9th Maths 2.4
NCERT Class 9th solution of Exercise 2.1
NCERT Class 9th solution of Exercise 2.2
NCERT Class 9th solution of Exercise 2.3
NCERT Class 9th solution of Exercise 2.5
Exercise 2.4
Q1. Determine each of the following polynomials has `(x+1)` a factor :
i) `x^3 + x^2 + x + 1` ii)`x^4 + x^3 + x^2 + x + 1`
iii) `x^4 + 3x^3 + 3x^2 + x + 1`
iv) `x^3 - x^2 - (2 + sqrt2)x + sqrt2`
Sol. :
i) Let `p(x) = x^3 + x^2 + x + 1`
and zero of `x + 1` is `- 1`
`p(-1) = (-1)^3 + (-1)^2 + (-1) + 1`
`p(-1) = - 1 + 1 - 1 + 1`
`p(-1) = 0`
Answer :
`(x+1)` is a factor.
ii) Let `p(x) = x^4 + x^3 + x^2+ x + 1`
and zero of `x+1` is `- 1`
`p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1`
`p(-1) = 1 - 1 + 1 - 1 + 1`
`p(-1) = 1 ≠ 0 `
Answer :
`(x+1)` is not a factor.
iii) Let `p(x) = x^4 + 3x^3 + 3x^2+ x + 1`
and zero of `x+1` is `- 1`
`p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2+ (-1) + 1`
`p(-1) = 1 - 3 + 3 - 1 + 1`
`p(-1) = 1 ≠ 0`
Answer :
`(x+1)` is not a factor.
iv) Let `p(x) = x^3 - x^2 - (2 + sqrt2)x + sqrt2`
and zero of `x+1` is `- 1`
`p(-1) = (-1)^3 - (-1)^2 - (2 + sqrt2)(-1)+ sqrt2`
`p(-1) = - 1 - 1 + (2 + sqrt2)+ sqrt2`
`p(-1) = - 2 + 2 + sqrt2 + sqrt2`
`p(-1) = 2sqrt2 ≠ 0`
Answer :
`(x+1)` is not a factor.
Q2. Use the Factor Theorem to whether `g(x)` is a factor of `p(x)` in each of the following cases :
i) `p(x) = 2x^3 + x^2 - 2x - 1, g(x) = x + 1`
ii) `p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2`
iii) `p(x) = x^3 - 4x^2 + x + 6, g(x) = x - 3`
Sol. :
i) `p(x) = 2x^3 + x^2 - 2x - 1`
and zero of `g(x) = x + 1` is `- 1`
`p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1`
`p(-1) = - 2 + 1 + 2 - 1`
`p(-1) = 0`
Answer :
`g(x)` is a factor.
ii) `p(x) = x^3 + 3x^2 + 3x + 1`
and zero of `g(x) = x + 2` is `- 2`
`p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1`
`p(-2) = - 8 + 12 - 6 + 1`
`p(-2) = - 14 + 13`
`p(-2) = - 1 ≠ 0`
Answer :
`g(x)` is a factor.
iii) `p(x) = x^3 - 4x^2 + x + 6`
and zero of `g(x) = x - 3` is `3`
`p(3) = (3)^3 - 4(3)^2 + 3 + 6`
`p(3) = 27 - 36 + 9`
`p(3) = 36 - 36`
`p(3) = 0`
Answer :
`g(x)` is a factor.
Q3. Find the value of `k`, if `x - 1` is a factor of `p(x)` in each of the following cases :
i) `p(x) = x^2 + x + k` ii) `p(x) = 2x^2 + kx + sqrt2`
iii) `p(x) = kx^2 - sqrt(2)x + 1` iv) `p(x) = kx^2 - 3x + k`
Sol. :
i) Let `p(x) = x^2 + x + k = 0`
and zero of `x-1` is `1`
`p(1) = (1)^2 + (1) + k = 0`
`1 + 1 + k = 0`
`2 + k = 0`
`k = - 2`
Answer :
`k = - 2`
ii) `Let p(x) = 2x^2 + kx + sqrt2 = 0`
and zero of `x-1` is `1`
`p(1) = 2(1)^2 + k(1) + sqrt2 = 0`
`2 + k + sqrt2 = 0`
`k = - 2 - sqrt2`
Answer :
`k = - (2 + sqrt2)`
iii) `let p(x) = kx^2 - sqrt(2)x + 1 = 0`
and zero of `x-1` is `1`
`p(1) = k(1)^2 - sqrt(2)(1) + 1 = 0`
`k - sqrt 2 + 1 = 0`
`k = sqrt2 - 1`
Answer :
`k = sqrt2 - 1`
iv) Let `p(x) = kx^2 - 3x + k = 0`
and zero of `x-1` is `1`
`p(1) = k(1)^2 - 3(1) + k = 0`
`k - 3 + k = 0`
`2k = 3`
`k = frac(3)(2)`
Answer :
`k = frac(3)(2)`
Q4. Factorise :
i) `12x^2 - 7x +1` ii) `2x^2 + 7x + 3`
iii) `6x^2+5x-6` iv) `3x^2 - x - 4`
Sol. :
i)`= 12x^2 - 7x + 1`
`= 12x^2 - 4x - 3x + 1`
`= 4x(3x - 1) - 1(3x - 1)`
`= (4x - 1)(3x - 1)`
Answer :
factors are `(4x - 1)(3x - 1)`
ii) `= 2x^2 + 7x + 3`
`= 2x^2 + 6x + x + 3`
`= 2x(x + 3) + 1(x + 3)`
`= (2x + 1)(x + 3)`
Answer :
factors are `(x + 3)(2x + 1)`
iii) ` = 6x^2 + 5x - 6`
`= 6x^2 + 9x - 4x - 6`
`= 3x(2x + 3) - 2(2x + 3)`
`= (2x + 3)(3x - 2)`
Answer :
factors are `(2x + 3)(3x -2)`
iv) `= 3x^2 - x - 4`
`= 3x^2 + 3x - 4x -4`
`= 3x(x + 1) - 4(x + 1)`
`= (x + 1)(3x - 4)`
Answer :
factors are `(x + 1)(3x - 4)`
Q5. Factorise :
i) `x^3 - 2x^2 - x + 2` ii) `x^3 - 3x^2 - 9x - 5`
iii) `x^3 + 13x^2 + 32x + 20` iv) `2y^3 + y^2 - 2y - 1`
Sol. :
i) `= x^3 - 2x^2 - x + 2`
`= x^2(x - 2) - 1(x + 2)`
`= (x - 2)( x^2 - 1)`
`= (x - 2)(x - 1)(x + 1)` `[ (a^2 - b^2) = (a + b)(a - b) ]`
Answer :
factors are `(x - 2)(x - 1)(x + 1).`
ii) `= x^3 - 3x^2 - 9x - 5`
`= x^3 + x^2 - 4x^2 - 4x - 5x - 5`
`= x^2(x + 1) - 4x(x + 1) - 5(x + 1)`
`= (x + 1)[ x^2 - 4x - 5]`
`= (x + 1)[x^2 - 5x + x - 5]`
`= (x + 1)[x(x - 5) + 1(x - 5)]`
`= (x + 1)(x - 5)(x + 1)`
Answer :
factors are `(x + 1)^2(x - 5)`
iii) `= x^3 + 13x^2 + 32x + 20`
`= x^3 + x^2 + 12x^2 + 12x + 20x + 20`
`= x^2(x + 1) + 12x(x + 1) + 20(x +1)`
`= (x + 1) [x^2 + 12x + 20]`
`= (x + 1) [x^2 + 2x + 10x + 20]`
`= (x + 1) [x(x + 2) + 10(x + 2)]`
`= (x + 1)(x + 2)(x + 10)`
Answer :
factors are `(x + 1)(x + 2)(x + 10)`
iv) `= 2y^3 + y^2 - 2y - 1`
`= y^2(2y + 1) - 1(2y - 1)`
`= (2y + 1)(y^2 - 1)`
`= (2y + 1)(y + 1)(y - 1)` `[ a^2 - b^2 = (a + b)(a - b)]`
Answer :
factors are `(2y + 1)(y + 1)(y - 1)`
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