find the zeroes of the quadratic polynomial:3x2 + 5x + 2
equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1.
find zeroes of the following polynomials (with steps ) (i) p(x)=6x^2-3-7x (ii) f(x)=2-x^2 without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 2x²+3x+5=0 has solutions by the quadratic formula x=(-3±√(9-4×2×5))/4 =(-3±√(-31))/4
there are no real solutions, since the discriminant is -31, whose square root is imaginary. zero solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` 2, 0 ( any two integers with one of them as 0). express as a sum of perfect square 16 a square + b square + 9 c square minus 4 a b minus 3 b c - 12 ca? f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer