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 `x^3 + 3x^2 + 3x + 1` is divided by :
i) `x+1` ii) `x- frac{1}{2}` 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 of``x + 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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