class 9
NCERT solutions
chapter 2 polynomials
please solve all the questions
i
ii
iii
iv
Answers
Answer:
(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)}
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