The value of `k`, if `x^2+2x+k` is a factor of `2x^4+x^3-14x^2+5x+6` i

1.	The value of `k`, if `x^2+2x+k` is a factor of `2x^4+x^3-14x^2+5x+6` is
Answer» Given that, `x^(2)+2x +k` is a factor of `2x^(4)+x^(3)-14x^(2)+5x +6`, then we apply division algorihm, `{:(" "ul(" "2x^(2)-3x+(-8-2k))),( {:x^(2)+2x+k) " "2x^(4)+x^(3)-14x^(2)+5x+6),(" "ul(underset(-)(2x^(4))underset(+)-4x^(3)underset(-)(+)2kx^(2)" ")),(" "-3x^(3)-(2k+14)x^(2)+5x+6),(" "ul(underset(+)(-3x^(3))underset(+)(-)6x^(2)underset(+)(-)3kx" ") ),(" "(6-2k-14)x^(2)+(3k+5)x+6),(" "ul(underset(-)((-8-2k))x^(2)underset(-)(+)2(-8-2k)xunderset(-)(+)k(-8-2k))),(" "(3k+5+16+4k)x+(6+8k+2k^(2))):}` Since, `(x^(2)+2x+k)` is a factor of `2x^(4)+x^(3)-14x^(2) +5x +6` So, when we apply division algorithm remainder should be zero. `:. (7k +21)x +(2k^(2)+8k+6) =0.x +0` `rArr 7k +21 = 0` and `2k^(2)+8k + 6 = 0` `rArr k =- 3` or `k^(2) +4k +3 = 0` `rArr k^(2) +3k +k +3 = 0` " "[by splitting middle term] `rArr k(k+3) +1 (k+3) = 0` `rArr (k+1) (k+3) = 0` `rArr k =- 1` or `-3` Here, if we take `k =- 3`, then remainder will be zero. Thus, the required value of k is -3 Now, Dividend = Divisor `xx` Quotient + Remainder `rArr 2x^(4) + x^(3) - 14x^(2)+5x +16 =(x^(2)+2x-3)(2x^(2)-3x-2)` Using factorisation method, `={x^(2)+3x -x -3) (2x^(2) -4x +x -2)` [by splitting middle term] `= {x(x+3)-1(x+3)} {2x(x-2)+1(x-2)}` `=(x-1)(x+3) (x-2) (2x+1)` Hence, the zeroes of `x^(2) +2x - 3` are 1,-3 and the zeroes of `2x^(4)+x^(3)-14x^(2)+5x+6` are `1,-3,2,(-1)/(2)`.

Discussion