9类RD Sharma解决方案–第六章多项式的因式分解-练习6.4 |套装1

在下面的每一个中，使用因子定理确定多项式g(x)是否是多项式f(x)的因子(问题1-7)：

问题1. f(x)= x ³ – 6x ² + 11x – 6，g(x)= x – 3

解决方案：

Given: f(x) = x³ – 6x² + 11x – 6, g(x) = x – 3

Here,

x – 3 = 0

x = 3 

To prove: g(x) is a factor of f(x), that is f(3) = 0 

On substituting the value of x in f(x), we get

 f(3) = 3³ – 6 × 3² + 11 × 3 – 6

= 27 – (6 × 9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题2。f(x)= 3x ⁴ + 17x ³ + 9x ² – 7x – 10，g(x)= x + 5

解决方案：

Given: f(x) = 3x⁴ + 17x³ + 9x² – 7x – 10, g(x) = x + 5

Here,

x + 5 = 0

x = -5

To prove: g(x) is a factor of f(x), that is f(-5) = 0

On substituting the value of x in f(x), we get

f(-5) = 3(-5)⁴+ 17(-5)³+ 9(-5)²– 7(-5) – 10

= 3 × 625 + 17 × (-125) + 9 × 25 + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 2135 – 2135

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题3. f(x)= x ⁵ + 3x ⁴ – x ³ – 3x ² + 5x + 15，g(x)= x + 3

解决方案：

Given: f(x) = x⁵ + 3x⁴ – x³ – 3x² + 5x + 15, g(x) = x + 3

Here,

x + 3 = 0 

x = -3

To prove: g(x) is a factor of f(x), that is f(-3) = 0

On substituting the value of x in f(x), we get

f(-3) = (-3)⁵ + 3(-3)⁴ – (-3)³ – 3(-3)² + 5(-3) + 15

= -243 + 243 + 27 – 27 – 15 + 15

= 0

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题4. f(x)= x ³ – 6x ² – 19x + 84，g(x)= x – 7

解决方案：

Given: f(x) = x³ – 6x² – 19x + 84, g(x) = x – 7

Here,

x – 7 = 0  

x = 7

To prove: g(x) is a factor of f(x), that is f(7) = 0

On substituting the value of x in f(x), we get

f(7) = 7³ – 6 × 7² – 19 × 7 + 84     

= 343 – (6 × 49) – (19 × 7) + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题5. f(x)= 3x ³ + x ² – 20x + 12，g(x)= 3x – 2

解决方案：

Given: f(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2

Here,

3x – 2 = 0  

x = 2/3

To prove: g(x) is a factor of f(x), that is f(2/3) = 0

On substituting the value of x in f(x), we get

$f(\frac23)=3(\frac23)^3+(\frac23)^2-20(\frac23)+12$

$= 3(\frac8{27})+\frac49-\frac{40}3+12$

$=\frac89+\frac49-\frac{40}3+12$

$= \frac{12}9-\frac{40}3+12$

Taking L.C.M 

 = $\frac{12-120+108}{9}$

 $=\frac{120-120}{9}$

 = 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题6. f(x)= 2x ³ – 9x ² + x + 13，g(x)= 3 – 2x

解决方案：

Given: f(x) = 2x³ – 9x² + x + 13, g(x) = 3 – 2x

Here,

3 – 2x = 0  

x = 3/2

To prove: g(x) is a factor of f(x), that is f(3/2) = 0

On substituting the value of x in f(x), we get

$f(\frac32)=2(\frac32)^3-9(\frac32)^2+(\frac32)+13$

$=2(\frac{27}8)-9(\frac94)+\frac34+13$

$=\frac{27}4-\frac{81}4 +\frac{3}{2}+12$

$=\frac{21-81+6+48}{4}$

 $= \frac{81-81}4$

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

问题7. f(x)= x ³ – 6x ² + 11x – 6，g(x)= x ² – 3x + 2

解决方案：

Given: f(x) = x³ – 6x² + 11x – 6, g(x) = x² – 3x + 2

Here,

x² – 3x + 2 = 0  

On factorizing the above, we get

⇒ x² – 3x + 2 = 0

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2)

⇒ (x – 1)(x – 2) are the factors 

To prove: g(x) is a factor of f(x), that is f(1) and f(2) should be 0

Let x = 1 

On substituting the value of x in f(x), we get

f(1) = 1³ – 6 × 1² + 11 × 1 – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0 

Let x = 2

On substituting the value of x in f(x), we get

f(x) = 2³ – 6 × 2² + 11 × 2 – 6 

= 8 – (6 × 4) + 22 – 6

= 8 – 24 +22 – 6

= 30 – 30

= 0

Since, the results are 0 g(x) is the factor of f(x)

问题8：证明(x – 2)，(x + 3)和(x – 4)是x ³ – 3x ² – 10x + 24的因数

解决方案：

Given:

 f(x) = x³ – 3x² – 10x + 24 

Factors given are (x – 2), (x + 3) and (x – 4)

To prove: g(x) is a factor of f(x), that is f(2), f(-3), f(4) should be 0

Here, x – 2 = 0

Let, x = 2

On substituting the value of x in f(x), we get

 f(2) = 2³ – 3 × 2² – 10 × 2 + 24 

= 8 – (3 × 4) – 20 + 24

= 8 – 12 – 20 + 24

= 32 – 32

= 0 

Here, x + 3 = 0

Let, x = -3

On substituting the value of x in f(x), we get

f(-3) = (-3)³ – 3 × (-3)² – 10 × (-3) + 24

= -27 -3(9) + 30 + 24

= -27- 27 + 30 + 24

= 54 – 54

= 0  

Here, x – 4 = 0

Let, x = 4

On substituting the value of x in f(x), we get

f(4) = (4)³ – 3 × (4)² – 10 × (4) + 24

= 64 – 3(16) – 40 + 24

= 64 – 48 – 40 + 24

= 84 – 84

= 0  

Since, the results are 0, g(x) is the factor of f(x)

问题9：证明(x + 4)，(x – 3)和(x – 7)是x ³ – 6x ² – 19x + 84的因数

解决方案：

Given:

f(x) = x³ – 6x²– 19x + 84

Factors given are (x + 4), (x – 3) and (x – 7)

To prove: g(x) is a factor of f(x), that is f(4), f(3), f(-7) should be 0

Here, x + 4 = 0

Let, x = -4

On substituting the value of x in f(x), we get

f(-4) = (-4)³ – 6(-4)² – 19(-4) + 84

= -64 – (6 × 16) – 19 × (-4) + 84

= -64 – 96 + 76 + 84

= 160 – 160

= 0

Here, x – 3 = 0

Let, x = 3

On substituting the value of x in f(x), we get

f(3) = (3)³ – 6(3)² – 19(3) + 84

= 27 – 6(9) – 19 × 3 + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0  

Here, x – 7 = 0

Let, x = 7

On substituting the value of x in f(x), we get

f(7) = (7)³ – 6(7)² – 19(7) + 84

= 343 – 6(49) – 19 × 7 + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0  

Since, the results are 0, g(x) is the factor of f(x)

问题10. a的值是(x – 5)是x ³ – 3x ² + ax – 10

解决方案：

Here, f(x) = x³ – 3x² + ax – 10

By factor theorem 

If (x – 5) is the factor of f(x) then, f(5) = 0 

⇒ x – 5 = 0

⇒ x = 5

On substituting the value of x in f(x), we get

f(5) = 5³ – 3 × 5² + a × 5 – 10

= 125 – (3 × 25) + 5a -10

= 125 – 75 + 5a – 10

= 5a + 40

Equate f(5) to zero

f(5) = 0 

⇒ 5a + 40 = 0

⇒ 5a = -40

⇒ a = -40/5

⇒ a = -8

When a = -8, then (x – 5) will be factor of f(x)

问题11。求出这样的值，即(x – 4)是5x ³ – 7x ² – ax – 28的因数

解决方案：

Given: f(x) = 5x³– 7x²– ax – 28

Using factor theorem

(x – 4) is the factor of f(x), then f(4) = 0 

⇒ x – 4 = 0

⇒ x = 4

On substituting the value of x in f(x), we get

f(4) = 5(4)³ – 7(4)² – a × 4 – 28

= 5(64) – 7(16) – 4a – 28

= 320 – 112 – 4a – 28

= 180 – 4a

Equate f(4) to zero, to find a

f(4) = 0

⇒ 180 – 4a = 0 

⇒ a = 180/4

⇒ a = 45     

When a = 45, then (x – 4) will be factor of f(x)

问题12。如果(x + 2)是4x ⁴ + 2x ³ – 3x ² + 8x + 5a的因数，请找到a的值

解决方案：

Given: f(x) = 4x⁴+ 2x³– 3x²+ 8x + 5a 

Using factor theorem 

If (x + 2) is the factor of f(x), then f(-2) should be zero

⇒ x + 2 =0 

⇒ x = -2

On substituting the value of x in f(x), we get

f(-2) = 4(-2)⁴+ 2(-2)³– 3(-2)²+ 8(-2) + 5a 

= 4(16) + 2 (-8) – 3(4) – 16 + 5a

= 64 – 16 – 12 -16 + 5a

= 5a + 20

Equate f(-2) to 0 

f(-2) = 4(-2)²+ 2(-2)³– 3(-2)²+ 8(-2) + 5a 

= 4(16) + 2(-8) – 3(4) – 16 + 5a

= 64 – 16 – 12 – 16 + 5a

= 5a + 20

Equate f(-2) to 0

f(-2) = 0

⇒ 5a + 20 = 0

⇒ 5a = – 20

⇒ a = -4

When a = -4, then (x + 2) is the factor of f(x)