CAN SOME ONE SEND CLASS 9 MATHS NCERT CHAPTER 2 EXERCISE 2.3
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Answers
(i) x+1
Solution:
x+1= 0
⇒x = −1
∴Remainder:
p(−1) = (−1)3+3(−1)2+3(−1)+1
= −1+3−3+1
= 0
(ii) x−1/2
Solution:
x-1/2 = 0
⇒ x = 1/2
∴Remainder:
p(1/2) = (1/2)3+3(1/2)2+3(1/2)+1
= (1/8)+(3/4)+(3/2)+1
= 27/8
(iii) x
Solution:
x = 0
∴Remainder:
p(0) = (0)3+3(0)2+3(0)+1
= 1
(iv) x+π
Solution:
x+π = 0
⇒ x = −π
∴Remainder:
p(0) = (−π)3 +3(−π)2+3(−π)+1
= −π3+3π2−3π+1
(v) 5+2x
Solution:
5+2x=0
⇒ 2x = −5
⇒ x = -5/2
∴Remainder:
(-5/2)3+3(-5/2)2+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1
= -27/8
2. Find the remainder when x3−ax2+6x−a is divided by x-a.
Solution:
Let p(x) = x3−ax2+6x−a
x−a = 0
∴x = a
Remainder:
p(a) = (a)3−a(a2)+6(a)−a
= a3−a3+6a−a = 5a
3. Check whether 7+3x is a factor of 3x3+7x.
Solution:
7+3x = 0
⇒ 3x = −7
⇒ x = -7/3
∴Remainder:
3(-7/3)3+7(-7/3) = -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 ≠ 0
∴7+3x is not a factor of 3x3+7x
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