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Polynomials : Exercise 2.4 (Mathematics NCERT Class 9th)


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Q.1     Determine which of the following polynomials has (x + 1) a factor :
              (i) {x^3} + {x^2} + x + 1
              (ii) {x^4} + {x^3} + {x^2} + x + 1
              (iii) {x^4} + 3{x^3} + 3{x^2} + x + 1
              (iv) {x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2  
Sol.

(i) In order to prove that x + 1 is a factor of  p\left( x \right) = {x^3} + {x^2} + x + 1, it is sufficient to show that p\left( { - 1} \right) = 0
Now, p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1
 = - 1 + 1 - 1 + 1 = 0
Hence, (x + 1) is a factor of   p\left( x \right) = {x^3} + {x^2} + x + 1

(ii) In order to prove that (x + 1) is a factor of   p\left( x \right) = {x^4} + {x^3} + {x^2} + x + 1 , it is sufficient to show that p (–1) = 0 .
Now,   p\left( { - 1} \right) = {\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1
 = 1 - 1 + 1 - 1 + 1 = 1 \ne 0
Therefore (x + 1) is not a factor of    {x^4} + {x^3} + {x^2} + x + 1

(iii) In order to prove that (x + 1) is a factor of p\left(x\right) = {x^4} + 3{x^3} + 3{x^2} + x + 1, it is sufficient to show that p(–1) = 0.
Now,  p\left({ - 1}\right) = {\left({ - 1}\right)^4} + 3{\left( { - 1}\right)^3} + 3{\left({- 1}\right)^2} +\left({- 1}\right)+1
 = 1 - 3 + 3 - 1 + 1 = 1 \ne 0
Therefore (x + 1) is not a factor of    {x^4} + 3{x^3} + 3{x^2} + x + 1 .

(iv) In order to prove that (x + 1) is a factor of   p\left( x \right) = {x^3} - {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 ,  it is sufficient to show that p(-1)= 0
 p\left( { - 1} \right) = {\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} - \left( {2 + \sqrt 2 } \right)\left( { - 1} \right) + \sqrt 2
 = - 1 - 1 + 2 + \sqrt 2 + \sqrt 2 = 2\sqrt 2 \ne 0
Therefore   (x + 1) is not a factor of    {x^3} - {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2


Q.2      Use the factor theorem to determine whether g(x) is a factor of p (x) in each of the following cases :
           
(i) p\left( x \right) = 2{x^3} + {x^2} - 2x - 1,\,g\left( x \right) = x + 1
           
(ii) p\left( x \right) = {x^3} + 3{x^2} + 3x + 1,\,g\left( x \right) = x + 2
           
(iii) p\left( x \right) = {x^3} - 4{x^2} + x + 6,\,g\left( x \right) = x - 3
Sol.

(i) In order to prove that g(x) = x + 1 is a factor of p\left( x \right) = 2{x^3} + {x^2} - 2x - 1, it is sufficient to show that p(–1) = 0.
Now,  p\left( { - 1} \right) = 2{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - 2\left( { - 1} \right) - 1
 = - 2 + 1 + 2 - 1 = 0
Therefore g(x) is a factor of p(x).

(ii) In order to prove that g(x) = x + 2 is a factor of   p\left( x \right) = {x^3} + 3{x^2} + 3x + 1, it is sufficient to show that p(–2) = 0
Now,  p\left( { - 2} \right) = {\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} + 3\left( { - 2} \right) + 1
 = - 8 + 12 - 6 + 1
 = - 1 \ne 0
Therefore g(x) is not a factor of p(x) .

(iii) In order to prove that g(x) = x – 3 is a factor of  p\left( x \right) = {x^3} - 4{x^2} + x + 6 . It is sufficient to show that p(+3) = 0
Now,   p\left( 3 \right) = {\left( 3 \right)^3} - 4{\left( 3 \right)^2} + 3 + 6
 = 27 - 36 + 3 + 6
36 - 36 = 0
Therefore g (x) is  a factor of p(x).


Q.3      Find the value of k, if x -1 is a factor of p (x) in each of the following cases :
           
(i) p\left( x \right) = {x^2} + x + k
           
(ii) p\left( x \right) = 2{x^2} + kx + \sqrt 2
           
(iii) p\left( x \right) = k{x^2} - \sqrt 2 x + 1
           
(iv) p\left( x \right) = k{x^2} - 3x + k
Sol.

(i) If (x -1) is a factor of  p\left( x \right) = {x^2} + x + k , then
 p\left( 1 \right) = 0
 \Rightarrow {\left( 1 \right)^2} + 1 + k = 0
 \Rightarrow 1 + 1 + k = 0
 \Rightarrow k = - 2
Hence, k = -2

(ii) If (x - 1) is a factor of    p\left( x \right) = 2{x^2} + kx + \sqrt 2 , then
p\left( 1 \right) = 0
 \Rightarrow 2{\left( 1 \right)^2} + k\left( 1 \right) + \sqrt 2 = 0
 \Rightarrow 2 + k + \sqrt 2 = 0
 \Rightarrow k = - \left( {2 + \sqrt 2 } \right)

(iii) If (x -1) is a factor of   p\left( x \right) = k{x^2} - \sqrt 2 x + 1 ,  then
p\left( 1 \right) = 0
 \Rightarrow k{\left( 1 \right)^2} - \sqrt 2 \left( 1 \right) + 1 = 0
 \Rightarrow k - \sqrt 2 + 1 = 0
 \Rightarrow k = \sqrt 2 - 1

(iv) If (x -1) is a factor of p\left( x \right) = k{x^2} - 3x + k, then
p\left( 1 \right) = 0
 \Rightarrow k{\left( 1 \right)^2} - 3\left( 1 \right) + k = 0
 \Rightarrow k - 3 + k = 0
 \Rightarrow 2k = 3
 \Rightarrow k = {3 \over 2}


Q.4      Factorise :
               (i) 12{x^2} - 7x + 1                            
            (ii) 2{x^2} + 7x + 3

               (iii) 6{x^2} + 5x - 6                          
            (iv) 3{x^2} - x - 4

Sol.

(i) Here p + q = coeff. of x = - 7
pq = coeff. of  {x^2} \times constant term
= 12 × 1 = 12
Therefore p + q = – 7 = – 4 – 3
and pq = 12 = (– 4)(–3)
Therefore 12{x^2} - 7x + 1 = 12{x^2} - 4x - 3x + 1
 = 4x\left( {3x - 1} \right) - 1\left( {3x - 1} \right)
 = \left( {3x - 1} \right)\left( {4x - 1} \right)

(ii) Here p + q = coeff. of x = 7
pq = coeff. of {x^2} \times constant term  
= 2 × 3 = 6
Therefore p + q = 7 = 1 + 6
and pq = 6 = 1 × 6
Therefore  2{x^2} + 7x + 3 = 2{x^2} + x + 6x + 3
 = x\left( {2x + 1} \right) + 3\left( {2x + 1} \right)
 = \left( {2x + 1} \right)\left( {x + 3} \right)

(iii) Here p + q = coeff. of x = 5
pq = coeff. of {x^2} \times constant term
= 6 × (– 6)  = – 36
Therefore p + q = 5 = 9 + (– 4)
and pq = – 36 = 9 × (– 4)
Therefore  6 {x^2} + 5x - 6 = 6{x^2} + 9x - 4x - 6
 = 3x\left( {2x + 3} \right) - 2\left( {2x + 3} \right)
 = \left( {2x + 3} \right)\left( {3x - 2} \right)

(iv) Here p + q = coeff. of x = –1
pq = coeff. of  {x^2} \times constant term
= 3 × (– 4) = – 12
Therefore p + q = –1 = 3 + (– 4)
and pq = – 12 = 3 × (–4)
Therefore  3{x^2} - x - 4 = 3{x^2} + 3x - 4x - 4
 = 3x\left ({x + 1} \right) - 4\left( {x + 1} \right)
 = \left( {x + 1} \right)\left( {3x - 4} \right)


Q.5      Factorise :
               (i) {x^3} - 2{x^2} - x + 2                                
            (ii) {x^3} - 3{x^2} - 9x - 5
           
(iii) {x^3} + 13{x^2} + 32x + 20                
            (iv) 2{y^3} + {y^2} - 2y - 1

Sol. 

(i) Let  f\left(x \right) = {x^3} - 2{x^2} - x + 2
The constant term in f(x) is + 2 and factors of + 2 are  \pm 1, \pm 2
Putting x = 1 in f(x),we have 
 f\left(1\right) = {\left(1\right)^3} - 2{\left(1\right)^2} - 1 + 2
 = 1 - 2 - 1 + 2 = 0
Therefore (x–1) is a factor of f(x) . 

putting x = – 1 in f(x) , we have 
 f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2  
 = - 1 - 2 + 1 + 2 = 0 
Therefore x + 1 is a factor of f (x). 
Putting x = 2 in f(x), we have 
 f\left( 2 \right) = {\left( 2 \right)^3} - 2{\left( 2 \right)^2} - \left( 2 \right) + 2
 = 8 - 8 - 2 + 2 = 0
Therefore (x + 2) is a factor of f(x)

Putting x = – 2 in f(x), we have
 f\left( { - 2} \right) = {\left( { - 2} \right)^3} - 2{\left( { - 2} \right)^2} - \left( { - 2} \right) + 2
 = - 8 - 8 + 2 + 2
 = - 12 \ne 0
Therefore x - 2 is not a factor of f (x).
Therefore The factors of f(x) are (x – 1), (x + 1) and (x – 2).
Let f(x) = k (x – 1) (x + 1) (x – 2)
\Rightarrow  {x^3} - 2{x^2} - x + 2 = k\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)
Putting x = 0 on both sides, we have
2 = k (–1) (1) (–2)  \Rightarrow k = 1
Therefore  {x^3} - 2{x^2} - x + 2 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)

(ii) Let  p\left( x \right) = {x^3} - 3{x^2} - 9x - 5
We shall look for all factors of – 5. Therefore are  \pm 1, \pm 5.

By trial, we find p (–1) = – 1 – 3 + 9 – 5 = 0 . So (x + 1) is a factor of p(x) .
Now, divide p (x) by (x + 1)
poly1

Therefore  p\left( x \right) = \left( {x + 1} \right)\left( {{x^2} - 4x - 5}\right)
 = \left( {x + 1} \right)\left( {{x^2} + x - 5x - 5} \right)
 = \left( {x + 1} \right)\left[ {x\left( {x + 1} \right) - 5\left( {x + 1} \right)} \right]
 = \left( {x + 1} \right)\,\left( {x + 1} \right)\left( {x - 5} \right)
Therefore {x^3} - 2{x^2} - x + 2 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right)

(iii) Let  p\left( x \right) = {x^3} + 13{x^2} + 32x + 20
We shall look for all factors of + 20, these are  \pm 1, \pm 2,\, \pm 4, \pm 5, \pm 10\,and\, \pm 20
By trial , we find 
p( - 2) = ( - 2)^3+ 13(2)^2+ 32( - 2) + 20
p(– 2) = – 8 + 52 – 64 + 20 = 0 
Therefore (x + 2) is a factor of p(x) 
Now, divide p (x) by x + 2
poly2
Therefore   p\left( x \right) = \left( {x + 2} \right)\left( {{x^2} + 11x + 10} \right)
 = \left( {x + 2} \right)\left( {{x^2} + x + 10x + 10} \right)
 = \left( {x + 2} \right)\left[ {x\left( {x + 1} \right) + 10\left( {x + 1} \right)} \right]
 = \left( {x + 2} \right)\,\left( {x + 1} \right)\left( {x + 10} \right)

(iv) Let  p\left( y \right) = 2{y^3} + {y^2} - 2y - 1
By trial we find p(1) = 2 + 1 – 2 – 1 = 0 
So, (y – 1) is a factor of p(y)
Now, divide p(y) by y – 1
poly3
Therefore  p\left( y \right) = \left( {y - 1} \right)\left( {2{y^2} + 3y + 1} \right)
 = \left( {y - 1} \right)\left( {2{y^2} + 2y + y + 1} \right)
 = \left( {y - 1} \right)\left[ {2y\left( {y + 1} \right) + 1\left( {y + 1} \right)} \right]
 = \left( {y - 1} \right)\,\left( {y + 1} \right)\left( {2y + 1} \right)