Question

"Question 5 Factorise: (i) x^3 − 2x^2 − x + 2 (ii) x^3 + 3x^2 −9x − 5 (iii) x^3 + 13x^2 + 32x + 20 (iv) 2y^3 + y^2 − 2y − 1

Class 9 - Math - Polynomials Page 44"

Answer 1

Factorization of a cubic polynomial:

First, find the factors of constant term and check at which factor given polynomial becomes zero by using hit and trial method and get one factor of given polynomial. Then divide the polynomial by this factor to get a quotient of second degree. Then factorise the second degree polynomial by middle term splitting to get required factors.

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Solution:

(i)

Let p(x)=x³– 2x² – x + 2

Factors of 2 are ±1 and ± 2

By trial method, we find that
p(1) = 0

So,(x+1) is factor of p(x)

Now,
p(x) = x³– 2x² – x + 2
p(-1)=(-1)³–2(-1)²–(-1)+ 2

= -1 -2 + 1 + 2 = 0
Therefore, (x+1) is the factor of  p(x)

On dividing p(x) by (x+1) ,we get

[division is in attachment]

 (x+1) (x²– 3x + 2)

= (x+1) (x²– x – 2x + 2)                [by middle term splitting]

= (x+1) {x(x-1) -2(x-1)}

= (x+1) (x-1) (x+2)

 

(ii) Let p(x) = x³ –3x² –9x – 5

Factors of 5 are ±1 and ±5

By trial method, we find that
p(5) = 0

So, (x-5) is factor of p(x)

 

Now,
p(x) = x³ – 2x² – x + 2
p(5) = (5)³ – 3(5)² – 9(5) – 5

 = 125 – 75 – 45 – 5 = 0

Therefore, (x-5) is the factor of  p(x)

On dividing p(x) by (x-5) ,we get

[division is in attachment]

 

 (x-5) (x² + 2x + 1)

= (x-5) (x² + x + x + 1)                 [by middle term splitting]  

= (x-5) {x(x+1) +1(x+1)}

= (x-5) (x+1) (x+1)

 

iii)

Let = x³ +13x²+ 32x + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that
p(-1) = 0

So, (x+1) is factor of p(x)

Now,
p(x) =  x³ + 13x² + 32x + 20

p(-1)= (-1)³+13(-1)²+32(-1) + 20

= -1 + 13 – 32 + 20 = 0

Therefore, (x+1) is the factor of  p(x)

On dividing p(x) by (x+1) ,we get

[division is in attachment]

 

 (x+1) (x² + 12x + 20)

= (x+1) (x² + 2x + 10x + 20)                    [by middle term splitting] 

= (x-5) {x(x+2) +10(x+2)}

= (x-5) (x+2) (x+10)

 

(iv)

 Let p(y) = 2y³ + y² – 2y – 1

Factors of ab = 2× (-1) = -2 are ±1 and ±2

By trial method, we find that
p(1) = 0

So, (y-1) is factor of p(y)

Now,
p(y) =  2y³ + y² – 2y – 1
p(1) = 2(1)³ + (1)² – 2(1) – 1

 = 2 +1 – 2 – 1 = 0

Therefore, (y-1) is the factor of  p(y)

On dividing p(y) by (y-1) ,we get

[division is in attachment]

 (y-1) (2y² + 3y + 1)

= (y-1) (2y² + 2y + y + 1)                        [by middle term splitting] 

= (y-1) {2y(y+1) +1(y+1)}

(y-1) (2y+1) (y+1)

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Answer 2

Answer:

I have answered the question.. Hpe it helps to u!!  :)

Step-by-step explanation:

actorization of a cubic polynomial:

First, find the factors of constant term and check at which factor given polynomial becomes zero by using hit and trial method and get one factor of given polynomial. Then divide the polynomial by this factor to get a quotient of second degree. Then factorise the second degree polynomial by middle term splitting to get required factors.

-----------------------------------------------------------------------------------------------------

Solution:

(i)

Let p(x)=x³– 2x² – x + 2

Factors of 2 are ±1 and ± 2

By trial method, we find that

p(1) = 0

So,(x+1) is factor of p(x)

Now,

p(x) = x³– 2x² – x + 2

p(-1)=(-1)³–2(-1)²–(-1)+ 2

= -1 -2 + 1 + 2 = 0

Therefore, (x+1) is the factor of  p(x)

On dividing p(x) by (x+1) ,we get

[division is in attachment]

(x+1) (x²– 3x + 2)

= (x+1) (x²– x – 2x + 2)                [by middle term splitting]

= (x+1) {x(x-1) -2(x-1)}

= (x+1) (x-1) (x+2)

 

(ii) Let p(x) = x³ –3x² –9x – 5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

 

Now,

p(x) = x³ – 2x² – x + 2

p(5) = (5)³ – 3(5)² – 9(5) – 5

= 125 – 75 – 45 – 5 = 0

Therefore, (x-5) is the factor of  p(x)

On dividing p(x) by (x-5) ,we get

[division is in attachment]

 

(x-5) (x² + 2x + 1)

= (x-5) (x² + x + x + 1)                 [by middle term splitting]  

= (x-5) {x(x+1) +1(x+1)}

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

 

iii)

Let = x³ +13x²+ 32x + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x) =  x³ + 13x² + 32x + 20

p(-1)= (-1)³+13(-1)²+32(-1) + 20

= -1 + 13 – 32 + 20 = 0

Therefore, (x+1) is the factor of  p(x)

On dividing p(x) by (x+1) ,we get

[division is in attachment]

 

(x+1) (x² + 12x + 20)

= (x+1) (x² + 2x + 10x + 20)                    [by middle term splitting]  

= (x-5) {x(x+2) +10(x+2)}

= (x-5) (x+2) (x+10)

 

(iv)

Let p(y) = 2y³ + y² – 2y – 1

Factors of ab = 2× (-1) = -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) =  2y³ + y² – 2y – 1

p(1) = 2(1)³ + (1)² – 2(1) – 1

= 2 +1 – 2 – 1 = 0

Therefore, (y-1) is the factor of  p(y)

On dividing p(y) by (y-1) ,we get

[division is in attachment]

(y-1) (2y² + 3y + 1)

= (y-1) (2y² + 2y + y + 1)                        [by middle term splitting]  

= (y-1) {2y(y+1) +1(y+1)}

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