If 7 + 4i is a zero of g(x)= x^3 - 24x^2 + 205x - 650 , find the real zero of a of g(x). a =.......?

Answer 1

easy, Because the expression has real coefficeint so 7-4i is also a zero . let 3rd root be a


sum of factors = a+7+4i+7-4i = 24 (-ve of coefficient of x^2)
solving this a = 10

now cross check

(x-10)(x-7-4i)(x-7+4i)
= (x-10)((x-7)^2 + 4^2)
= (x-10)(x^2-14x+65)
= x^3-24x^2+205-650

Answer 2

first the rules !

an equation in "real" coefficients will have conjugate complex roots

so here, if 7+4i is a root then so is 7-4i

so let the three roots be b=7+4i, c=7-4i and a (to be found)

we know, for a polynomial, the sum of roots = - coeff of x^2/coeff of x^3 = - -24/1 = 24

so a+b+c = 24
or, a+7+4i+7-4i = 24 implies that a = 10

you can verify this by taking the coeff of x

the sum of products of roots in pair = coeff of x/coeff of x^3
so 10(7+4i) + 10(7-4i) + (7+4i)(7-4i) = 205
or, 10(14) + (7^2-(4i)^2) = 205
or, 140 + 49 + 16 = 205
indeed true

even a simpler verification is to check the constant term

product of a,b,c = - constant / coeff of x^3 = 650
10(65:got above) = 650