问题1.使用因子定理,对多项式进行因子分解:x 3 + 6x 2 + 11x + 6
解决方案:
Given that, polynomial eqn., f(x) = x3 + 6x2 + 11x + 6
The constant term in f(x) is 6,
The factors of 6 are ± 1, ± 2, ± 3, ± 6
Let, x + 1 = 0
x = -1
Substitute the value of x in f(x) and we get,
f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6
= – 1 + 6 – 11 + 6 = 12 – 12 = 0
So, (x + 1) is the factor of f(x)
Similarly, (x + 2) and (x + 3) are also the factors of f(x)
Since, f(x) is a polynomial having a degree 3, therefore it cannot have more than three linear factors.
Hence, f(x) = k(x + 1)(x + 2)(x + 3)
x3 + 6x2 + 11x + 6 = k(x + 1)(x + 2)(x + 3)
Substitute x = 0 on both the sides
0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)
6 = k(1*2*3)
6 = 6k
k = 1
Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)
f(x) = (1)(x + 1)(x + 2)(x + 3)
f(x) = (x + 1)(x + 2)(x + 3)
Hence, x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3)
问题2.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 in f(x) is -2,
The factors of (-2) are ±1, ± 2,
Let, x – 1 = 0
x = 1
Substitute the value of x in f(x)
f(1) = (1)3 + 2(1)2 – 1 – 2
1 + 2 – 1 – 2 = 0
Similarly, the other factors (x + 1) and (x + 2) of f(x)
Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.
therefore, f(x) = k(x – 1)(x + 2)(x + 1 )
x3 + 2x2 – x – 2 = k(x – 1)(x + 2)(x + 1 )
Substitute x = 0 on both the sides
0 + 0 – 0 – 2 = k(-1)(1)(2)
– 2 = – 2k
k = 1
Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1)
f(x) = (1)(x – 1)(x + 2)(x + 1)
f(x) = (x – 1)(x + 2)(x + 1)
therefore, x3 + 2x2 – x – 2 = (x – 1)(x + 2)(x + 1)
问题3.使用因子定理,对多项式进行因子分解:x 3 – 6x 2 + 3x + 10
解决方案:
Given that, f(x) = x3 – 6x2 + 3x + 10
The constant term in f(x) is 10,
The factors of 10 are ± 1, ± 2, ± 5, ± 10,
Let, x + 1 = 0
x = -1
Substitute the value of x in f(x)
f(-1) = (−1)3– 6(−1)2 + 3(−1) + 10
-1 – 6 – 3 + 10 = 0
Similarly, the other factors (x – 2) and (x – 5) of f(x)
Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.
therefore, f(x) = k(x + 1)(x – 2)(x – 5)
Substitute x = 0 on both sides
x3– 6x2 + 3x + 10 = k(x + 1)(x – 2)(x – 5)
0 – 0 + 0 + 10 = k(1)(-2)(-5)
10 = k(10)
k = 1
Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)
f(x) = (1)(x + 1)(x – 2)(x – 5 )
therefore, x3 – 6x2 + 3x + 10 = (x + 1)(x – 2)(x – 5)
问题4.使用因子定理,对多项式进行因子分解:x 4 –7x 3 + 9x 2 + 7x –10
解决方案:
Given that, f(x) = x4–7x3 + 9x2 + 7x – 10
The constant term in f(x) is 10,
The factors of 10 are ± 1, ± 2, ± 5, ±10,
Let, x – 1 = 0
x = 1
Substitute the value of x in f(x)
f(x) = 14 – 7(1)3 + 9(1)2 + 7(1) – 10
1 – 7 + 9 + 7 – 10
10 – 10 = 0
(x – 1) is the factor of f(x)
Similarly, the other factors are (x + 1), (x – 2), (x – 5)
Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor.
therefore, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)
x4 –7x3 + 9x2 + 7x – 10 = k(x – 1)(x + 1)(x – 2)(x – 5)
Put x = 0 on both sides
0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5)
– 10 = k(-10)
k = 1
Substitute k = 1 in f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)
f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5)
(x – 1)(x + 1)(x – 2)(x – 5)
therefore, x4 – 7x3 + 9x2 + 7x – 10 = (x – 1)(x + 1)(x – 2)(x – 5)
问题5.使用因子定理,对多项式进行因子分解:x 4 – 2x 3 – 7x 2 + 8x + 12
解决方案:
Given that,
f(x) = x4 – 2x3 –7x2 + 8x + 12
The constant term f(x) is equal is 12,
The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12,
Let, x + 1 = 0
x = -1
Substitute the value of x in f(x)
f(-1) = (−1)4 – 2(−1)3–7(−1)2 + 8(−1)+12
1 + 2 – 7 – 8 + 12 = 0
therefore, x + 1 is factor of f(x)
Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)
Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.
f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)
x4 – 2x3 –7x2 + 8x + 12 = k(x + 1)(x + 2)(x – 3)(x – 2)
Substitute x = 0 on both sides,
0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3)
12 = 12K
k = 1
Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)
f(x) = (x – 2)(x + 1)(x + 2)(x – 3)
Hence, x4 – 2x3 – 7x2 + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3)
问题6.使用因子定理,对多项式进行因子分解:x 4 + 10x 3 + 35x 2 + 50x + 24
解决方案:
Given that, f(x) = x4 + 10x3 + 35x2 + 50x + 24
The constant term in f(x) is equal to 24,
The factors of 24 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24,
Let, x + 1 = 0
x = -1
Substitute the value of x in f(x)
f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24
1-10 + 35 – 50 + 24 = 0
(x + 1) is the factor of f(x)
Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x)
Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.
f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)
x4 + 10x3 + 35x2 + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4)
Substitute x = 0 on both sides
0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)
24 = k(24)
k = 1
Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)
f(x) = (1)(x + 1)(x + 2)(x + 3)(x + 4)
f(x) = (x + 1)(x + 2)(x + 3)(x + 4)
Hence, x4 + 10x3 + 35x2 + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)
问题7.使用因子定理,对多项式进行因子分解:2x 4 –7x 3 –13x 2 + 63x – 45
解决方案:
Given that, f(x) = 2x4–7x3–13x2 + 63x – 45
The factors of constant term – 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45,
The factors of the coefficient of x4 is 2.
Hence possible rational roots of f(x) are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2
Let, x – 1 = 0
x = 1
f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45
2 – 7 – 13 + 63 – 45 = 0
Let, x – 3 = 0
x = 3
f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3) – 45
162 – 189 – 117 + 189 – 45 = 0
therefore, (x – 1) and (x – 3) are the roots of f(x)
x2 – 4x + 3 is the factor of f(x)
Divide f(x) with x2 – 4x + 3 to get other three factors,
By using long division we get,
2x4 – 7x3 – 13x2 + 63x – 45 = (x2– 4x + 3) (2x2+ x – 15)
2x4 – 7x3– 13x2 + 63x – 45 = (x – 1) (x – 3) (2x2+ x – 15)
Now,
2x2 + x – 15 = 2x2 + 6x – 5x –15
2x(x + 3) – 5 (x + 3)
(2x – 5) (x + 3)
Hence, 2x4 – 7x3 – 13x2 + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5)
问题8.使用因子定理,对多项式进行因子分解:3x 3 – x 2 – 3x + 1
解决方案:
Given that, f(x) = 3x3 – x2 – 3x + 1
The factors of constant term 1 is ± 1,
The factors of the coefficient of x2 = 3,
The possible rational roots are ±1, 1/3,
Let, x – 1 = 0
x = 1
f(1) = 3(1)3 – (1)2 – 3(1) + 1
3 – 1 – 3 + 1 = 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,
3x3– x2 – 3x + 1 = (x – 1)( 3x2 + 2x – 1)
Now,
3x2 + 2x -1 = 3x2 + 3x – x – 1
3x(x + 1) -1(x + 1)
(3x – 1)(x + 1)
Hence, 3x3– x2– 3x + 1 = (x – 1) (3x – 1)(x + 1)
问题9.使用因子定理,对多项式进行因子分解:x 3 – 23x 2 + 142x – 120
解决方案:
Given that, f(x) = x3– 23x2 + 142x – 120
The constant term in f(x) is -120,
The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120,
Let, x – 1 = 0
x = 1
f(1) = (1)3– 23(1)2 + 142(1) – 120
1 – 23 + 142 – 120 = 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 we get,
x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Now,
x2 – 22x + 120 = x2 – 10x – 12x + 120
x(x – 10) – 12(x – 10)
(x – 10) (x – 12)
Hence, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12)
问题10.使用因子定理,对多项式进行因子分解:y 3 – 7y + 6
解决方案:
Given that, f(y) = y3 – 7y + 6
The constant term in f(y) is 6,
The factors are ± 1, ± 2, ± 3, ± 6,
Let, y – 1 = 0
y = 1
f(1) = (1)3 – 7(1) + 6
1 – 7 + 6 = 0
therefore, (y – 1) is the factor of f(y)
Similarly, (y – 2) and (y + 3) are also the factors
Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors
f(y) = k(y – 1)(y – 2)(y + 3)
y3 – 7y + 6 = k(y – 1)( y – 2)(y + 3) —————–(i)
Substitute k = 0 in eqn. 1
0 – 0 + 6 = k(-1)(-2)(3)
6 = 6k
k = 1
y3 – 7y + 6 = (1)(y – 1)( y – 2)(y + 3)
y3 – 7y + 6 = (y – 1)( y – 2)(y + 3)
Hence, y3–7y + 6 = (y – 1)( y – 2)(y + 3)