9类RD Sharma解决方案–第六章多项式的因式分解-练习6.5 |套装2

问题11：使用因子定理，对多项式进行因子分解：x ³ – 10x ² – 53x – 42

解决方案：

Given that,

f(x) = x³–10x² – 53x – 42

The constant in f(x) is – 42,

The factors of – 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, x + 1 = 0

x = – 1

f(-1) = (−1)³ –10(−1)² – 53(−1) – 42

-1 – 10 + 53 – 42 = 0

therefore, (x + 1) is the factor of f(x)

Now, divide f(x) with (x + 1) to get other factors

By using long division method we get,

x³ – 10x² – 53x – 42 = (x + 1) (x2 – 11x – 42)

Now,

x² – 11x – 42 = x² – 14x + 3x – 42

x(x – 14) + 3(x – 14)

(x + 3)(x – 14)

Hence, x³ – 10x² – 53x – 42 = (x + 1) (x + 3)(x – 14)

问题12.使用因子定理，对多项式进行因子分解：y ³ – 2y ² – 29y – 42

解决方案：

Given that, f(x) = y³ – 2y² – 29y – 42

The constant in f(x) is – 42,

The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, y + 2 = 0

y = – 2

f(-2) =  (−2)³ – 2(−2)²–29(−2) – 42

-8 -8 + 58 – 42 = 0

therefore, (y + 2) is the factor of f(y)

Now, divide f(y) with (y + 2) to get other factors

By using long division method we get,

y³ – 2y² – 29y – 42 = (y + 2) (y2 – 4y – 21)

Now,

y² – 4y – 21 = y² – 7y + 3y – 21

y(y – 7) +3(y – 7)

(y – 7)(y + 3)

Hence, y³ – 2y² – 29y – 42 = (y + 2) (y – 7)(y + 3)

问题13：使用因子定理对多项式进行因子分解：2y ³ – 5y ² – 19y + 42

解决方案：

Given that, f(x) = 2y³ – 5y² – 19y + 42

The constant in f(x) is + 42,

The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, y – 2 = 0

y = 2

f(2) = 2(2)³ – 5(2)² – 19(2) + 42

16 – 20 – 38 + 42 = 0

therefore, (y – 2) is the factor of f(y)

Now, divide f(y) with (y – 2) to get other factors

By using long division method we get,

2y³ – 5y² – 19y + 42 = (y – 2) (2y² – y – 21)

Now,

2y² – y – 21

The factors are (y + 3) (2y – 7)

Hence, 2y³ – 5y² -19y + 42 = (y – 2) (y + 3) (2y – 7)

问题14.使用因子定理，对多项式进行因子分解：x ³ + 13x ² + 32x + 20

解决方案：

Given that, f(x) = x³ + 13x² + 32x + 20

The constant in f(x) is 20,

The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20,

Let’s assume, x + 1 = 0

x = -1

f(-1) =  (−1)³+13(−1)² + 32(−1) + 20

-1 + 13 – 32 + 20 = 0

therefore, (x + 1) is the factor of f(x)

Divide f(x) with (x + 1) to get other factors

By using long division method we get,

x³ + 13x² +32x + 20 = (x + 1)( x2 + 12x + 20)

Now,

x² + 12x + 20 = x² + 10x + 2x + 20

x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, x³ + 13x² + 32x + 20 = (x + 1)(x + 10)(x + 2)

问题15：使用因子定理对多项式进行因子分解：x ³ – 3x ² – 9x – 5

解决方案：

Given that, f(x) = x³ – 3x² – 9x – 5

The constant in f(x) is -5,

The factors of -5 are ±1, ±5,

Let’s assume, x + 1 = 0

x = -1

f(-1) = (−1)³ – 3(−1)² – 9(-1) – 5

-1 – 3 + 9 – 5 = 0

therefore, (x + 1) is the factor of f(x)

Divide f(x) with (x + 1) to get other factors

By using long division method we get,

x³ – 3x² – 9x – 5 = (x + 1)( x² – 4x – 5)

Now,

x² – 4x – 5 = x² – 5x + x – 5

x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x + 1)

问题16：使用因子定理对多项式进行因子分解：2y ³ + y ² – 2y – 1

解决方案：

Given that, f(y) = 2y³ + y² – 2y – 1

The constant term is 2,

The factors of 2 are ± 1, ± 1/2,

Let’s assume, y – 1= 0

y = 1

f(1) = 2(1)³ +(1)² – 2(1) – 1

2 + 1 – 2 – 1 = 0

therefore, (y – 1) is the factor of f(y)

Divide f(y) with (y – 1) to get other factors

By using long division method we get,

2y³ + y² – 2y – 1 = (y – 1) (2y² + 3y + 1)

Now,

2y² + 3y + 1 = 2y² + 2y + y + 1

2y(y + 1) + 1(y + 1)

(2y + 1) (y + 1) are the factors

Hence, 2y³ + y² – 2y – 1 = (y – 1) (2y + 1) (y + 1)

问题17：使用因子定理对多项式进行因子分解：x ³ – 2x ² – x + 2

解决方案：

Given that, f(x) = x³ – 2x² – x + 2

The constant term is 2,

The factors of 2 are ±1, ± 1/2,

Let’s assume, x – 1= 0

x = 1

f(1) = (1)³ – 2(1)² – (1) + 2

1 – 2 – 1 + 2 = 0

therefore, (x – 1) is the factor of f(x)

Divide f(x) with (x – 1) to get other factors

By using long division method we get,

x³ – 2x² – y + 2 = (x – 1) (x² – x – 2)

Now,

x² – x – 2 = x² – 2x + x  – 2

x(x – 2) + 1(x – 2)

(x – 2)(x + 1)  are the factors

Hence, x³ – 2x² – y + 2 = (x – 1)(x + 1)(x – 2)

问题18：分解以下多项式：

1. x ³ + 13x ² + 31x – 45，因为x + 9是一个因子

2. 4x ³ + 20x ² + 33x + 18，因为2x + 3是一个因素

解决方案：

1. x³ + 13x² + 31x – 45

Given that, x + 9 is a factor

Let’s assume, f(x) = x³ + 13x² + 31x – 45

divide f(x) with (x + 9) to get other factors

By using long division method we get,

x³ + 13x² + 31x – 45 = (x + 9)( x2 + 4x – 5)

Now,

x² + 4x – 5 = x² + 5x – x  – 5

x(x + 5) -1(x + 5)

(x + 5) (x – 1) are the factors

Hence, x³ + 13x² + 31x – 45 = (x + 9)(x + 5)(x – 1)

2. 4x³ + 20x² + 33x + 18 

Given that, 2x + 3 is a factor

let’s assume, f(x) =  4x³ + 20x² + 33x + 18

divide f(x) with (2x + 3) to get other factors

By using long division method we get,

4x³ + 20x² + 33x + 18 = (2x + 3) (2×2 + 7x + 6)

Now,

2x² + 7x + 6 = 2x² + 4x + 3x + 6

2x(x + 2) + 3(x + 2)

(2x + 3)(x + 2) are the factors

Hence, 4x³ + 20x² + 33x + 18 = (2x + 3)(2x + 3)(x + 2)