9类RD Sharma解决方案–第六章多项式的因式分解-练习6.2

问题1.如果f(x)= 2x ³ -13x ² + 17x + 12，则找到

1. f(2)

2. f(-3)

3. f(0)

解决方案：

Given: f(x)=2x³-13x²+17x+12

1. f(2)

We need to substitute the ‘2’ in f(x)

f(2)=2(2)³-13(2)²+17(2)+12

= 2(8) – 13(4)+17(2)+12

= 16 – 52 +34+12

= 10

Therefore, f(2)=10

2. f(-3)

We need to substitute the ‘-3’ in f(x)

f(-3)=2(-3)³-13(-3)²+17(-3)+12

=2*(-27) – 13(9) – 17(3) + 12

= -54 -117-51+12

= -210

Therefore, f(-3) = -210

3. f(0)

We need to substitute the ‘0’ in f(x)

f(0)=2(0)³-13(0)²+17(0)+12

= 0+0+0+12

= 12

Therefore, f(0) = 12

问题2.在以下情况下，验证所指示的数字是否为对应于它们的多项式的零：

(1)f(x)= 3x + 1，x = -1/3

(2)f(x)= x ² -1，x =(1，-1)

(3)g(x)= 3x ² -2，x =( $\frac{2}{\sqrt3},\frac{2}{\sqrt3}$ )

(4)p(x)= x ³ -6x ² + 11x-6，x = 1,2,3

(5)f(x)=5x-π，x = 4/5

(6)f(x)= x ² ，x = 0

(7)f(x)= lx + m，x = -m / l

(8)f(x)= 2x + 1，x = 1/2

解决方案：

(1) f(x) = 3x+1, x=-1/3

We know that,

f(x) = 3x+1

Substitute the value of x = -1/3 in f(x)

f(-1/3) = 3(-1/3)+1

= -1+1

= 0

Since, the result is 0 x = -1/3 is the root of 3x+1

(2) f(x) = x² -1, x = (1,-1)

We know that,

f(x) = x² – 1

Given that x = (1,-1)

Substitute x=1 in f(x)

f(1) = 1² – 1

= 1-1

= 0

Now, substitute x = (-1) in f(x)

f(-1) = (-1)² – 1

= 1 – 1

= 0

Since, the results when x=(1,-1) are 0 they are the roots of the polynomial f(x) = x² – 1

(3) g(x) = 3x² – 2

Given that, x=($\frac{2}{\sqrt3}, -\frac{2}{\sqrt3}$)

substitute$x = \frac{2}{\sqrt3}$in g(x)

g($\frac{2}{\sqrt3}$) = 3($\frac{2}{\sqrt3}$)² – 2

= 3(4/3) – 2

= 4 – 2

= 2

= 2 ≠ 0

Now, substitute$x = -\frac{2}{\sqrt3}$in g(x)

g($-\frac{2}{\sqrt3}$) = 3($-\frac{2}{\sqrt3}$)² – 2

= 3(4/3) – 2

= 4 -2

= 2

= 2 ≠ 0

Since, the results when x = ($\frac{2}{\sqrt3},- \frac{2}{\sqrt3}$) are not 0, they are roots of 3x²-2.

(4) p(x) = x³-6x²+11x-6, x =1,2,3

Given that the values of x are 1,2,3

Substitute x = 1 in p(x)

p(1) = 1³-6(1)²+11(1)-6

= 1 -6 +11 -6

= 0

Now, substitute x= 2 in p(x)

p(2) = 2³-6(2)²+11(2)-6

= 8 -6(4) +22 – 6

= 8 -24 +22 – 6

= 0

Now, substitute x= 3 in p(x)

p(3) = 3³-6(3)²+11(3)-6

= 27 – 6(9) +33-6

= 27-54+33-6

= 0

Since, the result is 0 for x=1,2,3 these are roots of x³-6x²+11x-6

(5) f(x) = 5x – π

Given that, x = 4/5

Substitute the value of x in f(x)

f(4/5) = 5(4/5) – π

= 4 – π

≠ 0

Since, the result is not equal to zero, x=4/5 is not the square root of the polynomial 5x – π

(6) f(x) = x²

Given that, x = 0

Substitute the value of x in f(x)

f(0) = (0)²

= 0

Since, the result is zero, x=0 is the root of x²

(7) f(x) = lx +m

Given, x = -m/l

Substitute the value of x in f(x)

f(-m/l)= l(-m/l) + m

= -m + m

= 0

Since, the result is 0, x = -m/l is the root of lx+m

(8) f(x) = 2x+1

Given, x = 1/2

Substitute the value of x in f(x)

f(1/2) = 2(1/2) +1

= 1 + 1

= 2

=2 ≠ 0

Since, the result is not equal to 0, x=1/2 is the root 2x+1

问题3.如果x = 2是多项式f(x)= 2x ² -3x + 7a的根，则找到a的值。

解决方案：

We know that, f(x) = 2x² – 3x +7a

Given that x = 2 is the root of f(x)

Substitute the value of x in f(x)

f(2) = 2(2)² -3(2)+7a

= 2(4) -6 +7a

= 8 – 6 + 7a

= 2 +7a

Now, equate 7a+2 to zero

⇒ 7a + 2 =0

⇒ a = -2/7

The value of a = -2/7

问题4.如果x = -1 / 2是多项式p(x)= 8x ³ -ax ² -x + 2的零，则找到a的值。

解决方案：

We know that, p(x)=8x³-ax²-x+2

Given, x=-1/2

Substitute the value of x in f(x)

p(-1/2) = 8(-1/2)³-a(-1/2)²-(-1/2)+2

= 8(-1/8) – a(1/4) +1/2+2

= -1 – (a/4) + 5/4

= 3/2 – a/4

To, find the value of a, equate p(-1/2) to zero

p(-1/2) = 0

3/2 – a/2 = 0

On taking L.C.M

(6-a)/4 = 0

6-a = 0

a = 6

问题5.如果x = 0和x = -1是多项式f(x)= 2x ³ -3x ² + ax + b的根，请找到a和b的值。

解决方案：

We know that, f(x) = 2x³ – 3x² +ax +b

Given x = 0, -1

Substitute the value of x = 0 in f(x)

f(0) = 2(0)³ – 3(0)² +a(0)+b

= b —————— 1

Substitute the value of x = -1 in f(x)

f(-1) = 2(-1)³-3(-1)²+a(-1)+b

= -2 -3 -a +b

= -5 -a +b —————– 2

We need to equate equations 1 and 2 to zero

b = 0 and -5-a+b=0

Substitute value of b in equation 2

⇒ -5-a+0 = 0

⇒ a = -5

The values of a and b are -5, 0 respectively.

问题6.求多项式f(x)= x ³ + 6x ² + 11x + 6的整数根

解决方案：

Given:

f(x) = x³+6x²+11x +6

We can say that, the polynomial f(x) with an integer coefficient and the highest degree term coefficient, which is known as leading factor is 1.

So, the roots of f(x) are limited to integer factor of 6, they are .$\pm1,\pm2,\pm3,\pm6$

Let, x=-1

Substitute the value of x in f(x)

f(-1)=(-1)³+6(-1)²+11(-1)+6

= -1 +6 -11 +6

= -12 +12

= 0

let, x = -2

f(-2)=(-2)³+6(-2)²+11(-2)+6

= 8 + 6*4 – 22+6

= 8 +24 – 22+6

= 0

let, x = -3

f(-3)=(-3)³+6(-3)²+11(-3)+6

= -27 + 6(9) – 33 +6

= -27 + 54 – 33 +6

= 0

From all the given factors, only -1,-2,-3 gives the results as zero.

问题7：找到多项式f(x)= 2x ³ + x ² -7x-6的有理根

解决方案：

Given: f(x) = 2x³+x²-7x-6

f(x) is a cubic polynomial with an integer coefficient. If the rational root in the form of p/q, the values of p are limited to factors of 6 which are$\pm1,\pm2\pm3,\pm6,\pm{\frac12},\pm{\frac32}$

Let, x = -1

f(-1)=2(-1)³+(-1)²-7(-1)-6

= -2 +1 +7 – 6

= -8 +8

= 0

Let, x = 2

f(2) = 2(2)³+(2)²-7(2)-6

= 2(8) + 4 -14 – 6

= 16 +4 – 14 – 6

= 20 – 20

= 0

Let, x = -3/2

f(-3/2) = 2(-3/2)³ +(-3/2)-7(-3/2) – 6

= 2(-27/8) -3/2 +21/4-6

= -27/4 -3/2 +21/4 – 6

= -6.75+2.25+10.5-6

= 12.75 – 12.75

= 0

From all the factors only -1, -2 and -3/2 gives the result as zero. So, the rational roots of 2x³+x²-7x-6 are -1,2 and -3/2