x ³ + x ² + x +1除以x ³ +1

解决方案：

We have to divide x⁵ + x⁴ + x³ + x² + x + 1 by x³ + 1

x³ + 1 

So by using long division method we get

Quotient = x² + x + 1

Remainder = 0

问题16.将14x ³ – 5x ² + 9x – 1除以2x – 1

解决方案：

We have to divide 14x³ – 5x² + 9x – 1 by 2x – 1

So by using long division method we get

Quotient = 7x² + x + 5

Remainder = 4

问题17.将6x ³ – x ² – 10x – 3除以2x – 3

解决方案：

We have to divide 6x³ – x² – 10x – 3 by 2x – 3

So by using long division method we get

Quotient = 3x² + 4x + 1

Remainder = 0

问题18：将6x ³ + 11x ² – 39x – 65除以3x ² + 13x + 13

解决方案：

We have to divide 6x³ + 11x² – 39x – 65 by 3x² + 13x + 13

So by using long division method we get

Quotient = 2x – 5

Remainder = 0