Find all the zeros of f (x) = x³- 5x²- 7x + 51 given that 4 – i is a zero of f(x).

Step-by-step explanation:

complex roots always occur in pairs.

so two roots are 4-i and 4+i

[x-(4-i)] and [x-(4+i)] are factors of f(x)

or [(x-4)+i] and [(x-4)-i] are factors of f(x)

[(x-4)+i][(x-4)-i]=(x-4)²-i²=x²-8x+16-(-1)=x²-8x+17

x²-8x+17) x³  -5 x²  -7 x +51( x+3

               x³  -8x²  +17x

             -      +        -

                      3x²  -24x +51

                      3x²  -24x +51

                    -        +       -

                      0

so one more factor is x+3

or zero =-3

Hence zeros of f(x) are -3,4-i,4+i

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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