9th Maths 2.3
NCERT Class 9th solution of Exercise 2.1
NCERT Class 9th solution of Exercise 2.2
NCERT Class 9th solution of Exercise 2.4
NCERT Class 9th solution of Exercise 2.5
Exercise 2.3
Q1. Find the remainder when x3+3x2+3x+1 is divided by :
i) x+1 ii) x-12 iii) x iv) x+π v) 5+2x
Sol. :
i) Let p(x) = x^3 + 3x^2 + 3x + 1
and zero of x+1 is -1
p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1
p(-1) = - 1 + 3 - 3 + 1
p(-1) = - 4 + 4
p(-1) = 0
Answer :
The remainder is 0.
ii) Let p(x) = x^3 + 3x^2 + 3x + 1
and zero of x - frac{1}{2} is frac{1}{2}
p(frac{1}{2}) = (frac{1}{2})^3 + 3(frac{1}{2})^2 + 3(frac{1}{2}) + 1
p(frac{1}{2}) = frac{1}{8} + frac{3}{4} + frac{3}{2} + 1
p(frac{1}{2}) = frac{1+6+12+8}{8}
p(frac{1}{2}) = frac{27}{8}
Answer:
The remainder is frac{27}{8}
iii) Let p(x) = x^3 + 3x^2 + 3x + 1
and zero of x is 0.
p(0) = (0)^3 + 3(0)^2 + 3(0) + 1
p(0) = 0 + 0 + 0 + 1
p(0) = 1
Answer :
The remainder is 1.
iv) Let p(x) = x^3 + 3x^2 + 3x + 1
and zero ofx + π is - π`
p(-π) = (-π)^3 + 3(-π)^2 + 3(-π) + 1
p(-π) = -π^3 + 3π^2 - 3π + 1
Answer :
The remainder is p(-π) = -π^3 + 3π^2 - 3π + 1.
v) Let p(x) = x^3 + 3x^2 + 3x + 1
and zero of 5 + 2x is -frac{5}{2}
p(frac{-5}{2}) = (frac{-5}{2})^3 + 3(frac{-5}{2})^2 + 3(frac{-5}{2}) + 1
p(frac{-5}{2}) = frac{-125}{8} + frac{75}{4} + frac{15}{2} + 1
p(frac{-5}{2}) = frac{-125 + 150 - 60 + 8}{8}
p(frac{-5}{2}) = frac{-27}{8}
Answer :
The remainder is frac{-27}{8}
Q2. Find the remainder when x^3 - ax^2 +6x - a is divided by x - a.
Sol. :
Let p(x) = x^3 - ax^2 +6x - a
and zero of x - a is x = a
p(x) = (a)^3 - a(a)^2 + 6(a) - a
p(x) = a^3 - a^3 + 5a
p(x) = 5a
Answer :
The remainder is 5a.
Q3. Check whether 7 + 3x is a factor of 3x^3 + 7x ?
Sol. :
Let p(x) = 3x^3 + 7x
and zero of 7 + 3x is -frac{7}{3}
p(-frac{7}{3}) = 3(-frac{7}{3})^3 + 7(frac{7}{3})
p(-frac{7}{3}) = frac{-343 - 147}{9}
p(-frac{7}{3}) = frac{-490}{9}
Answer :
7 + 3x is not a factor.
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