第8类RD Sharma解决方案–第8章代数表达式的除法–练习8.5

问题1：在以下每个项中，将第一个多项式除以第二个多项式。另外，写出商和余数：

(i)3x ² + 4x + 5，x – 2

解决方案：

3x² + 4x + 5, x – 2

By using factorization method,

$⇒\:\frac{\left(3x^2+4x+5\right)}{x-2} = \:\frac{3x\left(x-2)+10(x-2\right)+25}{x-2}$                 

$=\:\frac{\left(x-2)(3x+10\right)+25}{x-2}$                                                     (Taking common (x-2) factor)

$=(3x+10)+\:\frac{25}{x-2}$

∴ the Quotient is 3x + 10 and the Remainder is 25.

(ii)10x ² – 7x + 8，5x – 3

解决方案：

10x² – 7x + 8, 5x – 3

By using factorization method,

$⇒\:\frac{\left(10x^2-7x+8\right)}{5x-3} = \:\frac{\left(2x(5x-3)-\:\frac{1}{5}(5x-3)+\:\frac{37}{5}\right)}{5x-3}$              

$=\:\frac{\left((5x-3)(2x-\:\frac{1}{5})+\:\frac{37}{5}\right)}{5x-3}$                                              (Taking common (5x-3) factor) 

$=(2x-\:\frac{1}{5})+(\:\frac{\:\frac{37}{5}}{5x-3})$

∴ the Quotient is (2x – 1/5) and the Remainder is 37/5.

(iii) ⁵年^{3 – 6年2} + 6年– 1年，5年– 1年

解决方案：

5y³ – 6y² + 6y – 1, 5y – 1

By using factorization method,

$⇒\:\frac{\left(5y^3-6y^2+6y-1\right)}{5y-1}= \:\frac{y^2\left(5y-1)-y(5y-1)+1(5y-1\right)}{5y-1}$             

$=\:\frac{(5y-1)(y^2-y+1)}{5y-1}$                                                    (Taking common (5y-1) factor)

$=(y^2-y+1)$

∴ the Quotient is (y² – y + 1) and the Remainder is 0.

(iv)x ⁴ – x ³ + 5x，x – 1

解决方案：

x⁴ – x³ + 5x, x – 1

By using factorization method,

$⇒\:\frac{\left(x^4-x^3+5x\right)}{x-1} = \:\frac{x^3\left(x-1)+5(x-1\right)+5}{x-1}$                             

$=\:\frac{\left(x-1\right)(x^3+5)+5}{x-1}$                                                       (Taking common (x-1) factor)

$=(x^3+5)+\:\frac{5}{x-1}$

∴ the Quotient is x³ + 5 and the Remainder is 5.

(v)y ⁴ + y ² ，y ² – 2

解决方案：

y⁴ + y², y² – 2

By using factorization method,

$⇒\:\frac{\left(y^4+y^2\right)}{y^2-2}=\:\frac{y^2(y^2-2)+3(y^2-2)+6}{y^2-2}$                               

$=\:\frac{(y^2-2)(y^2+3)+6}{y^2-2}$                                                          (Taking common (y²-2) factor)

$=(y^2+3)+\:\frac{6}{y^2-2}$

∴ the Quotient is y² + 3 and the Remainder is 6.

问题2：找出第一个多项式是否是第二个多项式的因数：

(i)x + 1、2x ² + 5x + 4

解决方案：

x + 1, 2x² + 5x + 4

Let us perform factorization method,

$⇒\:\frac{\left(2x^2+5x+4\right)}{x+1}= \:\frac{2x\left(x+1)+3(x+1\right)+1}{x+1}$               

$=\:\frac{(x+1)(2x+3)+1}{x+1}$                                                                   (Taking common (x+1) factor)

$=(2x+3)+\:\frac{1}{x+1}$

Since remainder is 1, therefore the first polynomial is not a factor of the second polynomial.

(ii)y – 2，3y ³ + 5y ² + 5y + 2

解决方案：

y – 2, 3y³ + 5y² + 5y + 2

Let us perform factorization method,

$⇒\:\frac{(3y^3+5y^2+5y+2)}{y-2}=\:\frac{3y^2(y-2)+11y(y-2)+27(y-2)+56}{y-2}$       

$=\:\frac{\left((y-2)(3y^2+11y+27)+56\right)}{y-2}$                                                        (Taking common (y-2) factor)

$=(3y^2+11y+27)+\:\frac{56}{y-2}$

Since remainder is 56 therefore the first polynomial is not a factor of the second polynomial.

(iii)4x ² – 5，4x ⁴ + 7x ² + 15

解决方案：

4x² – 5, 4x⁴ + 7x² + 15

Let us perform factorization method,

$⇒\:\frac{\left(4x^4+7x^2+15\right)}{4x^2-5}=\:\frac{x^2(4x^2-5)+3(4x^2-5)+30}{4x^2-5}$        

$=\:\frac{(4x^2-5)(x^2+3)+30}{4x^2-5}$                                                                      (Taking common (4x²-5) factor)

$=(x^2+3)+\:\frac{30}{4x^2-5}$

Since remainder is 30 therefore the first polynomial is not a factor of the second polynomial.

(iv)4 – z，3z ² – 13z + 4

解决方案：

4 – z, 3z² – 13z + 4

Let us perform factorization method,

$⇒\:\frac{\left(3z^2-13z+4\right)}{4-z} = \:\frac{\left(3z^2-12z-z+4\right)}{4-z}$

$=\:\frac{3z\left(z-4)-1(z-4\right)}{4-z}$                                                                     (Taking common (z-4) factor)

$=\:\frac{\left(z-4)(3z-1\right)}{4-z}$

$=\:\frac{\left(4-z)(1-3z\right)}{4-z}$

$=1-3z$

Since remainder is 0 therefore the first polynomial is a factor of the second polynomial.

(v)2a – 3，10a ² – 9a – 5

解决方案：

2a – 3, 10a² – 9a – 5

Let us perform factorization method,

$⇒\:\frac{\left(10a^2-9a-5\right)}{2a-3}=\:\frac{5a\left(2a-3)+3(2a-3\right)+4}{2a-3}$

$=\:\frac{\left(2a-3)(5a+3\right)+4}{2a-3}$                                                                    (Taking common (2a-3) common)

$=(5a+3)+\:\frac{4}{2a-3}$

Since remainder is 4 therefore the first polynomial is not a factor of the second polynomial.

(vi)4年+1、8年² – 2年+1

解决方案：

4y + 1, 8y² – 2y + 1

Let us perform factorization method,

$⇒\:\frac{\left(8y^2-2y+1\right)}{4y+1}=\:\frac{2y\left(4y+1)-1(4y+1\right)+2}{4y+1}$

$=\:\frac{\left(4y+1)(2y-1\right)+2}{4y+1}$                                                                   (Taking common (4y+1) factor)

$=(2y-1)+\:\frac{2}{4y+1}$   

Since remainder is 2 therefore the first polynomial is not a factor of the second polynomial.