Polynomials
Mathematics Class Tenth
NCERT Exercise 2.3
Question: 1. Divide the polynomial `p(x)` by the polynomial `g(x)` and find the quotient and remainder in each of the following:
(i) `p(x) = x^3-3x^2+5x-3`, `g(x) = x^2-2`
Solution:
Thus, Quotient `=x-3`
And Remainder `=7x-9`
(ii) `p(x) = x^4-3x^2+4x+5`, `g(x) = x^2+1-x`
Solution:
Thus, Quotient `=x^2+x-3` and Remainder `=8`. Answer
(iii) `p(x) = x^4-5x+6`, `g(x) = 2-x^2 =-x^2+2`
Solution:
Thus, Quotient `=-x^2-2` and Remainder `=-5x+10`. Answer
Question: 2. Check whether the first polynomial is a factor of the second polynomial by dividing second polynomial by the first polynomial:
(i) `t^2-3`, `2t^4+3t^3-2t^2-9t-12`
Solution:
Since, here remainder `=0`. Thus, given first polynomial is a factor of given second polynomial. Answer
(ii) `x^2+3x+1`, `3x^4+5x^3-7x^2+2x+2`
Solution:
Since, here remainder `=0`. Thus, given first polynomial is a factor of given second polynomial. Answer
(iii) `x^3-3x+1`, `x^5-4x^3+x^2+3x+1`
Solution:
Since, here remainder `!=0`. Thus, given first polynomial is not a factor of given second polynomial. Answer
Question: 3. Obtain all other zeroes of `3x^4+6x^3-2x^2-10x-5`, if two of its zeroes are `sqrt(5/3)` and `-sqrt(5/3)`.
Solution:
Given, `p(x) = 3x^4+6x^3-2x^2-10x-5`
And given two zeroes are `sqrt(5/3)` and `-sqrt(5/3)`
Therefore,
`(x-sqrt(5/3))(x+sqrt(5/3))=(x^2-5/3)` is a factor of `3x^4+6x^3-2x^2-10x-5`
Therefore, after dividing `3x^4+6x^3-2x^2-10x-5` by `(x^2-5/3)` we can get the third factor.
Thus,
`3x^4+6x^3-2x^2-10x-5` `=(x^2-5/3)(3x^2+6x+3)`
`=3(x^2-5/3)(x^2+2x+1)`
Thus, After factorizing `x^2+2x+1` we get
`=(x+1)^2`
`=(x+1)(x+1)`
Thus, zeros of this polynomial `x^2+2x+1` will be given by `x+1=0`
i.e. `x=-1`
Since, it has the term `(x+1)^2`, therefore, there will be two zeroes, which are `-1` and `-1`.
Thus, the zeroes of the given polynomial are `sqrt(5/3)`, `-sqrt(5/3)`, `-1` and `-1` Answer.
Question: 4. On dividing `x^3-3x^2+x+2` by a polynomial `g(x)`, the quotient and remainder were `x-2` and `-2x-4`, respectively. Find `g(x)`.
Solution:
Given,
Dividend `=x^3-3x^2+x+2`
Quotient `=x-2`
Remainder `=-2x+4`
Thus, Divisor `g(x)=?`
We know that,
Dividend = Divisor `xx` Quotient + Remainder
∴ Divisor `xx` Quotient = Dividend - Remainder
`=>g(x) xx (x-2)` `=x^3-3x^2+x+2-(-2x+4)`
`=>g(x) xx (x-2)` `=x^3-3x^2+x+2+2x-4`
`=>g(x) xx (x-2)` `=x^3-3x^2+3x-2`
`:. g(x) =(x^3-3x^2+3x-2)/((x-2)`
Thus, `g(x) = x^2-x+1` Answer
Question: 5. Give examples of polynomial `p(x), g(x), q(x)` and `r(x)`, which satisfy the division algorithm and
(i) deg `p(x) =` deg `q(x)`
(ii) deg `q(x)=` deg `r(x)`
(iii) deg `r(x) =0`
Solution:
According to division algorithm, if `p(x)` and `g(x)` are two polynomials with `g(x) !=0`, the we can find polynomials `q(x)` and `r(x)` such that
`p(x) = g(x) xx q(x)+r(x)`
Where `r(x) =0` or degree of `r(x) \<` degree of `g(x)`
[Ref: NCERT Book class X math page number: 34]
(i) deg `p(x)=` deg `q(x)`
Solution:
Degree of quotient will be equal to the degree of dividend when divisor is constant (i.e. when any polynomial is divided by a constant)
Let, the division of `6x^2+2x+2` by `2`
Here, `p(x) = 6x^2+2x+2`
And, `g(x) =2`
Thus, `q(x) = 3x^2+x+1`
And `r(x) =0`
Here degree of `p(x) =` degree of `q(x)=2`
Now, checking for division algorithm,
`p(x) = g(x) xx q(x) + r(x)`
`=>6x^2+2x+2 = (2)(3x^2+x+1)+0`
Thus, the division algorithm is satisfied.
(ii) deg `q(x)=` deg `r(x)`
Solution:
Let the division of `x^3+x` by `x^2`
Here `p(x) = x^3+xg(x) = x^2 q(x)` `=x and r(x) =x`
Clearly, the degree of `q(x)` and `r(x)` is the same, i.e. equal to 1.
Checking for division algorithm,
`p(x) = g(x) xx q(x) + r(x)`
`x^3+x = (x^2) xx\x+x`
`=>x^3+x = x^3+x`
Thus, division algorithm is satisfied.
(iii) deg `r(x) =0`
Solution:
Let us assume the division of `x^3+1` by `x^2`
Here, `p(x) = x^3+1`
And, `g(x) = x^2`
`q(x) = x ` and `r(x) =1`
Clearly, the degree of `r(x) =0`
Checking for division algorithm,
`p(x) = g(x) xx q(x) + r(x)`
`=>x^3+1 = x^2xx\x+1`
`=>x^3+1 = x^3+1`
Thus, division of algorithm is satisfied.
Reference: