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Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.2
  • Last Updated : 21 Feb, 2021

Question 1. If f(x) = 2x3-13x2+17x+12, find

1. f(2)

2. f(-3)

3. f(0)

Solution:

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

Question 2. Verify whether the indicated numbers are zero of the polynomial corresponding to them in the following cases:

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

(2) f(x) = x2-1, x=(1,-1)

(3) g(x) = 3x2-2, x=(\frac{2}{\sqrt3},\frac{2}{\sqrt3} )

(4) p(x)=x3-6x2+11x-6, x=1,2,3

(5) f(x)=5x-π, x=4/5

(6) f(x) = x2, x=0

(7) f(x)=lx+m, x=-m/l

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

Solution:

(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})

substitutex = \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, substitutex = -\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

Question 3. If x=2 is a root of the polynomial f(x)=2x2-3x+7a, find the value of a.

Solution:

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

Question 4. If x=-1/2 is zero of the polynomial p(x)=8x3-ax2-x+2, find the value of a.

Solution:

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

Question 5. If x=0 and x=-1 are the roots of the polynomial f(x)=2x3-3x2+ax+b, find the value of a and b.

Solution:

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.

Question 6. Find the integral roots of the polynomial f(x)=x3+6x2+11x+6

Solution:

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.

Question 7. Find the rational roots of the polynomial f(x)=2x3+x2-7x-6

Solution:

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

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